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Transcript
COLLEGE
ALGEBRA
\
COLLEGE
ALGEBRA
~~b\
REVISED EDITION
8y PA U L R. RID ER Ph. D.
I
Chief Statistician, Aeronautical Research Laboratory
Wright-Patterson Ai r Force Base
ANGRAU
,..... Central library
'lilt Rajendranaga,
,.
1111111111111111111
Professor Emeritus of Mathematics
Washington University
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'
l
." ..
~~--------------~~
APAU CENTRAL LItl/<AI\ Y
1
.
Ace: No;
'41
Date: 2. '6" c...ce,-:s.
THE MACMILLAN COMPANY. NEW YORK
COlliER-MACMILLAN LIM ITED. LON DON
MARUZEN COMPANY LIMITED.
TO KYO
ANGRAU
512
N55RID
Acc No 147
'\991
~C:~t1::
C~
C()pyright 1938. 1940. 1947. 1955 in the United
States of America hy The \lacmil1an Company
All rights reserved-no part of this book may be
reproduced in any form without permission in
writing from the publisher, except by a reviewer
who wishes to quote brief passages in connection with a review written for inclusion in
magazine or newspaper.
,\uthorized reproduction from the Revised Enl(lish-lan~uagl'
. l'c1ition. published in the United Statl's of ,\m .. rica by
Tlw :\1a.. millan Company and simultanl'ol"ly puhlislwd
in Cana.!..,
RINTED IN JAPAN
Preface to Revised Edition
The principal change in this revised edition is in the
introduction. A more complete review of elelnentary
algebra is provided, thus making possible an easier and
more gradual approach to the subject, an approach
designed especially for the student who is not so well
prepared. .More detailed consideration and fuller discusaion are given to basic ideas and fundamental principles; also the logical structure has been improved in
certain places. A more complete discussion of the number system of algebra is provided. Completely ne.w sets
of exercises are furnished.
The style of presentation which was used in the original
edition, and which apparently has been favorably received,
has been retained, although several sections have been
rewritten with the object of making them even clearer.
The manuscript was critically read by Professor Franz
E. Hohn of the University of Illinois, who made many
valuable suggestions for improvement. He also checked
the answers to the exercises. For this assistance I wish
to express my sincere thanks.
P. R.. R.
DAYTON, OHIO
Preface to First Edition
This textbook has been prepared for the use of first-year
students in colleges and technical schools. The methods
of presentation are those that have been found most successful during many years of teaching experience with
groups and individuals having varying degrees of preparation. A preliminary multigraphed edition was used in
more than twenty sections of students in various divisions
of Washington University-the College of Liberal Arts,
including pre-business and other pre-professional courses,
the Schools of Engineering and Architecture, and the
evening classes.
Although the primary objective has been clarity of
explanation, the question of logical rigor has been kept
constantly in mind, and an effort has been made to distinguish carefully between assumptions and proofs. It is
hoped that an appreciation of rigor will be developed in
students of the book, particularly those students who will
later take more advanced courses in mathematics.
The first part of the book constitutes a thorough review
of the topics of elementary and intermediate algebra. It
was written with the idea in mind that many students
have little skill in algebraic manipulation, although it does
assume an intelligence level that may reasonably be
expected of college or university freshmen. Well prepared classes can cover this part quite rapidly, or even
omit it altogether, and proceed to the later portions.
Mature students will find enough material for a very
complete course containing all of the topics usually taught
in College Algebra, also certain additional topics which,
because of their importance, it ha~ seemed desirable to
vii
viii
PREFACE
include. Those not so well prepared can cover the first
part more slowly and then be given such chapters and
topics in the latter part of the book as the instructor may
deem appropriate.
The arrangement of chapters and topics was decided
upon only after members of the Macmillan staff had consulted a considerable number of teachers of the subject in
various institutions in different parts of the country.
Although the present order is a composite of suggestions,
and represents the arrangement which seemed most desirable to the majority, the individual chapters have been
rendered to a large extent independent by means of cross
references and a certain amount of repetition of important
connecting ideas. Because of this, the instructor 'who
prefers a different sequence of chapters, or who desires to
omit certain material, will find that the book can readily
be adapted to his needs.
The book contains a liberal supply of selected and
graded exercises. Answers to the odd-numbered exercises
are printed at the back of the book, answers to the evennumbered exercises are printed in a separate pamphlet.
This pamphlet is available to students upon request of the
instructor.
Among the special features may be mentioned the treatment of determinants. This follows the method outlined
by G. D. Birkhoff in his article, "Determinants," in the
Encyclopedia Britannica. It is believed to be somewhat
simpler than the usual method based upon the idea of
inversions in order. My thanks are due to my colleague,
Professor H. R. Grummann, for calling my attention to
this method.
The ehapter on the theory of equations contains more
material than is to be found in many texts on college algebra. One important feature is the inclusion of Descl),rtes'
rule of signs in extended form, which states that the number of positive roots is equal to the number of variations
PREFACE
ix
in sign or is less by an even number. In this form it is much
more useful and but slightly more difficult to prove. In
addition to Horner's method for the solution of equations,
another approximate method, which is essentially linear
interpolation, is given. This alternate method, in addition to its simplicity, has the advantage of being applicable
to any type of equation. The material is so arranged that
either method may be omitted at the option of th"
instructor.
Because of its usefulness, the binomial formula for other
than positive integral exponents is given, without prooL
however.
A decided simplification is made in the rules for determining the characteristic of a logarithm and for pointing
off an antilogarithm, whereby the four rules ordinarily
given are replaced by a single rule. The question of the
accuracy of results obtained by using approximate numbers is discussed.
In connection with the chapter on compound intereiit
and annuities there are given tables of compound amounts,
present values, amounts and present values of annuities.
Since anyone having much to do with financial transactions will use tables of this sort, it seems desirable for the
student to learn to employ such tables in the solution of
problems.
The application of probability to mortality tables i:::;
emphasized in the chapter on probability.
Because of the recent rapid growth of the subject of
statistics and because many students may later be engaged
in statistical work, it has seemed advisable to include an
elementary chapter on finite differences. This subject,
long omitted from books on algebra, deserves to he
included again.
I wish to take this opportunity to thank my eolle:.tglle~
in Washington University for using the preliminary
multigraphed edition of the book and for making; varion:::;
x
PREFACE
suggestions for its improvement. I wish also to express
my appreciation to The Macmillan Company for the very
valuable editorial assistance which they have rendered
during the process of revision and publication. The
revision of the multigraphed edition was critically read
by three of their advisers, and after another very thorough
revision, in which the criticisms of these advisers were
taken into ac~ount and their suggestions incorporated, it
was again read by three advisers, after which further
revisions were made.
P. R. R.
ST. LOUIS, Mo.
February, 1940
\.
Contents
I. Review of Elementary Algebra.
1.1 Introduction, 1. 1.2. Fundamental operations, 1. 1.3. Fundamental assumptions, 2.
1.4. Signed numbers, 3.
1.5. Operations
with signed numbers, 4.
1.6. Algebraic expressions, 6.
1.7. Exponents, 7.
1.8. Addition and subtraction of similar terms, 7.
1.9. Multiplication of terms, 8.
1.10. Powers of terms, 9.
1.11. Division of terms, 10.
1.12. Symbols of grouping, 11.
1.13. Polynomials, 14.
1.14. Addition and subtraction of polynomials, 15.
1.15. Multiplication of polynomials, 16.
1.16. Division of polynomials, 18.
1.17. Constants and variables. 20.
1.18. Functions, 21. 1.19. The axis of real numbers, 24. 1.20. Rectangular coordinates, 25. 1.21. Graphs, 26.
1
',J
II. Linear Equations.
2.1. Equations, 30. 2.2. Equivalent equations, 31. 2.3. Opera2.4. Linear equations in
tions yielding equivalent equations, 31.
2.5. Worded problems, 34.
2.6. Linear equaone unknown, 32.
tions in two unknowns, 40.
2.7. Linear equations in three or more
unknowns, 42. 2.8. More worded problems, 45.
III. Factoring.
S3
3.1. Polynomials, 53. 3.2. Factoring, 53. 3.3. Type products,
54. 3.4. Elementary factor types, 55. 3.5. More complicated types,
3.7. Suggestions for factoring, 60.
57.
3.6. Other types, 59.
3.8. Highest common factor and lowest common multiple, 62.
IV. Fractions.
64
4.1. Fundamental principles, 64. 4.2. Addition and subtraction
of fractions, 66. 4.3. Mixed expressions, 67. 4.4. MUltiplication of
fractions, 70. i.5. Powers of fractions, 70. 4.8. Division of fractions,
71.
4.7. Complex fractions, 72.
4.8. Extraneous roots, 76.
4.9. Equations involving fractions, 77.
V. Exponents and Radicals.
6.1. Laws of exponents, 80. 6.2. Roots, 80. 6.3. Radicals, 81.
5.4. Fractional exponents, 82.
5.6. Zero exponent, 83.
6.6. Xcgative exponents, 84.
6.7. Rational and irrational numbers, 87.
6.8. Laws of radicals, 87. 6.9. Removing factors from and introducing
them into radicals, 88.
6.10. Reducing the index of a radical, 88.
6.11 Rationalizing the denominator, 89. &,12. Reduction of radical
,,;
80
.<lii
CONTENTS
--------------------------------------------------___.
expr('ssions to simplest form, 92. 5.13. Addition and subtrantion of
radicals. 93. 5.14. Multiplication of radicals, 94. 5.15. Division of
radicals, 96. 5.16. Imaginary and complex numbers, 97. 5.17. Operations with complex numbers, 98.
/
:
101
VI. Quadratic Equations.
6.1. Quadratic equations, 101.
6.2. Solution by factoring, 101.
6.4. Completing the square,
6.3. First-degree term missing, 102.
102.
6.5. Solution by formula, 103.
6.6. Equations in quadratic
form, 109. 6.7. Equations involving radicals, 110. 6.8. Character of
the roots, 114. 6.9. Sum and product of the roots, 117. 6.10. Formation of an equation with given roots, 118. 6.11. Factoring by solving a quadratic, 120.
6.12. Graphic represcntation of a quadratic
function, 122. 6.13. Maximum or minimum value of a quadratic function, 123.
VII. Systems of Equations Involving Quadratics.
126
7.1. Quadratic equations in two unknowns, 126.
7.2. One
equation linear, one quadratic, 126.
7.S. Equations linear in the
squares of the unknowns, 128. 7.4. All terms involving the unknowns
of second degrpe, 130. 7.5. Symmptric equations, 13l. 7.8. 1\liscellaneous methods and types, 134. 7.7. Graphic representation, 136.
145
VIII. Inequalities.
8.1. Inequalities, 145.
8.3. Solution of inequalities, 148.
function, 149.
" ,
I'~
8.2. Fundamental properties, 146.
8.4. The general quadratic
IX. Variation.
154
9.1. Ratio and proportion, 154.
9.2. Direct variation, 154.
9.3. Inverse variation, 155. 9.4. Joint variation, 155.
X. Mathematical Induction and the Binomial Formula.
160
10.1. 1\Iathematical induction, 160. 10.2. The binomial formula,
10.4. The general term of the
164.
10.3. Pascal's triangle, 166.
binomial formula, 167. 10.5. The binomial theorem for positive integral exponents, 168. 10.6. The binomial series, 170.
XI. Progressions.
173
11.1. Arithmetic progression, 173. 11.2. The nth term of an
arithmetic progression, 173. 11.3. The sum of an arithmetic progres11.5. Geometric progression, 173.
11.4. Arithmetic means, 175.
sion, 176.
11.6. The nth term of a geometric progression, 176.
11.8. Geometric
11. 7. The sum of a geometric progression, 177.
11.10. Remeans, 178.
11.9. Infinite geometric progression, 179.
peating decimals, 181. 11.11. Harmonic progression, 183.
!
CONTENTS
XII. Complex Numbers.
xii;
188
12.1. Imaginary and complex numbers, 188. 12.2. The number
system of algebra, 189. 12.3. Operations with complex numbers, 190.
12.4. Geometric representation of complex numbers, 192. 12.6. Geometric addition and subtraction of complex numbers, 192. 12.6. Trigonometric form of complex numbers, 194. 12.7. Multiplication and
division of complex numbers in trigonometric form, 195. 12.8. Powers
of complex numbers, 196. 12.9. Roots of complex numbers, 197.
201
XIII. Theory of Equations.
13.1. Polynomial,201.
13.2. Remainder theorem, 201.
13.3. Factor theorem, 202. 13.4. Converse of the factor theorem. 202.
13.6. Location of the real zeros of a
13.5. Synthetic division, 203.
polynomial, 205.
13.7. Upper and lower bounds for roots, 207.
13.9. Imaginary roots. 211.
13.8. ~umber of roots, 209.
13.10. Quadratic surd roots, 212. 13.11. Graph of a factored polynomial, 213. 13.12. Descartes' rule of signs, 214. 13.13. Rational
roots, 217.
13.14. Irrational roots, 221.
13.16. Transformation to
diminish the roots of an equation by a fixed amount, 224. 13.16. Horner's method, 225. 13.17. Suggestions for finding the real roots of a
numerical equation, 228. 13.18. The general cubic, 230. 13.19. The
gen~ral quartic, 233.
13.20. Algebraic solution of equations, 236.
13.21. Coefficients in terms of roots, 236.
239
.. XIV. Logarithms.
14.1. Logarithm, 239. 14.2. Mantissa, 240. 14.3. Characteristi(', 241. 14.4. Finding the mantissa, 243. 14.6. Finding the antilogarithm, 246. 14.6. Laws of logarithms, 247. 14:i. Computation
with logarithms, 249.
14.8. Cologarithm, 253.
14.9. Exponential
equations, 256.
14.10. Other bases than 10, 257.
14.11. Common
and natural logarithms, 258.
14.12. Change of base. 259.
12.13. Miscellaneous equations, 260.
XV. Compound Interest and Annuities.
263
16.1. Compound amount, 263.
15.2. Present value, 264.
16.3. Annuity, 267. 16.4. Amount of an annuity, 268. 15.5. Present
value of an annuity, 269.
XVI. Permutations and Combinations.
274
16.1. Fundamental principle, 274. 16.2. Permutations, 275.
16.3. Permutations of things some of which are alike, 276. 16.4. Combinations, 279. 16.5. Binomial coefficients, 280. 16.6. Total number
of combinations, 280.
XVII. Probability.
17.1. Probability, 285.
17.3. Expectation, 287.
285
17.2. Mortality tables, 286.
17.4. Mutually exclusive events, 290.
xiv
CONTENTS
17.6. Independent events, 291.
17.6. Dependent events, 292.
17.7. Probability in repeated trials, 295.
XVIII. Determinants.
301
18.1. Determinants of order two, 301. 18.2. Determinants
of order three, 303.
18.3. Determinants of any order, 30S.
18.4. Properties of determinants, 309.
18.6. Expansion of a determinant by minors, 312.
18.6. Application to the solution of linear
equations, 316.
18.7. Inconsistent and dependent equations, 31S.
18.8. Linear systems with more equations than unknowns, 320.
18.9. Homogeneous equations, 322.
XIX. Partial Fractions.
326
19.1. Partial fractions, 326. 19.2. Case 1. Factors of the
denominator linear, none repeated, 327. 19.3. Case 2. Factors of the
denominator linear, some repeated, 32S. 19.4. Case 3. Factors of
the denominator quadratic, none repeated, 329. 19.6. Case 4. Factors of the denominator quadratic, some repeated, 331.
XX. Infinite Series.
334
20.1. Sequences and series, 334. 20.2. Limit, 335. 20.3. Converg~nce and divergence, 336. 20.4. Necessary condition for conver-'
gence, 337. 20.6. Fundamental assumption, 33S. 20.6. Comparison
test, 33S. 20.7. Useful comparison series, 340. 20.8. Ratio test, 341.
20.9. Series of negative terms, 345.
20.10. Alternating series, 346. (
20.11. Absolute and conditional convergence, 34S. 20.12. Ratio test
extended, 350. 20.13. Power series, 351.
Tables
355
Index
367
Answers to Odd-Numbered Exercises
375
•
_ . COLLEGE
ALGEBRA
CHAPTER I
Review of Eleqtentary Algebra
.;
:
1.1. Introdudion.
Algebra deals with numbers and operations on them.
The principal feature of algebra which distinguishes it
from arithmetic is the use of letters to represent numbers.
A number represented by a letter is called a literal number.
Since a letter, such as x or a, ml1y represent different
numbers, depending on circumstances, a literal number is
sometimes referred to as a general number.
The product of two literal numbers, for example, a
and b, is ordinarily written ab, without a ~ultiplication
sign, although it may also be written a t><. b or a' b.
Similarly, the product of an arithmetic number, such as
3, and a literal number, such as x, is usually written 3x.
In this chapter we shall state some of the fundamental
assumptions of algebra and shall recall certain principles
which result from them. Examples are given which
illustrate the application of these principles, and exercises
are provided foc the purpose of enabling the student to
review various techniques of algebraic manipulation.
1.2. Fundamental operations.
The fundamental operations of algebra are addition,
subtraction, multiplication, and division.
To subtract b from a is to find a number c such t.hat
a = b + c.
To divide a by b is to find a number c such that a = bc.
(The number c is called the quotient
of a divided by b.)
.an___..
2
REVIEW OF ELEMENTARY ALGEBRA
[Ch. I
Division by zero cannot consistently be defined to yield
a unique result. For if any number a (different from zero)
were divided by zero and the result were c (i.e., if a-+-O
were c), it would follow that a = Oc = 0, which contradicts the hypothesis that a is different from zero.
On the other hand, if the number a were zero and we
attempted to assign a value c to the quotient 0 -+- 0,
it would follow that Oc = 0, and c could be any number
whatever.
It is possible, however, to divide zero by any number
other than zero. In fact, 0 -+- b = 0, provided b ~ O.
(The symbol ~ means "is not equal to," or "is different
from.")
With the exception just noted, each of the four fundamental operations is always possible and always yields a
II ,_:, v'i] >v .);jjJ l!i~ ;'t :_;,;.,
unique result.
,j,j
1.3. Fundamental assumptions.
D.
;j~l'lI* \~Ii'!IUliJ)~rt) ~1 .,\~
"['V"
Tr
rf'9.!iCH(;j;;
Lj,
.rn.
Some of the fundamental assumptions are the following:
The sum of two numbers is the same in whatever order
they are added. That is,
a
+b =
b
+ a.
1LUi
This is the commutative law for addition.
The sum of three or more numbers is the same in whatever
way the numbers are grouped. Thus,
a
+b +c =
(a
+ b) + c = a + (b + c)
= (a
+ c) + b.
This is the associative law for addition.
The product of two numbers is the same in whatever order
.
they are multiplied. That is,
ab = ba.
This is the commutative law for multiplication.
§1.41
SIGNED NUMBERS
3
The product of three or more numbers is the same in
whatever way the numbers are grouped. Thus,
abc = (ab)c = a(bc) = (ac)b.
This is the associative law for multiplication.
The product of a number and the sum of other numbers
is the same as the sum of the products obtained by multiplying each of the other numbers by the first number. In
symbols,
a(b
+ c)
= ab
+ ac.
This is the distributive law of multiplication with respect
to addition.
Other fundamental assumptions are the following
axioms:
Quantities equal to the same quantities or to equal quantities are equal to each other.
If equal quantities are added to equal quantities the sums
are equal.
If equal quantities are subtracted from equal quantities
the remainders are equal.
If equal quantities are divided by equal quantities the
quotients are equal. *
A quantity may be substituted for its equal in any algebraic
expression. t
1.4. Signed numbers.
In § 1.1 was noted an essential difference between
algebra and arithmetic. Another difference between the
two subjects is that while arithmetic deals with positive
numbers only! algebra deals with both positive and
negative numbers (as well as certain other new types of
number).
Negative numbers arise quite naturally as a convenience in dealing with quantities which are opposite
* Division by zero is excluded. See § 1.2.
t See § 1.6.
4
REVIEW OF ELEMENTARY ALGEBRA
[(h. I
in character. Thus, assets may be regarded as positive
quantities, liabilities or debts as negative quantities.
Temperatures above zero are designated as positive,
temperatures below zero as negative. The student should
easily be able to think of other examples.
Positive and negative numbers are sometimes called
signed numbers, that is, numbers having a plus (+) or
a minus ( - ) sign. Consequently, in algebra the symbols
+ and - may be used to indicate whether a quantity is
positive or negative as well as to denote the operations of
addition and subtraction. If it is necessary to distinguish
between the + and - signs denoting positive and negative
quantities and the + and - signs denoting operations,
we may enclose the quantities, together with their signs,
in parentheses. Thus, (+8) + (-5) means that a negative 5 is to be added to a positive 8; (-3) - (+4) means
that a positive 4 is to be subtracted from a negative 3.
If no sign precedes a number the + sign is to' be
understood.
The absolute value of a positive number, or of zero,
is the number itself. The absolute value of a negative
number is found by changing the sign of the number.
Thus, the absolute value of either +5 or -5 is 5 (i.e., +5).
The symbol for absolute value is 1 I; e.g., 1-31 = 3.
1.5. Operations with signed numbers.
Certain rules, or laws, are used in operating with signed
numbers. Like the fundamental assumptions of § 1.3,
they are to be accepted as axioms. However, they could
be proved, on the basis of even more fundamental axioms,
in a more advanced treatment of the number system.
They lead to consistent results and their application
enables us to obtain correct and useful answers to practical problems. These laws for the four fundamental
operations follow.
§ 1.5]
OPERATIONS WITH SIGNED NUMBERS
5
To add two numbers with like signs (i.e., both + or
both -), add their absolute values and prefix their common
sign to the result.
To add two numbers with unlike signs, subtract the smaller
absolute value from the larger and prefix to the result the
sign of the number having the larger absolute value.
Examples of addition.
+7
+3
+10
+7
-3
+4
-7
+3
-4
J,~~,
-7
-3
-10
~_d
To subtract one signed number from another, change the
8ign of the number to be subtracted and proceed as in addition.
Examples of subtraction.
+7
+7
-7
-7
+3
-3
+3
+4
+10
-10
-3
-4
To multiply two I>igned numbers, multiply their absolute
values. If the numbers have like signs prefix a plus sign
to the product, if they have unlike signs prefix a minus
sign to the product.
Examples of multiplication.
+7
-3
-21
-7
-7
+3
-3
+21
-21
To divide one signed number by another, divide the
absolute value of the first by the absolute value of the second.
If the numbers have like signs prefix a plus sign to the
quotient, if they have unlike signs prefix a minus sign to the
quotient.
Examples of division.
+3)+21
+7
-3)+21
-7
-3)-21
+7
•
REVIEW OF ELEMENTARY ALGEBRA
6
[Ch.1
EXERCISES I. A
Add:
1. +8
+5
2.
6. +32
+27
7.
+9
.r
-4
-1
+17
11. -45
+45
12. +16
-32
16.
17.
0
+14
·d.,:\1bhr,.~
"j;)"
:>:
:i._1
4. -13
3. -7
+5
-2
~)a;
y'f«
- 1':iotJ
~5'~'
-15
Ij
8. +3
-3
\}.,
,,13. +8
'1
o
18.
0
-20
"
0
o
.:
9. -10
+6
14.-8
0
19. +2.5
-0.8
10. -20 "
-20 .
15. -66
-82
20. -8.7
-9.3
."(")
21-40. Subtract the lower number from the upper number in
exercises 1-20.
41-60. Find the products of the numbers in exercises 1-20.
Find the quotients, where possible, of the following pairs
of numbers. (The division is indicated in fractional
form.)
+16
61. +20. ", 62.
-4
+4 ;,'\
.,
-15
" 67. +15'
-21
63. +7'
-32
64. -8 .
66 +119.
. +17
68 +44.
. -44
-82
69. -82'
70 +1.2,
. +0,3
0
71. +75'
73.
+10
0
-11
74. -0'
'76.
66. -=_9t
0
72. -75'
~,
1.6. Algebraic expressions.
A symbol or combination of symbols representing a
number may be called an expression. If the only operations involved in the expression are those of addition,
subtraction, multiplication, division, raising to powers
(see § 1.7), and extraction of roots (see §§ 5.2 and 5.3) the
~xpression is an algebraic expression. For example,
5c 2
3ab - a
+ Vd
§ 1.8J
ADDITION AND SUBTRACTION OF SIMILAR TERMS
7
is an algebraic expression. (The meanings of the symhols
c2 and Vd will be understood by those who have studied
elementary algebra; they will be explained in the sections
just cited.)
When an expression is composed of parts connected
by plus or minus signs, each part, taken with the sign
immediately preceding it (+ being understood if no
\vritten sign precedes it), is called a term. The expression
in the preceding paragraph is thus composed of three
terms.
When an expression is composed of parts multiplied
together, anyone of these parts, or the _product of any
of them, is called a factor of the expression.
Any factor of a term is called the coefficient of the
remaining factor. The coefficient of theliteral part of 3
term is the numerical coefficient of the term. For
example, in the term 5axy, 5a is the coefficient of xy,
and the number 5 is the numerical coefficient of the term.
The sign is included in the coefficient. Thus, in the
expression - 3a, the coefficient of a is - 3.
If no numerical coefficient is written it is understood to
be 1. Thus, the coefficient of x is 1; the coefficient of
-y is -1.
~~~
1.7. Exponents.
3
The expression a means a' a' a. That is, it indicates
the third power of a. The number 3 is called the eyonent
of the power. The number a may be called the ~.
In general, an means a . a . . • to n factors and is called
the nth power of a.
If no exponent is written it is understood to be 1.
Thus, in the expression 5a 2 bc 3 the exponent of b is 1.
1.8. Addition and subtraction of similar terms.
By similar, or like, terms we mean numbers such as
3a and 8a, or 5X 2 y4 and _7x2y4, which are products of
REVIEW OF ELEMENTARY ALGEBRA
[Ch. I
II
the same letters with the same exponents. The numerical iI
coefficients mayor may not be the same.
To add or subtract ;:,imilar terms, add or subtract their
coefficients. This will give the coefficient of the sum or
difference, which will be similar to the added or subtracted
terms.
Examples of addition.
i';J
,'J
"'-7x 2
x2
-Gx 2
.5a 2b
-3a 2b
--_
2a 2b
lOx
3x
13x
-4abc
-4abc
---Sabc
"
,
I
I
Examples of subtraction.
,-
-7x~
5a 2b
-3a 2 b
8a 2b
lOx
3x
7x
-4abc
-4abc
--0
2
x
-8x 2
,~
,"
",
-1"
EXERCISES I. B
t"
~
>'. '
;'1'
Add:
1. 13a
9a
6. -32z2
-17z 2
9. -4U 3V 2
-9U 3V 2
13. -15t 2w 3
-42t 2w 3
I
'It
I
J
'-
2.
13a
-9a
3.
6. -27abc
7.
-72abc
1D. -20r 2 s
-19r
_2s
11.
!',
4. -7xy
4x
7x 3
3
5xy
8. -2xyz
IGmn 2
-3mn 2
-2xyz
30cx
-30cx
14. O_3z 5
2_8z 5
12.
16.
0
--6abc
~ ~',
75a 3
-19a 3
16-30. Subtract the lower number from the upper number in
exercises 1-15_
1.9. Multiplication of terms.
The product of a and a is a 2 •
Similarly,
a 2 • a 3 = (a' a) . (a - a' a)
=
a 5•
I
§ 1.10]
POWERS OF TERMS
9
In general,
That is, to multiply powers of the same base, add the
exponents of the powers to obtain the exponent to be
used in the product.
..
Note that we cannot multiply a 2 and b3 , for example,
except to indicate the product as a 2b3 •
To multiply terms, multiply their numerical coefficients
to obtain the numerical coefficient of the product. Multiply
the literal parts as explained above. (The rules of sIgns
for the multiplication of signed numbers hold.)
Exampl..s.
(-2a 2 ) • (-7a 3) = 14a 5 ,
8ab 2c3 • (-3abd 4 ) = -24a 2b*c'd'.
3a . 5b = 15ab,
3x 2 • 4 y 5 = 12x 2 y 5,
1.10. Powers of terms.
Raising to a power is merely a continued multiplicatiom
Thus,
c,
.1
$'<.
and in general
it being assumed that m and n are positive whole numbers.
Also,
.
.~
3
3
a • b = (ab) 3.
Y.\
and in general,
am . bm = (ab)m,
am. b m • em ••• = (abc . • . )m.
i --
,
Examples.
(a 3)4
(3X 3)2 = 9x 6,'
(-2X
(-4a 2b3 )3 = -64a 6b9 ,
3
(3x y)2. (2xy2)2 = (3x 3y. 2 xy 2)? = (6;r.4 y 3) 2.
4 )4
-------
•
---~
= a 12 ,
= 16x 16,
REVIEW OF ELEMENTARY ALGEBRA
10
[Ch.1
1 .11. Division of terms.
Since a 2 X as = aHa = as, it follows that, inversely,
aD + a 3 = a 5-
3
a 2•
=
In general,
,I'
I.~
where for the present we assume that m is greater than n.
That is, to divide powers of the same base, subtract the
exponent of the divisor from that of the dividend to
obtain the exponent to be used in the quotient.
To divide terms, divide the n1,merical coefficient of the
dividend by that of the divisor to obtain the numerical
coefficient of the quotient. Divide the literal parts of the
numbers as explained above. (The rules of signs for the
division of signed numbers hold.)
tc.~!':~J
Examples.
8a S + 2a 2
(-7ab 3)
2 3
-;-
28a b
= 4a 6,
= -4a,
_{r;~i'~';
2X2
- 6,
- 32xy3~5IrJig,J t:)~yz2)\,~ 8y2z~'r
EXERCISES I. C. """'/':"('"
" [';
,l.."lI'" ,.1"1 n)
t
Find the following products:
1.
4.
7.
10.
13.
15.
17.
19.
21.
23.
24.
~
j ....
'J'f;::,:,:,q....
(."1J.~.• uL,...,t
)~JldT
-3x' 6y.
~j) , ::):3. 7p' (-9q).
3a' 7a.
6. 8a' 15a 2•
2
3
3x • 3x •
/"9. 5xy· lOxy2. ',8
9xt 2 • 3x 3t3•
12. 8x 2 ·2 y 8.
14. 8uv' 9vw.
7X 2 y 3. 7 yz 6.
'i ,')',.!~ ~.16. -17 xy 2z 3. 3X 3y 2Z •
ti
22a 6x 6 y · 21ax 6 y 6.
18. a' 3ab . 20abc.
32axy· 8bxy.
. 20. 4ab . ( - 6a 2b3c4) • bcd.
3
4
(-3uv)( _2U V2)( -9v ).
22. -2a 2 b· (-3a 3 b)( -4a 4b) •.
: ".,,"
2
2
3
12m n • (-2mn8)( -38 t4).
-llx2y. ( - 9xy 2Z ) ( - 4yz 2).
8a . 2x.
2.
-5x' (-16y).
6.
-4a 3 • 3a 4 •
8.
3r 2s' (-7rs 2 ).
11.
-5a 3b2 • (-6ab 2c).
Find the following powers:
25. (X 3)6.
26. (2x2) 2.
27. (3a 3)3.
29. (a2b3) 4.
30. (-2a 2b)3. 31. (-3r 3s 6)4.
33. (10Xl0)6.
34. (-8a 2x 3 )S. 36. (9m 3n 3)3.
28. (-4X 5)3.
32. (5x6) 6.
36. (4X 4 )4.
§ 1.12]
SYMBOLS OF GROUPING
11
Find the following products:
37.
39.
41.
43.
46.
46.
Find the following quotients:
47.
60.
63.
66.
67.
68.
69.
60.
as -+ a 3•
48. 6x 6 -+ 3x 2.
49. 14x 6 -+ (-2x 2).
-50a 9 -+ 25a 6• 61. -36t2 -+ (-9t). 62. 6m 6 -+ 2m 3 •
72x 6 y 4z 4 -+ 18xy 3z2.
64. 13u 5v6w 7 -+ UV 2W 2 •
(3a 2b3c4 )7 -+ (3a 2b3c4)4.
66. (_7x2y)9 -+ (7x 2y)7.
6m 3n 6 . 2m 2n 3 -+ (-4mn 6).
(2abc 2)6. (-3a 2bc)4 -+ (12a 2b2c)2.
(-27x 2y 3)2. (-5X 2y 4Z6)3 -+ (-3xYZ)6.
(25x 3 y 4z 5)10. (-25x 3y 4Z6)6 -+ (-25x 3 y4z 6)13.
1.12. Symbols of grouping.
It is frequently convenient to group together several
terms to indicate that they are to be considered as a
single number. The symbols used for this purpose are:
parentheses ( ), brackets [ ], braces { }, and the
vinculum - - .
When an expression composed of a sum * of terms is
enclosed within symbols of grouping which are preceded
by a plus sign, the symbols may be removed without changing
the value of the expression..
. ; '.1>0'_': ,
Example.
3a
+ (b
:~ f::\ P H 11_~.;·;:
- 5)
= 3a
+b-
5.
An expression composed of a sum of terms may always
be enclosed within symbols of grouping preceded by a plus
sign, the sign of every term being retained.
* In algebra an expression such as 4a
- 3b - c
+ 7d is referred to as a
sum even though subtractions as well as additions are involved, for the
Bubtraction of a number is equivalent to the addition of the negative of
.
the number.
[(h. I
REVIEW OF ELEMENTARY ALGEBRA
----------------------------~~~-----
12
Example.
x -
y
+
z = x
+
(-y
+ z).
When an expression composed of a sum of term,~ ~s
enclosed within symbols of grouping which are preceded
by a minus sign, the symbols may be removed provided the
sign of every term is changed.
, l
Example.
4x - (3y - z
+ 2)
= 4x - 3y
+z
- 2.
An expression composed of a sum of terms may be enclosed
within symbols of grouping with a minus sign prefixed
provided the sign of every term enclo~;ed is changed.
Examples.
a
+
X -
b - c = a - (- b + c),
Y - z = x - (y + z).
<~
.J:
A quantity multiplying or dividing an expression within
symbols of grouping multiplies or divides every term within
the symbols.
~. ~
'f·
'.,
• '"
Examples.
+
4x
2(5y - 3z) = 4x + lOy - 6z,
4x - 2(5y - 3z) = 4x - lOy + 6z.
This is merely the distributive law of multiplication
with respect to addition. (See § 1.3.) The rule relating
to the change of signs when removing symbols of grouping
preceded by a minus sign is a special case, since the minus
sign is equivalent to the multiplier -1.
The foregoing rules apply when symbols of grouping are
contained within other such symbols. In removing symbols of grouping there is ordinarily less likelihood of mistakes if the innermost symbols are removed first, although
if care is exercised they may be removed in any order
whatever. Similar terms should be combined at each
stage.
EXERCISES
13
Examples.
7a - [6b - (c - d)
+ e]
= 7a - 6b
+ (c
- d) - e
7a - 6b + c - d - e,
2[x - 4y - (c - d)] - 5}
= 3a - {b + 2[x - 4y - c + d] - 5}
= 3a - {b + 2x - 8y - 2c + 2d - 5}
= 3a - b - 2x + 8y + 2c - 2d + 5.
=
3a -
{b
+
The correctness of the operations in an example such
as the foregoing can be checked, although not proved, by
assigning numerical values to the literal quantities and
comparing the value of the original expression with that
of the final form. * Do not assign the values 0 or 1 to any
of the letters.
Thus, in the first example of the last pair, let us set a -= 2,
b = 3, c = 4, d = 5, e = 6. We find
7a - [6b - (c - d)
7a - 6b
+c-
+ e]
= 7·2 _: [6' 3 - (4 - 5)
= 14 - [18 + 1 + 6]
=
14 - 25
=
d - e = 7· 2 - 6·3
= 14 - 18
-
+
6]
11.
+4 -
+4 -
5 - 6
5 - 6 = -11.
EXERCISES I. D
Remove the symbols of grouping and combine like terms:
1. 8 - [7 - (3 - 5) + (10 - 4)].
2. 15 - 2[5 - 3(13 - 9) + 11].
3.
4.
6.
6.
7.
8.
9.
10.
lOa - [3a + 2(b - 3e) - 6].
3x + 2[5x - 6y - (x + y - z)].
a(b + c - d) - b(a - e + d).
2{7a 2 - [a 2 + b2 - (3a 2 + b2 - 5)] + e 2 + 6}.
3(a - 5b) - {7a - [3(a - b - 2) + 4]}.
a(b - c) + b(e - a) + e(a - b).
x(x - 4) + 8{3x - 2x[x + 5y - (3x - x - 2)]{.
9{a - [2b - 7(a - e)Jl - 6[3(a - b) - 4(a - e)].
* See last
paragraph of § 1.15.
ft
t
[(h. I
REVIEW OF ELEMENT AIW ALGEBRA
14
Enclose the last three terms of the following expressions in
parentheses preceded by a minus sign:
11. a - 2b - 3c - 4d. ,.'
13. 5p
6q - 7r
88~;;~
16. x 2 - y2 + 2yz - Z2:·'~.:
+
+
;{
12. 3x ~ y + z - 10.
14. m 2 + n 2 + 3mn - 2m - 4n.
16. 4a 2 - 9b 2 - 12bc - 4c 2•
Enclose the coefficient of x in parentheses preceded by a plus
,.(i{ :
sign:
d,H;
17. ax - bx + 3x.
.; !;,! 18. ax ~ bx + 3.
~Jf!tc!~H·~;:~g
19. x + ax + bx - ex. If"
20. 2x ~ 4ax + 3xy. ;!,! nI1nq l1!(r~)
21. - 3xy + y2 - 5x + 7 y + 4.
Jf'niJ ~)I~J io
22. xy2 + xz - 3x + lOy - 11.
',j-jEll
'edit 30
Find the value of each of the following expressions for the
given values of the literal numbers irlvolved:
Hi <,."lifT '
23. 5(2x - 3y + 7) - 4{x - 2y[2x ~ 3 + 4(1 - x)J ~'iyr; 'J
x = 2, y = 3.
24. 10{7a - [6(a - 2b - 3) + 4ab)}; a = -6, b = -7.
25. 8{a - [3b - 5(a - 2c)]} - 7[4(a - b) - 3(a - c»); a = 3,
b = -2, c = 5.
26. 5{2a - b + c - 3[2a - \b - c\ ~ 2(a - c»)l; a = 1,
b = 2, c = 3.
27. x 2 - y(x - z)[3x - 2(y - Z)2 - 7x 3 y 2z1; x = 4, y = -3,
z = 7.
./
(j(
1.13. Polynomials.
'.
h,.,.-8f4
d
i rn ': .?
An expression composed ~f only one t~rm 'is a monomial,
an expression composed of two terms is a binomial, an
expression composed of three terms is a trinomial.
An expression composed of m()re than one term is a
polynomial, or a multinomial.
NOTE. The word "polynomial" iig sometimes restricted to
mean an expression such as
5x 3
+ 3x
2
-
4x
+ 9,
which is composed of positive inte~ral (i.e., positive wholenumber) powers of a literal number or literal numbers. (See
.
EXERCISES
15
§ 3.1.) In this restricted case it is sometimes convenient to
classify a monomial as a polynomial.
1.14. Addition and subtraction of polynomials.
Polynomials are. added or subtracted by combining
similar terms.
.f t ._. ::,.f)
Example 1.
Add 3a - 2b
+
e2 and 4a
+ 6b
',~"
- 9c 2 •
'0 ,\! ....
'SOLUTION. Arrange the expressions so that simil~'ierms will
be in vertical alignment. Then add as in § 1.8. ..
+
3a - 2b
e2
4a
6b - 9c 2
+
7a + 4b
: r ,", --
- 8e\
~":t
Example 2.
Subtract 3a - 2b
+
5d from 5a
+
7b - 8c . .?
SOLUTION. Arrange the expressions so that similar terms will
be in vertical alignment, placing the expression which is to be
subtracted below the other expression. Then subtract as in
§ 1.8.
+
5a
7b - 8c
3a - 2b
5d
2a + 9b - 8c - 5d
+
EXERCISES I. E
"
Add the following polynomials:
1. 16a + 9b, 8a + 7b.
3. x 3 - 3x 2 + 7x + 2, 5x 3
4.
5.
6.
7.
8.
'9.
+x
'r
2. 21x - 30y, 19x
2
+
lOx
24.
-3a
7b - 31e, 6a - 14b - 44c.
7a~b .~ 3ab 2 , 12a 2b - 18ab 2 - a 3b2 •
3x 4
5x 3 - 7x
2, x 3
2x 2
7x - 11.
3
2
3
5x - 6x - 3x
4, 5x
6x 2
3x - 4.
x 3 - llx 2
22x - 14, 3x 4 - 2x 3
7x 2 - 10.
ab
be
cd, ae
be
bd.
+
+
+
+ +
+
+
-
+ +
+ +
+ +
+
+ 20y.
REVIEW OF ELEMENTARY ALGEBRA
16
+ 41b 2c + 28c d4, 24ab
10. 27ab 2
3
3
2
57b 2c 3
-
+
22c 3d 4•
-
11. 3p2 - 2pq + 17q2, 2q2 - 18qp - 5p2.
12. 3a 3b 2e2 - 32a 2b 3e 2 - 24ab 4c 2 + 18abe 5 ,
[Ch.1
1 I
8a b c
50a b c - 66ab c 2 - 68abc 5 •
2
2
13. 7a
3b - 5c 2
x 2 - 5y2 - 6z 2,
,~
2
2
x - 2y2
3z - 4a 2 - 8b 2
9c 2 •
140 6a
b - 4e, 3a - 6b - 5e, -7 a - 3b
4e.
15. x 3 - 64x 2 - 99x
18, 5x 3 + 9x 2 - 14,
4x 8 - 4x 2 - 75x - 6.
\ ,
~1r5!~' ~" '1
16. 4a + 3b + 2e, 6b - 8e - 7d, 14e - 18d - lle:"' .
17. p - q, q - r, r - s, s - t, t - p.
18. x - a, x + a, a - x, a
x.
19. 2lm 2n - 31mn2, 18mn 2
22m2n, m 2
n 2.
3
2
2
20. 32x + 53x y - 73xy 2, 5x y - 71xy2 + 26 y 3, x 3 _ y3. ;!' A
21-40. Subtract the second expression from the first in ekellcises 1-20.
41. Subtract 5x 2 - 17xy
22y2 from O.
42. Subtract x 3 - 3x 2
7x
4 from 1.
43. Subtract the sum of
3
2 2
+
2 3 2
+
+
+
4
:f
+
+
+
+
+
+
13a
+ 22b
+
+
+
- 8lc and 25a - 18fi+ a
from lOa - 20b + 35e.
.>" •.~ /'"
2
44. Subtract 42x2 + 68y2 + 72z from the sum of
86x 2
-
t
37z 2 anti 51 y 2
+ 94z
2
.?
)'" "iJ
•
45. Subtract the sum of
x3
+
17x 2
+ 3Gx
- 15 and 3x 3
-
13x
+ 33
from the sum of
-2x 3
-
8x 2
+ 79x + 8 and 5x
3
-
73x 2
+ 99x + 67.
1.15. Multiplication of polynomials.
To multiply a polynomial by a monomial, multiply each
term of the polynomial by the monom1:al.
To multiply one polynomial by another, multiply each
term of either by each term of the other. Combine similar
terms.
.r.
i
EXERCISES
11
Example.
Multiply 2a 2
SOLUTION.
3ab
-
2a 2
a2
-
+
2
by a 2
3ab +
2ab -
+
3
15a 2b 2
a b -
CBECJ[. Let a = 3, b = 2.
2a 2 - 3ab + b 2
a 2 + 2ab - 5b 2
2a 4
+ aSb -
15a 2 b 2
+ 17ab
3
+ 2ab
b2
5b 2
3a 3b +
a 2b 2
3
4a b 6a 2b 2
- lOa 2b 2
2a 4 -
2a 4
+b
-
.~,.
+
+
+
- 5b 2
".1'
2(!b 3
15ab 3
17ab
3
-
5b 4
-
5b 4
= 18 - 18 + 4 = 4
= 9 + 12 - 20 = 1
Product = 4.
5b 4
,<"
, .~.' "
,rJl
','
= 162 + 54 - 540 + 408 - 80 = 4.
The first line below the first horizontal rule, viz.,
2a 4 - 3a 3b + a 2b2, is the product of a 2 and 2a 2 - 3ab
+ b2 , and so on. Similar terms, such as -3a 3b and
4a 3 b, have been placed in columns and added. This is
permissible, since the order in which numbers are added is
imma terial.
The work may be checked algebraically by multiplying
with multiplier and multiplicand interchanged, or by
dividing (see next section) the product by either of the
polynomials to obtain the other. Checking by the substitution of numbers .does not prove that the algebraic
work is correct. If the numbers check it is probable that
the work is correct; on the other hand, if the numbers
fail to check there is certainly an error.
EXERCISES I. F
Find the products of the following polynomials:
1.
2.
3.
4.
3x 2
8x 2
6x 2
9x 2
+
-
-
61: - 5.
7x - 9,
X + 12,
4x - 3,
x + 7.
x - 5.
3x - 5.
7x + 8.
REVIEW OF ELEMENTARY ALGEBRA
18
[(h. I
5..r2 + g.t + 10, x 2 + X + 4.
6. 2a 2 - 3a + 5, 3a 2 + 15a + 20.
7. 5x 2 - 8xy + 2y2, x 2 - 2xy + 3y2.
8. a + b - c, a - b + c.
9. 2a - 3b - 8e, 4a + 7b - 3c.
10. 2x 3 - 5x 2 + 6x - 2, 2x - 5.
'
11. 7x 3 - 2X2 - 8x + 11, x 2 - 4x + 3.
12. 2X4 - 3x 2 5x - 4, 2x 3 - 5x - 1.
13. 2x 3 - x2y + y3, 3x 2y - 4xy2 + 5y3.
14. 2X2 - x 3 - 4x + 5, 3x - 2x 2 + 7.
)
,,
I
I
+
SUGGESTION. Rearrange both expressions according to
scending powers of x.
15. 10x 2 - 3 X 3 - 4x - 3, x 3 - 2x 2 - 5.
16. x - x 3 - x 2 + 9, 2x2 + 7x - 8.
17. x - 2, 2x + 3, x 2 - 7x + 6.
18. x 2 - 2x + 3, 3x 2 - 4x - 7, x 2 + 5. i
[I
19. a 2 - 2ab - b2, 4a 2 - ab - 3b 2, 9a 2 - 4ab - 2b 2•
de~
Square:
21. a + b + (!,.
23. x 2 - 2xy .:.... 3y2.
20. 3a 2 - 2b 2 •
22. 2X2 - 3x + 5.
Cube:
+
24. a + b.
26. x 2 - 2y2.
25. x
2y.
27. 2x2 + 5y2.
Perform the indicated g~erations:
28. (a + b)2(a - b).
30. (x - y)3(X + y).
+ 2y)2(2x
29. (3:e
31. (a 2
-
2ab
- 3y)2.
+ 3b 2)(2a -
b) II.
1.16. Division of polynomials.
To divide a polynomial by a monomial, divide each term
of the polynomial by the monomial.
Example.
(6a 3b2 - 12a 2b3
-
3ab 4 )
-;-
3ab
= (6a 3b2 -;- 3ab) - (12a 2b3
= 2a 2 b - 4ab 2 - b3 •
-;-
3ab) - (3ab 4 + 3ab)
DIVISION OF POLYNOMIALS
~1.161
19
The method of dividing one polynomial by another is
illustrated by the following example.
Example.
Divide 5x 2
SOLUTION.
powers of x.
2x 3
-
+ 3 by x
2 -
4 - 2x ..
,
,~
'".. '
\ ... .l. '
} ;
Arrange bo~h polynomials according ~~~aui.
The work IS shown below.
;~: '. ;,.. ' ~)
(Divisor =) x 2
-2x + 1( = Quotient) jl!; •.• '
2x - 4) - 2x 3 + 5x 2
+ 3 ( ""'" Dividend) ;
-2x 3 + 4x 2 + 8x
,&0 .
x 2 - 8x + 3
x 2 - 2x - 4
- 6x + 7 (= Remainder)
-
It is found, by dividing the first term of the divisor into the
first term of the dividend, that the first term in the quotient
is -2x (i.e., -ZX 3 /X 2 = -2x). Multiply the divisor x 2 - 2x 4 by -2x, obtaining the polynomial -2x 3 + 4x 2 + 8x, which
is written below the dividend and subtracted from it. Continue the process until the remainder is of lower degree* than
the divisor.
Division may be checked by using the relation
Dividend
=
Quotient X Divisor
+ Remainder
(1)
either algebraically or by substituting numbers. A number which makes the divisor equal to zero must not be
used. (See § 1.2.)
We shall check the foregoing division by letting x = 2.
find
Dividend
Divisor
Quotient
Remainder
We
= -16 + 20 + 3 = 7.
2x - 4 = 4 - 4 - 4 = -4.
= -2x + 1 = -4 + 1 = -3.
= -6x + 7 = -12 + 7 = -5.
= -2x 3 + 5x 2 + 3
= x2
-
* The degree of a polynomial consisting of positive integral (Le., wholenumber) powers of x is the exponent of the highest power of x occurring in
the polynomial. Thus, the polynomial 2x.3 - 5x 2 + 4 is of degree three.
REVIEW OF ELEMENTARY Af-GEBRA
20
[Ch.1
Substituting in (1), we get
,
7 = -3(-4) - 5,
t!
7 = 7.
or
EXERCISES I. G
Divide the first polynomial by the second:
1.
2.
3.
4.
5.
7.
9.
10.
11.
12.
13.
14.
15.
17.
19.
21.
23.
24.
25.
26.
27.
x3
+ 5x 2 -
x3 -
8x - 12, x - 2.
'L
+ 3.
- 25x + 18x + 7x 20x + 3x + 28, 4x
1 ,
llx - 6, x
2
3
12x4
20, 3x - 4.
4
2
2
8x 6x - 7.
as - b3 , a-b.
6. a 3
b3 , a + b.
3
3
a - b , a + b.
8. as + b3 , a-b.
2
3
x - 7x + 2x + 4, x - 5.
Dxa + 5x~ - 2x - 8, 2x
3.
X4 + 5 X 3 - 13x 2 + lOx - 8, x 2 + 7x + 3.
6x 4 - llx 3
26x 2 - 7x
7, 2X2 - :3x
7.
J
4
8
5x - 13x y
4x2y2 ,- 2xy s - 7y\ x~ - 3xy
2y2.
6a 4 - 2a 3b - 15a 2b2
12ab 3 - 5b4, 3a 2 - 4ab
b2 . " "
4
a 4 - b4, a
b.
16. a - b4, a-b.
a 4 - b4 , a 2
b2 •
18. a 4 - h4, a 2 - b2•
a 4 + b4, a
b.
20. a 4
b4 , a-b.
aD
bD, a
b.
22. as - hO, a-h.
x 3 - X4
x2 - X
2688, x
7.
x 2 - 3x 4 - 3x 3
7x
8, x 2 - X - 2.
XO + 2X4 - 3 X 3 - 8X2 - 2x
18, X2 - 3x
2.
12x 5 - 2X4
3x 3
70x - 7, 3x 2 + X - 7.
8x 5 - 6x 4 - llx 3
3x 2
17x
15, 2x 3
3x 2
5x - 2.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Perform the following operations:
28. [(x 3 + y3) -+- (x + y)](x - y).
29. (x 4 - y4) -++ y)(x - y)].
30. (x 6 - y6) -+- [(x + y)(x - y)].
31. (x 6 + y6) -+- [(x + y)(x - y)].
rex
1.17. Constants and variables.
A. symbol which, throughout a discussion, does not
change in value is called a £2nstant. (There are, of course,
some constants, such as 3, -2, and 7r, which never change
in value.)
/
§ 1.18J
FUNCTIONS
21
A symbol which may change in value during a discussion is called a variable.
In the formula for the area of a triangle,
1
A = - bh
2
'
t
is a never changing constant. If in a certain problem we
regard only the base b of the triangle as fixed in value,
then for that problem, b is a constant, and thf) altitude h
and the area A are variables.
1.18. Functions.
When two variables are so related that to each of a set
of values of one there correspond one or more values of
the other, the second variable is said to be a function of
the first. The first variable is called the independent variable, the second the dependent variable. For example, if
. y = X2, then y is a function of x, and we may regard x as
the independent variable, to which values are arbitrarily
assigned, and y as dependent on x.
This type of relation is often written in the form
f(x) =
X2,
which may be read" the f function of x equals x 2 ," or,
more briefly, "f of x equals x 2." If f(x) = X2, then f(a)
= a 2 , f(3) = 3 2 = 9, f( -3) = (-3)2 = 9, and so on.
In another problem or discussion we might have f(x)
= 2x + 3 or f(x) = !/x.
That is, f(x) does not always
represent the same function of x. N or is f the only letter
used to represent functions; we may have F(x), g(x),
cp(x), etc.
The area of a circle is obviously a function of its radius,
and we may write A = fer). In this instance ,ve know the
form of the function, namely, fer) = 7rr2. Similarly, the
edge of a cube is a function of its volume. Here fey)
= ~V.
I
[(h. I
REVIEW OF ELEMENTARY ALGEBRA
22
i
I
NOTE. It is assumed that the student is familiar with the
ideas of square root, cube root, etc., and with the corresponding
symbols. However, these topics will be treated later in detail.
Often one variable is a function of two or more other .
variables. For example, the volume of a right circular
cone is a function of its base and its altitude, since V
= pr 2h. This may be written
1
V = f(r,h) = 311"r 2 h.
i,
\ j
If the
radius is 2 and the altitude 3, we have
V = f(2,3) =
1
311"
. 22. 3
=
411".
Example 1.
Given f(x) = 2x 2
SOLUTION.
f(3x
-
3x - 35; find f(3x
+ 5).
Replace x everywhere by 3x
+ 5)
= 2(3x
+ 5)2 -
3(3x
+ 5.
+ 5)
J • ,
I
i
~
l'
!':jd ~ ': i,
- 35
= 2(9x2 + 30x + 25). - 3(3x + 5) = 18x 2 + 60x + 50 - 9x - 15 - 35
= 18x 2
f
+ 51x.
.j
C
,
35 ',-:
'
Example 2.
f
11
,"
Given lex) = I
x
2
+ x' g(x)
= x2
-
2; find f[g(x)J.
SOLUTION. Replace x, wherever it appears in f(x)~
then simplify.
f[(l(x)] = I
[g(X)]2
(x 2 - 2)2
g(x) = 1
X2 _ 2
+
+
by ,,(21),
;'"
/1,...
I
Example 3.
Given f(x,y) == x 2
2).
+ 4y2 + 6x -
16y
+ 9; find f(x
- 3, y
+
/
EXERCISES
SOLUTION.
I(x - 3, y
23
Replace x by x - 3 and y by y
+ 2)
= (x - 3)2
= x 2 - 6x
+ 2.
+ 4(y + 2)2 + 6(x
+ 9 + 4y2 +
= x 2 + 4y2 - 16.
- 3)
- 16(y + 2) + !)
16y + 16 + 6x - 18
- 16y - 32 + \)
EXERCISES I. H
1. Given f(x) = 2x + 3; find f(O), f(1), f(2), f( -1), f( -2),
fC!), f(a), f(a 2) , f(1 - y).
2. Given f(x) = 7x - 2; find f(O), f(I), f(2), f( -1), f( -2),
f(a + b), f(3 - 2e).
3. Given f(x) = 3 - x; find f(I), f(3), f( -3), f(10), f(x 2),
f( -x 3 ).
4. Given f(x) = x 2 - 7x + 12; find f(O) , 1(1), f(2), f(3), f(4),
f(5),j(-I),f(-2),j(-3),f(-4),j(-5),j(a + 3),f(a - 4).
6. Given f(x) = x 2 + 2, g(x) = 3x - 5; find f(2) + g(4),
f(3) . g(5), f[g(2)], g[f(I)]' [f(-2))2· g(-3), [g(a)]2 - 9f(a)
- 7.
6. Given F(x) = 5x - 3; find [F(z)]2 - F(Z2).
7. Givenf(x) = 3x
7; findf(a - 3)[f(a) - 3].
8. Given hex) = (x
a) (x + b); find h(a), h(b), h(a
b),
h(a - b).
9. Given f(x) = 3x 4 + 5x 2 - 18; show that f( -x) = f(x).
10. Givenf(x) = 7x 3 - 2x; show thatf(-x) = -f(x).
11. Givenf(x) = x 2 - 3x + 2; findf[f(x)].
12. Givenf(x,y) = x 2 - 3xy + 4y2; findf(2,3),f(3,2),f(2,-3);
f( -2,3), f( -2,-3), f( -3,2), f( -3,-2), f(a,a).
13. Given F(x,y) = 3x 2 .;_ 2xy - 4y2 - 6x + 3y + 7; find
F(y,x), F( -x,y), F(x,-y), F( -x,-y).
14. Given f(x) = ex; show that f(a) + feb) = f(a + b).
16. Given f(x) = ex; show that f(a) - feb) = f(a - b).
16. Given f(x) = x-I; show that [f(a)F + f(a 2) ~ 2af(a).
+
+
+
Express the following in functional notation and then giye
the particular form of the function:
17. The circumference e of a circle in terms of its radius r.
SOLUTION.
e = fer) = 211'r.
REVIEW OF ELEMENTARY ALGEBRA
[(h. I
,
18. The area A of a circle in terms of its radius r.
19. The volume V and the surface S of a cube in terms of its
side a.
20. The distance d, in miles, covered by an automobile traveling
at the rate of 55 miles an hour, in terms of the time t, in
hours.
21. The number of gallons g of gasoline used by an automobile
in traveling a distance of m miles, the mileage of the car
being 18 per gallon.
22. The monthly cost C, in dollars, for telephone service, in
terms of n, the number of calls made during the month,
there being a basic charge of $3.00 per month and an additional charge of 5 cents for each call made.
23. The volume V and the total surface T of a right circular
cylinder in terms of its radius r and height h.
24. The total surface T of a rectangular parallelepiped in terms
of its length l, width w, and height h.
25. The lateral surface L and the total surface T of a right
cin'ular cone in terms of h, its height, and r, the radius of
its base.
1.19. The axis of real numbers.
On a straight line X' X, of unlimited extent in both
directions, let us select an arbitrary reference point O.
(See Fig. 1.1.) This point will be called the origin and
will be used as a zero
o
~~6--~4~-~3--~2---1~+O~1--2~+3~4~~5~X point from which to
measure distances.
FIG. 1.1.
Next let us choose a
convenient unit of measurement. If we regard distances
measured to the right as positive and those measured to the
left as negative, we may mark off the numbers 1, 2, 3, . . .
to the right of 0 and -1, -2, -3, . . . to the left of O.
~OTE. The symbol . . . is the symbol of continuation and
may be read" and so on."
The numbered scale thus obtained is called the axis
of real numbers, or the real axis, and it will be assumed
!
t
I
§ 1.20]
RECTANGULAR COORDINATES
25
that to every real number there corresponds one and
only one point on the axis and that to every point on the
axis there corresponds one and only one real number.
(For a discussion of real numbers and other kinds of
numbers see § 12.2.)
1.20. Rectangular coordinates.
Let us now take two axes X' X and Y' Y intersecting
at right angles at a common origin O. (See Fig. 1.2.)
On each axis we choose a
y
convenient unit of measureP(-,...3__
,4.;_)_:
ment, which mayor may
:I
not be the same for both
2
axes.
:t __ 3 1
The horizontal line X'X
is called the x-axis, the X'-5 -4 -3 -2 -1 -10 1 2 8
vertical line y' Y is called
-2
-s
the y-axis. Positive direc-4
tions will be chosen to the
y'
right for the former and
FIG. 1.2.
upward for the latter, as
is indicated by the arrow tips. This choice is arbitrary,
but customary.
Now take any point P. The distance of the point from
the y-axis is called the abscissa of the point and is denoted
by x. Its distance from the x-axis is called its ordinate
and is denoted by y. The abscissa and ordinate together are called the coordinates (more specifically,
the rectangular coordinates) of the point. The point
P whose abscissa is x and whose ordinate is y is designated by (x,y) or P(x,y), the abscissa always being
named first.
Obviously, if a point is to the right of the y-axi~
its abscissa is positive, if it is to the left its abscissa
is negative. If a point is above the x-axis its ordif'
26
[Ch. I
REVIEW OF ELEMENTARY ALGEBRA
nate is positive, if it is below the x-axis its ordinate is
negative.
The point P( -3,4) in Fig. 1.2 has the abscissa x = -3
and the ordinate y = 4.
1.21. Graphs.
The process of locating and marking a point whose
coordinates are given is called plotting the point. Plotting
y
y
= 2x
x
I
-1
0
1
0
2
--- +-
I
I
y
-3
-2
,
+3
3
-3
-1
1
3
5
7
9
y,
is facilitated by the use of
coordinate paper, or graph
paper, that is, paper having hoth horizontal and vertical
rulings. (See Fig. 1.3.)
Let us now consider the function
FIG. 1.3.
y = 2x + 3
y = f(x) = 2x
+ 3.
By assigning values to x and finding the corresponding
values of y we can construct a table such as the accompanying one. If the pairs of numbers found in this table
are taken as coordinates of points and these points are
plotted, they will be found to lie on a straight line. (See
Fig. 1.3.) The line is called the graph of the function f(x)
= 2x + 3 or of the equation y = 2x + 3.
Proceeding in the same way with the function
y = f(x)
=
X2,
,,
i
EXERCISES
27
we find that the points which we obtain will lie on a
curve (Fig. 1.4).
If the relation between x and y is given in the form
3x + 2y = 6, for example, in which neither variable is
y
x
y
-4
16
9
4
1
0
1
4
9
16
-3
-2
-1
0
1
2
3
4
II
\
,
expressed explicitly as a function of the other, we may
solve for one of them before
attempting to construct the graph.
equation we should find
0
FIG. 1.4.
y = x2
Thus, in the above
,
,
befo~ ~btaining
3x,
or
x
y'
'I'
= 6 -
J
iI
1\
x
2y
I
iT
\
y=
6 - 3x
2
values and plotting.
EXERCISES I. I
Plot and label the following points:
1. (5,8), (5,-8), (-5,8), (-5,-8), (8,5), (8,-5), (-8,5),
(-8,-5).
2. (6,0), (-6,0), (0,6), (0,-6), (0,0).
3. (211',3), (211',311'), (2 + 11',3), (2,3 - 11'), (-311',11' - 6).
,4. (7,7), (7,-7), (-7,7), (-7,-7), (11',11'), (11',-11'), (-11',11'),
(-11',-11").
.
6. (t,-4), (i,2i), (-1.8,-3.2), (3.25,-6.75), (-8.3,0).
28
REVIEW OF ELEMENTARY ALGEBRA
[(h. I
6. Find the distance from the origin to each of the following
points:
(5,12), (3,-4), (-8,15), (-24,-7), (9,0), (0,-10),
(6,03), (-8,-6), (0,yl1), (-2,7).
SUGGESTION.
Use the theorem of Pythagoras.
Draw the graphs of the following equations. If the graph is
not a straight line, enough points should be plotted to give a
smooth appearance to the curve. Give fractional values to one
of the variables if necessary.
7. !J = .r - 4.
9. !J = 2x.
11. y = -x.
13. y = 2x - 5.
8.
10.
12.
14.
16.
18.
20.
22.
24.
+
15. y = i-x
4.
17. y = 3x - 6.
19. y = x 2 - 9.
21. y = tx2 - 2x.
23. y = 4 - x 2•
25. y = 6x - x 2 •
27. y = x 2 - 6x + 5.
29. y = x 2 - 6x + 9.
31. y = (x - 2) (x
3).
33. y = x(x - 6).
S
+
35. y = 0.lx 3 •
37. x = y2 - 3y.
39. x 2 + 3y - 9 = O.
26.
,T::<
28.
30.
32.
y
y
y
y
y
y
Y
y
= -x - 4.
= x + 3.
=
3 - x.
+
= 2x
5.
= -tx
5.
= - 4x
7.
= iX2
1.
= X2 - X
1.
y = 6
2x - x 2 •
+
+
+
+
+
y = 2x 2 + 5x.
y = x + 7x + 10.
y = x 2 + 4x + 4.
y = (x + 1) (x - 5).
y = x(x + 6).
2
34.
36. y = t x 3 - x.
38. x 2 = 2y - 6.
40. x 2 + X + y + 1 = O.
41. A farmer wishes to enclose a rectangular plot of ground.
A stone wall is available for one side of the enclosure, and
he has 50 yards of fence to use for the other three sides.
Taking x as the length of that side of the enclosure which
is perpendicular to the wall, express the area of the enclosure as a function of x. Draw a graph of the function and
from it estimate the value of x which will give the greatest
area for the enclosure.
42. A lidless box is constructed by cutting equal squares from
the corners of a sheet of cardboard 16 by 16 inches and
/
I,
EXERCISES
29
bending up the sides. Express the volume of the box as a
function of the side x of the squares cut out. Draw a graph
of the function and estimate the value of x that will yield
the box of greatest volume.
43. Work the preceding exercise for the case in which the sheet
of cardboard is 10 by 16 inches.
44. A box with a square base is to have a capacity of 54 cubic
feet. The cost of material for the top and bottom is 20
cents per square foot, for the sides it is 10 cents per square
foot. Express the total cost of material as a function of x,
the side of the base. Draw a graph of the function and estimate the value of x which will make the cost a minimum.
CHAPTER II
Linear EquatioO:$:d
i
I
(I
"
l,i.
2.1. Equations.
.,' ~ • \ (~J
1
;>
An equation is a statement that two expressions are
equal, the expressions being called the members, or sides,
of the equation. These members will be called left or
right according as they precede or follow the sign = in
the equation. An equation ordinarily contains one or
more literal quantities, which are called unknowns, and
it is the object of the processes of algebra to find the values
of these quantities. Thus,
3x-7=5
is an equation'in which the unknown is x.
An equation whose members are equal for' all admissible* values of the unknown or unknowns which it
contains is an identical equation, or an identity. An
equation whose members are equal for certain values (or
possibly for no values), of the unknown or unknowns
which it contains, but not for all admissible values, is a
conditional equation. The word" equation," when used
without a modifier, will ordinarily be interpreted to mean
"conditional equation."
If an equation contains only one unknown, any number
which, when substituted for the unknown, makes the
members of the equation equal to each other is called a
root, or solution, of the equation, and is said to satisfy
the equation. If an equation contains more than one
* An admissible value is one for which all of the expressions Gontained
in the equation have meaning,
30
',2.3J
OPERATIONS YIELDING EQUIVALEN7 EQUATIONS
31
Jnknown, any set of numbers which, \\Then substituted
the unknowns, makes the members equal to each othel
\s a solution. '
The equations
f
'~or
2(x - 3) = 2x - 6,
x
2
+2
x-I
+ y)2 = x + 2xy + y2,
x + 1 + _3_
(x
=
2
x-I
"re identities, being satisfied by all admissible values of
the unknowns. (Note that in the last identit.y, 1 is not
an admissible value for x; the members are not defined
for x = 1, since this value would involve division by
zero.)
The equation 3x - 7 = 5, on the other hand, is 8
conditional equation, since it is satisfied only by x = 4.
An identity is often written with the sign == instead
of = to emphasize that it is an identity rather than 3
conditional equation. Thus,
2(x - 3) == 2x - 6.
2.2. Equivalent equations.
Two equations are equivalent if they have exactly the
same solutions. Thus, the equations
x-5=0
llnd
2x
are equivalent; each has the solution x
= 10
=
5 and no other.
2.3. Operations yielding equivalent equations.
The following operations on an equation will always
lead to an equivalent equation:
Adding the same number or expression to both sides or
subtracting the same number or expression from both sides.
Multiplying or dividing both sides by the same number or
expression, provided that this number or expression is not
zero and docs not contain an 'unknown.
[Ch. II
LINEAR EQUATIONS
32
NOTE. We shall at times use such phrases as "multiplying
an equation by 2" instead of the longer form" multiplying both
sides of an equation by 2."
2.4. Linear equations in one unknown.
An equation of the form
ax + b = 0
in which a ~ 0*, is a linear equation in the unknown :to
Examples of linear equations in one unknown are
",I'
'I~',
[I·
3x
+2
=
x - 6 = 0,
0,
2x
= O.
)1
To solve an equation, we perform the same allowable
operations (see preceding section) on both sides, with the
ultimate purpose of getting the unknown alone on one
side of the equation.
Example 1.
Solve the equation x
+2 =
7.
SOLUTION. Subtract 2 from both sides:t
x
or
+
2 - 2
x
= 7 - 2,
= 5.
CHECK. Substitute x = 5 in the original equation~ ,'.
5
,
{~
1
,"
+2 =
7,
7 = 7.
Example .2.
Solve the equation 4x - 5 = x
+ 1.
SOLUTION. Subtract x from both sides to get rid of the x on
the right side, and add 5 to both sides to get rid of the - 5 I.m
* The symbol
;;<!
is read "is not equal to," or "is different from."
t We may occasionally phrase a statement such as this in the form
"Transpose the 2." By the first fundamental principle of § 2.3, a numl;ler
may be transposed from one side of an equation to the other if its sign is
changed.
EXERCISES
the left.
33
These operations can be performed simultaneously:
4x - 5 - x
or
+
5 = x
3x = 6.
+1-
x
+ 5,
Divide both sides by 3:
x = 2.
CHECK.
r
Substitute x = 2 in the original eQuation:
4· 2 - 5 = 2
8 - 5 = 3,
3 = 3.
.H.
+ 1,
.i <\
:
Example 3.
Solve the equation Ix - 1 = 2x
+ 9.
'
SOLUTION. Multiply both sides by 4 to "clear of fractions"
(i.e., to get rid of the fractional coefficient on the left):
3x - 4 = Sx
+ 36.
Subtract Sx from both sides and add 4 to both sides:
3x - 4 - Sx
+ 4 = Sx + 36 - 8x
-5x = 40.
+ 4,
Divide both sides by - 5:
..,
x
CHECK.
.C
= -S.
Substitute x = -S in the original equation:
+
+
I(-S) -1 = 2(-S)
9,
-6 - 1 = -16
9,
-7 = -7.
JiA'
//
EXERCISES II. A
Solve for x if possible:
1.
3.
5.
6.
7.
7x + 42 = O.
2. 2x
5x - 8 = 2x + 7.
4. 6x
3x + 2 - 4(x - 3) = 2(5x - 4).
X2 + 4x + 5 = (x + 2) (x + 7).
(x - 2)(x - 4) = (x - 3)(x + 6).
+ 19 =
+ 13 =
35.
11x - 10.
{Ch. II
LINEAR EQUATIONS
34
8. (2x
+ 3)(3x + 2)
9. 3ax - 2 = 5bx
+
+
+
+
+
= (x - 5)(6x
+ 7.
+
+
+ I),
,
(x
a) (x
b) = (x
c)(x
d),
(ax
c)(bx
d) = (bx
c)(ax
d).
(2x
5)2 - x(4x - 5) = 100,
(3x - 2)2 - 9x 2 = 4(1 - 3x).
(2x - a)2 - 4x2 = X
4a.
x 2 + a(a - 2x) = (x - 2a)2,
x2
a(a - 2x) = a 2
x(x - 2a).
(x - 5)(5 - x) = 5(2x - 3) - x 2•
(2x - 9)(2x
7) = (2x - 5)(2x
3).
19. Solve for r: P(l
rn) = A.
20. Solve for d and for n: l = a
(n - l)d.
10.
11.
12.
13.
14.
15.
16.
17.
18.
+
I
+
!
+
+
+
+
. J
I
+
+
+
n
21. Solve for l: S = '2 (a
+ l).
,I
•
.
22. Solve for r: V = i'nN(3r - h).
2
23. Solve for d: p + ghd + -§-v d = c.
24. The formula for changing a temperature reading frotndegrees Fahrenheit F to degrees centigrade C is
,',~,
C = t(F - 32).
(a) Solve this equation for F. (b) For what temperature
will the Fahrenheit and centigrade readings be the same?
2.5. Worded problems.
Many problems which are stated in words have to be
translated into mathematical symbols before they can
be solved. All answers should be checked by testing
them in the original worded problem.
Example 1.
How many pounds of water must be evaporated from 50
pounds of a 3-per cent salt solution so that the remaining solution
will be 5 per cent salt?
SOLUTION. The number of pounds of salt in the solution is
0.03 X 50 = 1.5. Let x = the number of pounds of water
which must be evaporated. Then 50 - x = the number of
pounds of solution remaining, and the amount of salt will be
,
I
!'
I
"
'--I
EXERCISES
unchanged.
we have
35
Since 5 per cent of the solution will now be salt,
'.
0.05(50 - x) = 1.5,
2.5 - 0.05x = 1.5,
-0.05x = -1.0,
x = 20.
That is, 20 pounds of water must be evaporated.
CHECK. If 20 pounds of water are evaporated, the number of
pounds of solution remaining will be 50 - 20 = 30, of which
5 per cent must be salt. But 0.05 X 20 = 1.5, which is the
actual number of pounds of salt.
Example 2.
Two airplanes left airports which are 600 miles apart and
flew toward each other. One planE: flew 20 miles per hour faster
than the other. If they passed each other at the end of an
hour and 12 minutes, what were their rates?
SOLUTION. Let x be the rate, in miles per hour, of the slower
plane. Then x + 20 will be the rate of the faster plane. The
distance that each flew is found by mUltiplying its rate by the
time, It hours. Therefore the total distance that they flew was
Itx + li(x + 20), and this must be 600 miles. Thus,
!x + Ix
+!
. 20 = 600,
¥x + 24
.
I
= 600,
-1fx = 600 - 24 = 576,
12x = 5 X 576 = 2880,
x = 2880 + 12 = 240,
x + 20 = 260.
i
That is, the rates of the two planes were 240 miles per hour and
260 miles per hour, respectively.
CHECK. The distance that the slower plane flew was It X 240
miles, or 288 miles; the distance that the faster plane flew was
Ii X 260 miles, or 312 miles. The sum of these distances is
288 + 312 = 600 miles.
EXERCISES II. B
1. One number is 17 less than another; their sum is 125.
What are the numbers?
'
36
LINEAR EQUATIONS
[Ch. II
2. Find three consecutive integers whose sum is 378.
3. The length of a rectangle is 13 inches greater than its
width. Its perimeter is 8 feet. Find its dimensions.
4. Each equal side of an isosceles triangle is 3f inches longer
than the base. The perimeter of the triangle is 2 feet.
Find the length of its base.
5. A vertical pole was broken by the wind. The upper part,
still attached, reached a point on the level ground 15 feet
from the base. If the upper part is 9 feet longer than the
lower part, how tall was the pole?
6. The width of a rectangle is 9 inches. The length is 1 inch
shorter than the diagonal. Find the length of the diagonal.
7. A man is three times as old as his son. Four years ago
he was four times as old as his son WaS at that time. How
old is the son?
8. A sum of $2.45 is composed of nickels and quarters, there
being twice as many nickels as quarters. How many
quarters are there?
9. An organization sold 591 tickets to a benefit performance
for a total of $848.25. Some of the tickets sold for $1.25
each, some for $1.75 each. How many of each kind were
sold?
10. How many ounces of pure nickel must be added to 150
ounces of alloy 70 per cent pure to make an alloy which
is 85 per cent pure?
11. How many pounds of cream containing 12 per cent butterfat must be added to 1800 pounds of milk containing 2 per
cent butterfat to obtain milk having 3 per cent butterfat?
12. How much water must be added to 200 gallons of mixture
which is 80 per cent alcohol to reduce it to a 75 per cent
mixture?
13. How much coffee costing 85 cents a pound must be added
to 150 pounds costing 90 cents a pound to make a mixture
costing 87 cents a pound?
14. The attendance at a moving picture theater on a certain
day was 737 persons. If there were 289 more adults than
children, how many children were there?
16. A man has $16,000 which he places at interest, part at
3t per rent, the rest at 4 per cent. His total annual income
EXERCISES
16.
17.
18.
19.
20.
21.
22.
23.
37
from this investment is $597.50. How much is invested
at each rate?
A man walks a certain distance at the rate of 5 miles an
hour and returns at the rate of 4 miles an hour. If the
total time that it takes him is 3 hours and 9 minutes, what
is the total distance that he walks?
A man rides up a ski lift at the rate of 5 miles an hour and
skis back down the hill at the rate of 50 miles an hour.
If the complete trip requires 22 minutes, how far is it from
the bottom to the top of the hill?
An airplane flew in a direct line from its base at the rate
of 300 miles an hour and returned over the same route at
the rate of 250 miles an hour, requiring 15 minutes longer
on the return trip. How far from its base did it fly?
Two automobiles set out simultaneously from places 100
miles apart. They traveled toward each other and passed
at the end of 1 hour and 40 minutes .. The speed of one
automobile was 5 miles per hour greater than that of the
other. Find the speed of each.
An airplane climbed at an average vertical speed of 300
feet per minute and descended immediately at an average
speed of 25 miles per hour. The flight was performed in
25 minutes. What altitude did the plane reach?
A car 15 feet long overtakes a truck 30 feet long which is
traveling at the rate of 45 miles per hour. How fast must
the car travel to pass the truck in 3 seconds?
A bullet is fired at a target and the sound of its impact is
heard 6 seconds later. If it travels at the rate of 2200 feet
per second and sound travels at the rate of 1100 feet per
second, how far away is the target?
How soon after six o'clock will the hands of a clock be
together?
Let x equal the number of minute spaces
that the minute hand travels. Then x/12 will be the
number of minute spaces that the hour hand travels.
SUGGESTION.
24. How soon after noon will the hands of a clock extend in
opposite directions?
38
LINEAR EQUATIONS
I
[Ch. II
!
25. How soon after one o'clock will the hands of a clock form
a right angle?
26. * Mary is twenty-four. She is twice as old as Ann was when ,
Mary was as old as Ann is now. How old is Ann?
!
I
A lever is a rigid bar having a single point of support,
called the fulcrum. If a weight w is attached to the lever at
a distance d from the fulcrum then d is called the lever arm
of the weight w, and the product of the weight and lever arm,
wd, is called the moment of the force about the fulcrum.
In physics it is shown that when two or more weights are
attached to a lever, then, if the lever balance8, the sum of
the moments on one side of the
fulcrum
is equal to the sum of
Ai
di F
tVa the moments on the other side.
Thus, for the lever in Fig. 2.1
FIG. 2.1.
we have wld l = W-z(i2.
Unless the contrary is stated, the weight of the lever itself
will be disregarded. If the weight of the lever is to be taken
into consideration, it will be assumed that it is distributed
uniformly (i.e., each linear foot of it weighs the same
amount). In such a case the entire weight of any part of the
lever may be regarded as concentrated at the mid-point of that
part.
A
6
27. Two children are balanced on a
F
seesaw. The smaller, weighing 40 40
60
pounds, is 6 feet from the balancFIG. 2.2.
ing point. How far from the
balancing point must the other sit if he weighs 60 pounds?
6
!
1
SOLUTION. Let x = no. of ft. from balancing point.
Moment of 1st child = 40 . 6 = 240.
Moment of 2nd child = 60x.
60x = 240,
x = 4.
28. A board 10 feet long weighs 8 pounds per foot. It is supported at a point 3 feet from one end. How large a weight
* This problem, for some unaccountable reason, attracted a greal deal of
attention in the early years of the present century and the question, "How
old is Ann ?," was a slang expression of the day, apparf:'ntly meaningless.
EXERCISES
39
must be attached at the extremity of the short end of the
board to make it balance?
SOLUTION. Method 1. Let x = weight (in lb.). Short
end of board weighs 3 X 8 = 24 lb. Since this may be
considered as concentrated at its midpoint, its lever arm
is 1.5 ft. and its moment
3
7
is 24 X 1.5. Long end of
board weighs 7 X 8 = 56
lb., its lever arm is 3.5
ft., and its moment is
FIG. 2.3.
56 X 3.5.
,
Equating the sum of moments on the left of the balancing
point to the moment on the right, we have
.}
3x
+ 24 X 1.5 =
56 X 3.5,
x = 53t.
Method 2. Weight of board is 10 X 8 = 80 lb. This may
be regarded as concentrated at its midpont, which is 2 ft.
to right of fulcrum. Its moment is therefore 80 X 2 = 160,
and we have
3x = 160, x = 53t.
29. Two children weighing 65 pounds and 45 pounds respec-
30.
31.
32.
33.
34.
tively are on opposite ends of an ll-foot board. Where
must the point of support be placed if the board is to
balance? Neglect the weight of the board.
If, in the preceding exercise, the board weighs 2 pounds
per foot, where must the point of support be placed?
A laborer uses a crowbar 4 feet long. If the fulcrum is
7 inches from one end of the bar, how heavy a weight can
he move by exerting a force of 175 pounds?
If the fulcrum in the preceding exercise is 8 inches from one
end of the bar, how much force must be exerted to move a
weight of 1000 pounds?
A crowbar is 4t feet long. Where must the fulcrum be
placed so that a weight of 1600 pounds can be moved by
exerting a force of 200 pounds?
A rod has a weight of 45 pounds attached to one end, e
weight of 65 pounds to the other .. It balances at a point
----
.
LINEAR EQUATIONS
[Ch. II
8 inches from the middle. How long is it? (Assume that
the rod itself is weightless.)
35. A rod is 8 feet long. A weight of 100 pounds is fastened
to one end, a weight of 75 pounds to the other. At a
distance of 2 feet from the 100-pound weight a third weight
of 65 pounds is fastened. Where is the balancing point
(a) if the weight of the rod can be neglected? (b) if the
rod weighs 10 pounds per foot?
2.6. linear equations in two unknowns.
A single linear equation in two unknowns, such as
2x - y = 3, is satisfied by an unlimited number of pairs
of values. We sometimes say that it has an infinite
number of solutions. However, when we have a pair of
linear equations in two unknown quantities, there is
commonly a single pair of values of the unknowns which
will satisfy both of the equations at the same time (i.e.,
a single solution), although there may be an infinite number of solutions, or there may be none. 'Vhen the
equations are considered together and our interest is in
solutions which will satisfy both of them, the equations
are referred to as simultaneous equations, or they are
said to form a set, or system, of equations.
A method of solving a system of two linear equations
is shown in the following example.
Example.
.f
Solve the equations
2x - 3y
5x
2y
+
= 16,
=
2.
SOLUTION. Multiply the first equation by 2 and the second by
3, in order to make the coefficients of y numerically equal. We
have
4x -. 6y = 32,
15x + 5y = 5.
Adding, we get
19x
= 38.
Dividing by 19 gives
x = 2.
LINEAR EQUATIONS IN TWO UNKNOWNS
§2.61
41
From the second of the original equations we find
2y = 2 - 5x = 2 - 10 = -8,
y = -4.
or
NOTE. The value of y could of course have been obtained by
multiplying the first of the original equations by 5, the second
by 2, and then subtracting.
It is readily seen that the values x = 2, y = -4 satisfy both
equations.
The equations may also be solved by solving one of them for
y in terms of x, say, then substituting this expression in the
other and solving the resulting equation for x.
Y
'l~
In Fig. 2.4 the lines which
~~t
\~
are the graphs of the two .'
:i
equations are shown. It
\'
V
will be observed that they
t..;
X'
X
intersect at the point x = 2,
\.:;
:I \
~~
y = - 4, which is the solu~.,.,
\
tion of the two equations.
Thus, the solution might
1/
\
have been obtained graphi- '.
V
cally by plotting the two
Y
lines and reading off the
FIG. 2.4.
coordinates of the point at
which they intersect. Ordinarily this method gives only an
approximate result, whereas the algebraic process given above
y
is exact.
~
Two equations such as
~
,
. 2x
2x
'I'.
~
Xl
X
0
y,
FIG. 2.5.
+y
+y
=
4,
= 8,
have no solution. They are said
to be inconsistent. Their graphs
(Fig. 2.5) are parallel lines and
consequently do not intersect.
[Ch. II
LINEAR EQUATIONS
42
Two equations such as
2x
4x
+y
+ 2y
= 4,
= 8,
\,l.
in which the second can be obtained by multiplying both
sides of the first by 2, are said to be dependent. Any pair
of values of x and y which satisfies one will satisfy the
other. Their graphs coincide.
2.7. Linear equations in three or more unknowns.
To solve linear equations involving three or more
unknowns, we take them in pairs and eliminate one of the
unknowns, reducing the equations to a set in which there is
one less equation and one less unknown. This process is
repeated until we have a single equation with only one
unknown, which can be solved.
Example.
, Solve the simultaneous equations: ,..,
5x - Y
2x + 3y
7x - 2y
it'
+ 4z = 5,
+ 5z
+ 6z
(1)
(2)
(3)
=' 2,
= 5.
SOLUTION . We shall first combine equations (]) and (2),
eliminating y, and shall then combine (1) and (3), also eliminating y.
Multiply (1) by 3. This gives
15x - 3y
Add (2) and (4).
+ 12z
= 15.
,~-,-."
.•..
,
(4)
Result:
17x
or
+ 17z = 17,
x+z=1.
(5)
Multiply (1) by 2, getting
lOx - 2y
Subtract (3) from (6).
+ 8z
= 10.
(6)
We get
3x
+ 2z = 5.
(7)
I
EXERCISES
43
We are now ready to combine (5) and (7).
2, we obtain
2x + 2z = 2,
Multiplying (5) by
(8)
and subtracting (8) from (7) we find
3.
x =
Substituting this value of x in (5), we get
Finally, by putting x = 3,
y
=
5x
z
=
-2.
Z
=
-2 in (1) we obtain
+ 4z -
5
=
15 - 8 - 5
=
2.
By substituting x = 3, y = 2, Z = -2, in the original equations
(1), (2), (3), we verify the correctness of the solution.
(For a method of solving sets of linear equations by means of
determinants see Chapter XV I I I.)
EXERCISES II. C
Solve the following simultaneous equations:
2.
1. 3x + y = 44.
2x + y = 3l.
.~;e
'\
3. 4x- 6y = 35,
., 4.
7x + 11y = 29.
' ... , 1:
6.
6. 8x + 7y = 46,
,V, , )
12x - 5y = 100.
':C
8.
7. 56x + 84y = 29,
42x + 28y = 13. f' :;. .t:-e ..
10.
9. 17x - 3y = 0', + ',l'
.
8x + l8y = 275.
11. 14x - 12y = 7, + 'is
12.
21x - 18y = 21.
13. 14x - 12y = 0,
14.
49x - 42y = O.
;"t
16. 7x + 12y = 240,
16.
17x - 13y = 35.
17. 2lx + 8y = 47a,
18.
25x - 6y = 87a.
19. 4x + 15y = a,
20.
6x - 25y = b.
..
J
2x + 8y = 79,
2x-5y= -38. "
t&
3x - 6y = -38,
6x - 9y = 44.
51x - 42y = -14,
17x + 34y =
20.
52x - 28y = 5,
4x+ 4y = 5. - ',f,; .et
~1'~J,'
13x + 15y = 0,
"'
"
,
..
~'tJ~
22x - 19y = O.
~..";:'r"~
22x + 16y = 6,
33x + 24y = 9.
34x - l3y = -237,
51x + 17y =
119.
32x + 33y = -2,
36x + 9y = Ii.
21ax + 8y = 47,
35ax - 6y = 117.
ax + by = c,
dx + ey = f.
\
+
5
2
22.
21. x- - -y = 29 ,
4
3
= 37.
x
y
- +-
SUGGESTION.
13. 4x
7x
+ -2y =
+ ~y
3,
24.
+
+
-9x + lly - 10z =
29. 3x + 2y
7z = 161,
2x + 7y
4z = 43,
7x + 4y + 5z = 155.
+
y
4
3
Y+ Z =
i +~ =
z
x
26';'
00'
= 29.
,
I
1
x12 + 7y =
17
2'
13y = -15.
+ 8y + 2z =
+ 5y + 3z =
3x - 6y + 4z =
26. 5x
2x
28. x
+ y = 2,
-14,
3,
- 4.
+
30. 6x
21y - 19z = -102,
4x - 9y - 5z =
24,-
((_
7x
+
6y -
z
=
20
i + ~ - ~ = 22 .f:
x
y
z
'''if.
Z _ ~ + i = 51 n
x
y
z
'
33.
~- ! - I =
I,4
32.
..g,.
1
x
10
33. 3x - 2y + 5z + 8w =
2x + y + 7z - 2w =
4x - 3y - 8z + 3w =
5x - 4y - 4z - W =
;
t~
y+z=4,
z + x = 9.
1.
+
~+~ =
+ ~y
18
+
x
~x
Z
=
y
- x
37x - 23y
17z = -32,
14x - 22y - lIz = -91,
·27.
-
1
= 5.
+
~x
Solve first for - and -.
x
y
26. 3x
2y
z = - 4,
5x - 6y
z =
92,
~x + 7y - z = -67.
31.
leh. II
LINEAR EQUATIONS
44
,t:
35.
+ + z = 6,
+ z + W = 7,
z + W + x = 8,
W + x + y = 9.
99, 34. x
32,
y
-9,
-8.
z
y
y
O.
:\
N aTE. Sometimes a special method will effect a neater and
quicker solution. For example, exercise 34 can be solved as
follows: Add the four equations and divide by 3. Subtract
each equation in turn from the result.
36. x
y
z
W
+
+
y
=
2,
z = 7,
+W
+x
=
=
i3,
8.
36. x
y
+
y
+
z
=
a
+ b,
+ z + w = b + c,
z + w + x = c + d,
w + x + 1/ = d + a.
MORE WORDED PROBLEMS
§2.81
37.
x+y
x +y
x - y
-x + y
+ y+ zz+w -
+z-
= a,
38. x
-
= b,
= c,
= d.
y
+
+
w
z+w
z+w
z+w
45
z = a,
w
= b,
x = c,
x - y = d.
w+
:J.
2.8. More worded problems. ~
Many worded problems lead to linear equations, as was
seen in the study of linear equations in one unknown.
Obviously there is no mechanical method which can be
used for setting up the equations for all such problems.
Each problem must be analyzed independently. The
meaning of each statement must be clearly understood and
the various statements must then be translated into mathematical equations. However, for certain problems involving mixtures, rates of motion, rates of interest,etc., a tabular
arrangement such as the one used in the following illustrative example may be of assistance in determining the
equations whose solution gives the answer to the problem.
Example 1.
A seed company wishes to use bluegrass seed selling at $1.30
per pound and clover seed selling at $1.80 per pound to make 100 .
pounds of mixture to sell at $1.38 per pound. How much of
each kind of seed must be used?
SOLUTION.
No. of X Price
lb.
per lb.
=
Total
price
Bluegrass
x
1.30
1.30x
Clover
y
1.80
1.80y
100
1.38
138.00
E-:,
,.
jl·
Mixture
1.30x
x +y =
+ 1.80y =
100,
138.
These equations have the solution x = 84, y = 16. That is,
84 pounds of bluegrass seed and 16 pounds of clover seed must
be used.
[Ch. II
LINEAR EQUATIONS
46
CHECK.
84 X $1.30 = $109.20
16 X $1.80 = $ 28.80
100 pounds for $138.00
= $1.38 per pound. -,
It will be noted that for any row in the table the product
of the numbers in the first two columns gives the number
in the third column, or that the number in the third
column divided by the number in the first or second
column gives the number in the other column.
In problems involving rates of motion we have the
fundamental relation
'1
Time X Rate = Distance.
In problems involving simple interest we have the ~lati<)q,;
I.
.
l
I
Principal X Rate = Annual Interest.
Consequently the above tabular arrangement can often
be used to advantage in such problems.
NOTE. Care must be taken to employ the same units throughout a given problem. Thus, in the foregoing illustrative example we must either use cents throughout or dollars throughout.
In problems dealing with boats in rivers we make use
of the principle
Rate with current
=
Rate in still water
+ Rate of
current,
Rate against current = Rate in still water - Rate of
current.
Example 1 can be solved by using only one unknown
quantity, although with two unknowns the solution is
perhaps neater. In some instances it is almost essential
to use two, or even more, unknowns.
In a certain type of problem it is convenient to use the
reciprocals of the unknown quantities. For example,
47
MORE WORDED PROBLEMS
§2.8]
if it requires a man x days to do a piece of work, the part
of the work which he can do in 1 day is l/x. Similarly,
if it takes a pipe x hours to fill a tank, the part of the tank
that it can fill in 1 hour is l/x.
Example 2.
Two pipes running simultaneously can fill a tank in 2 hours
and 40 minutes. After the larger pipe had run for 3 hours the
smaller pipe was also turned on and the tank was full 40 minutes later. How long would it take each pipe alone to fill the
tank?
SOLUTioN. Let x = the number of hours it would take the
larger pipe to fill the tank and y = the number of hours it would
take the smaller pipe. Then
!x
= the part of the tank that the larger pipe can fill in 1 hour,
!y
= the part of the tank that the smaller pipe can fill in 1 hour.
Since the two pipes running simultaneously can fill the tank in
2i hours, the part of the tank that they can fill in 1 hour is
1
1
3
2j = ~ = S·
Consequen tly
-x1 + -y1 =-.83
'''§F.;
j"
(1)
The part of the tank that the larger pipe fills in 3 hours is
~,
x
and the part that both pipes fill in 40 minutes (i.e., in i of an
hour) is i . t = t. It follows that
3
1
-+-=1
x 4
(2)
(since the entire tank may be denoted by 1). Equations (1) and
(2) can be solved for 1/x and 1jy, and it will then be found that
x = 4, y = 8. That is, it will take the larger pipe 4 hours an~
the smaller pipe 8 hours to fill the tank.
~'b.:'
LINEAR EQUATIONS
48
[Ch. II
The part of the tank that both pipes can fill in 1 hour
is i + k = l Therefore, to fill the tank it will take them t
hours = 2 hours 40 minutes.
CHECK.
EXERCISES II. 0
1. A candy manufacturer has two grades of candy, selling at
75 cents and 90 cents a pound respectively. How many
pounds of each must he use to make a mixture of 150 pounds,
to sell at 78 cents a pound?
2. A tobacconist wishes to blend two grades of tobacco,
costing 8 cents an ounce and 10 cents an ounce respectively,
and to sell the mixture at 12 cents an ounce with a profit
of 25 per cent of the cost. How many ounces of each kind
must he use to the pound? (1 lb. = 16 oz.)
3. A fruit dealer sold 9 dozen lemons and 16 dozen oranges
for $17.40. The following day he sold, at the same prices,
11 dozen lemons and 20 dozen oranges for $21.60. Find
!!
the prices of his lemons and oranges.
I
~
A
vat
contains
a
mixture
of
acid
and
water.
If
25
gallons
1.$,
of acid are added the mixture will be 80 per cent acid, if I
25 gallons of ,vater are added the mixture will be 60 per
cent acid. Find the percentage of acid in the mixture.
'''I. How much tin and how much lead must be added to 700
pounds of an alloy containing 50 per cent tin and 25 per.
cent lead to make an alloy which is 60 per cent tin and
20 per cent lead?
6. A receptacle contains x ounces of acid. A second receptacle
contains x ounces of water. From the acid y ounces are
removed, placed in the water, and the contents thoroughly
mixed. Then, from the second receptacle, y ounces of the
mixture are removed and placed in the acid. Find (a)
the concentraton of acid in the second receptacle, and (b)
the concentration of water in the first receptacle.
7. If the length of a rectangle is increased by 8 feet and its
width by 3 feet its area will be increased by 200 square feet.
If its length is increased by 3 feet and its width by 8 feet
its area will be increased by 255 square feet. What are
its dimensions?
8. If the altitude and the base of a triangle are each increased
I
EXERCISES
49
by 4 inches the area of the triangle will be increased by 42
square inches. If the altitude is inceased by 3 inches and
the base is decreased by 2 inches the area will be increased
by 5 square inches. Find the base and the altitude of the
triangle.
9. A man wishes to fence in a rectangular lot. If he uses
material costing $1.70 a foot for the front of the lot and
material costing $1.40 a foot for the other three sides the
fence will cost $890. If he uses the cheaper material for
all four sides of the lot the fence will cost $854. Find the
dimensions of the lot.
10. Tickets to a benefit performance were sold at $1.50 and
$0.85 apiece respectively, the total receipts being $444.70.
If twice as many of the $1.50 tickets and half as many of
the cheaper tickets had been sold the receipts would have
been $593.60. How many of each were sold?
11. One angle of a triangle is twice as large as another and 25<>
more than the third angle. Find the angles.
12. A man has two sums of money at interest, one at 3 per cent,
the other at 3t per cent. His annual income from the two
investments is $99.90. If he could get 3t per cent on all
of his money this income would be $109.20. How much
does he have invested at each rate?
13. A man has three sums of money invested, one at 4 per cent,
one at 3t per cent, and one at 3 per cent. His total annual
income from the three investments is $416. The first
investment yields $104 more per year than the other two
combined. If he could receive t per cent more on each
investment his annual income would be increased by $55.75.
How much does he have invested at each rate?
14. Ten years ago a man was three times as old as his son.
In 6 more years he will be twice as old as his son. How
old are they?
15. In 2 years a boy will be three-fourths as old as his sister.
Two years ago he was two-thirds as old as she. How old
are they?
16. Three circles are tangent externally. The distances
between their centers are 23 inches, 25 inches, and 20 inches.
Find their radii.
50
LINEAR EQUATIONS
[Ch. "
17. Two circles are tangent to a third circle internally and are
tangent to each other externally. The distances between
their centers are 14 inches, 7 inches, and 5 inches. Find
their radii.
18. Two pipes running simultaneously can fill a swimming pool
in 6 hours. If both pipes run for 3 hours and the first is
then shut off, it requires 4 hours more for the second to fill
the pool. How long does it take each pipe running separately to fill the pool"?
19. A pipe can fill a tank in 3 hours if the drain is open. If the
pipe runs \\'ith the drain open for 1 hour and the drain is
then closed, the tank will be fi~led in 40 minutes more.
How long does it take the pipe to fill the tank if the drain
is closed'?
20. Two airplanes working together can take the necessary
photographs to map a certain region in 5 hours. After
'f.~ they had worked for 3 hours the first plane developed
engine trouble and the second had to finish the job, requiring
, ., 4t hours of additional work. How long would it take each
plane alone to do the job '?
21. A, B, and C can do a piece of work in 10 days. A and B
can do it in 12 days, A and C in 20 days. How many day.s
would it take each to do the work alone'?
22. The difference between the digits of a two-place number
is 3. If the digits are reversed and the resulting number
added to the original number the sum is 121. What is the
original number'?
Sl.'GGESTION.
If x is the digit in tens' place and y the
digit in units' place, the number is lOx + y.
23. The sum of the digits of a three-place number is 16. If
the digits are reversed and the resulting number added to
the original number the sum is 1049. If the resulting
number is subtracted from the original number the difference is 297. What is the number?
24. A rod 7 feet long is supported at a point 6 inches from the
center. A weight of x pounds is attached at the end of the
ter lever arm, a weight of y pounds at the other end.
rder to make the rod balance it is necessary to attach a .
i
EXERCISES
51
10-pound weight at a distance of I foot from the weight of
If the weights x and yare interchanged it i8
necessary to place a 2-pound weight with the y-pound
weight to effect a balance. Find x and y (a) neglecting the
weight of the rod, (b) assuming that the rod weighs k
pounds per foot.
An airplane made a trip of 350 miles against the wind in an
hour and 10 minutes. Returning with the wind it took
only an hour. Find the speed of the plane in still air and
the rate of the wind.
A motorboat traveling at full speed against a current goes
14 miles an hour. Traveling at half speed with the current
it goes 10 miles an hour. Find the rate of the current and
the maximum speed of the boat.
A train is moving along a straight stretch of track parallel
to a highway. An automobile traveling 60 miles an hour
in the same direction as the train can pass it in 30 seconds.
Traveling in the opposite direction the automobile requires
only 7t seconds to pass the train. How long is the train
and how fast is it moving? (Neglect the length of the
automobile.)
A and B are running on a track having 8 laps to the mile.
H they run in opposite directions they meet every 20 seconds. If they run in the same direction it requires 3
minutes for A to overtake B. Find the rate of each in
feet per second.
An army officer made the first part of a trip in a plane
which flew at the rate of 300 miles an hour. At the landing
field he was met by a car which took him the rest of the
way to his destination at the rate of 55 miles an hour. The
trip required 3 hours and 42 minutes. On his return trip
the car traveled at the rate of 60 miles an hour and the
plane which he took flew at the rate of 250 miles an hour.
The return journey required 4 hours and 6 minutes. Find
the total distance that he flew and the total distance that
he traveled by car.
A commuter returning by train from his office ordinarily
reaches his suburban station at 5 o'clock. His chauffeur
leaves his home with a car just in time to meet him when
x pounds.
25.
26.
27.
28.
29.
30.
\
\
51
LINEAR EQUATIONS
[Ch. II
the train arrives, and drives him back home. On a certain
day the man takes a train which arrives at the station an
}J
hour earlier. He walks toward home until he meets his
chauffeur and car. The chauffeur turns and drives the
man the rest of the way home, arriving home 10 minutes
earlier than usual. Find the time that the man walks.
31. A man arrives at the railroad station nearest his country
home an hour and 5 minutes before the time at which he
~
had ordered his chauffeur to meet him with his car. He
starts to \valk at the rate of 4 miles an hour and meets his
car when it has traveled 8 miles. It turns back and takes
him the remainder of the way home, where he arrives just
10 minutes ahead of the time he had originally planned to
get there. Find the distance from the station to his home
and the rate of the car.
.Vi:'
]C;1
Vlt1H
',(
/,.
,t; ,
<;.,kr~r· <\.d
CHAPTER III
Factoring
f- n,i
3.1.
Poynomias.
I
I
<,,,t.",,,,
L
",\
,
t'.,) .:
It""....
f.,.,
A polynomial, or integral rational expression, in certain
variables, for example, x, y, z, is a sum of terms such as
kXaybzc, in which a, b, c, are positive whole numbers and k
is a constant. It may contain a term not involving
x, y, z. The expressions
5x 3
-
7x 2
-
4x
+ 3,
rn
are polynomials.
The degree of a term of a, polynomial is the sum of the
exponents of the letters in the term. Thus, 8X 2y 3z is of
degree 6 in x, y, and z, although it is of degree 2 in x
alone, degree 3 in y alone, and degree 1 in z alone. A
term not involving the letters of the polynomial is said
to be of degree zero. The degree of a polynomial IS
that of its term or terms of highest degree.
3.2. Factoring.
When an algebraic expression is the product of two
or more other expressions, each of these other expressions
is said to be a factor of the given expression. The
process of finding the factors of which a given expression
is composed is called factoring the expression. An
expression will be called prime if it has no other factors
than itself, or its negative, and 1.
Skill in factoring depends upon the ability to recognize
certain type products. Several important type products
are listed in the next section.
53
FACTORING
54
[Ch. '"
3.3. Type produds.
The following important type products can be verified
by actual multiplication according to the methods
explained in Chapter I:
+ b)(a - b) = ai - bt •
(a + b)t = at + 2ab + b t .
(a - b)t = at - 2ab + b
(x + a)(x + b) = Xi + (a + b)x + abo
(ax + b)(cx + d) = acx + (ad + bc)x + bd.
(a + b + C)I = at + bl + cI + 2ab + 2ac + 2bc.
(a
l
•
l
(1)
(2)
(3)
(4)
(5)
(6)
It may be helpful to memorize some of these formulas
in words. The first three may be expressed respectively
as follows:
(1) The product of the sum and the difference of two quantities is equal to the difference of their squares.
(2) The square of the sum of two quantities is equal to the
square of the first, plus twice the product of the first
and second, plus the square of the second.
(3) The square of the difference of two quantities is equal to
the square of the first, minus twice the product of the
first and second, plus the square of the second.
Formula (6) may be stated in more general fashion as
follows:
The square of a polynomial is equal to the sum of the
squares of the terms and twice the product of each term
by each term, taken separately, that follows it.
NOTE. In formulas (1)-(6) the literal quantities are not
restricted to be monomials. For example, in (1) a might be
x y, in which case we should have
+
(x
+ y + b)(x + y
- b) = ((x
=
(x
+ y) + bJ[(x + y)
+
y)2 - b2.
- bJ
,
ELEMENTARY FACTOR TYPES
§ 3.4J
55
EXERCISES III. A
Write out the following products:
+
+
+
1. (m + n)(m - n).
3. (x + 3y)(x - 3y).
+
+
+ +
25.
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
49.
+
g)(2a + g).
b2 )(a 2
b2 ).
uV)(6xy - uv).
b)
c][(a
b) - c}.
v)
6][(u - v) - 6].
y)
5][3(x
y) - 5}.
(Ca
b)
(x - y)][(a
b) - (x - y)}.
[4 - 3(m
n)J[4
3(m
n)}.
(p
q
7r)(p
q - 7r).
(a - 2b
3c)(a - 2b - 3c).
(a
b
c
d)(a
b - c - d).
[5(p - q)
6(r
s)J[5(p - q) - 6(r
s)}. '1 '\())':\'I:'l'
[(3ab - 2c) - (4w
xy)}[(3ab - 2c)
(4w
zy)}.
(u
V)2.
22. (u
3V)2.
(2a
7b)2.
;)
24. (3x - 2y)2.
(5p2q - 4rs 2)2.
(): ;,
26. (gmn - 8p 2q2) 2.
(4ab 3 - lld 2e)2.
28. (25u 2v - 6WX 5)2.
(x
7)(x
11).
30. (x
10)(x - 5).
(x - 15)(x - 3).
32. (x - 13)(x
7).
(x
2b)(x
3b).
34. (x - 8y)(x - 4y).
(5x - 2)(3x
7).
36. (4x - 3)(lOx
9).
(9x - 4)(7x - 6).
38. (12x - 5)(2x
25).
(5x
2)(3x
7).
40. (x
2y)(2x
y).
(3x
2y)(5x - y).::t;f
42. (4a - 3b)(7a - 8b). f
(x - y
Z)2.
44. (a
b - 2C)2.
(3x
2y
Z)2.
46. (5x - 2y - 3Z)2.
(a
b
c
d)2.
48. (5a - 2b - c
4d)2.
(r - 2s
3t - 4u
5v)2.
5. (2a 7. (a 2 9. (6xy
11. [(a
12. [(u 13. [3(x
14.
15.
16.
17.
18.
19.
20.
21.
23.
2. (x
5)(x - 5).
4. (3x - 2y)(3x
2yJ.
6. (2X2
3a)(2x2 - 3a).
8. (m 3
n 3)(m 3 - n3).
10. (abc + d)(abc - d).
+
+
+ +
+ +
+ +
+
+
+
+
+
+
+
+ + +
+
+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1>
>
; :••
+
+
+
+
+
+ + +
+
+
+
+
3.4. Elementary factor types.
In this section and several following sections will be
given some of the types of factoring which are of most
frequent occurrence. Many of. these are simply the
[Ch. III
FACTORING
56
reverse asper,t of the type products given in the preceding
section.
i
I
Common factor:*
+ ay =
ax
Examples.
+ y) •.
" (1)
''
= 4x 2(2x - 3y),
4ab - 8a x = 2a(a + 2b - 4a:c))
8x 3
+
+ q) -
2a
rep
a(x
2
-
12x2y
2
3(p
+ q)
= (r - 3)(p
+ q).
Difference of two squares:
a'J. - b'J
+ b)(a -
= (a
b).
(2)
E~ample.
16x 2
--
25y2 = (4x
+
5y)(4x - 5y).
Trinomial perfect square:
+ 2ab + b
2ab + b
a'J
a2
l
l
-
= (a + b)I,
= (a - b)2.
(3)
(4)
For a trinomial to be a perfect square, one term must be
numerically equal to twice the product of the square
roots of the other two terms.
Examples.
4m 2
+ 20mn + 25n 2 = (2m + 5n)2,
49x
42x + 9 = (7x - 3)2.
2
-
Trinomial of the form
Xl
+ (a + b)x + ab =
(x
+ a)(x + b).
Examples.
/
X2
x2
x2
x2
* Cf.
§ 1.3.
+ 5x + 6 =
5x + 6
+ 5x - 6 =
;=
-
-
5x - 6
=
+
(5)
-
+
(x
3)(x
2),
(x - 3)(x - 2),
(x
6)(x - 1),
(x - 6)(x
1).
+
+
the distributive law of multiplication with respect to addition,
MORE COMPLlCATED TYPES
§3.51
57
It will be noted that in factoring this type we must
find two numbers whose product is the constant term
(i.e., the term not involving x) and whose algebraic sum
is the coefficient of the first-degree term (i.e., the term
in x).
If the constant term is positive, these two numbers
will have the same sign, namely, that of the first-degree
term.
If the constant term is negative, the numbers will have
opposite signs, and the numerically greater will have the
sign of the first-degree term.
Trinomial of the form
acx
ll
" '{
I
+ (ad + bc)x + bd ,., (¢J + b) (ex
'*
d).
(6)
This is similar to (5).
Example.
6x 2
+ 7x
= (2x
- 20
+ 5)(3x
- 4).
In factoring this type by inspection one can write
down various possibilities until he discovers the right combination. For instance, in the present example he might
try several combinations, such as (6x - l)(x + 20), before
finding the correct one.
(In Chapter VI it will be shown how types (5) and (6)
can be factored by solving a quadratic equation.)
Grouping:
ax
+ ay + bx + by
= a(x
=
(a
+ y) + b(x + y)
+
b)(x
+ y).
(7)
Example.
ax - 2bx - 3a
+ 6b
= x (a - 2b) - 3 (a - 2b)
= (x - 3)(a - 2b).
3.5. More complicated types.
Sum of two cubes:
a3
+b
3
= (a
+ b) (at -
ab
+ lJS).
(1)
FACTORING
58
[Ch. III
Example.
x3
.!
+8
= (x
+ 2)(x
2
+ 4) .
2x
-
Difference of two cubes:
a3
b 3 = (a -
-
b) (a 2
+ ab + b2 ).
(2},
Example.
27m 3
8n 3 = (3m - 2n)(9m 2
-
+ 6mn + 4n
2 ).
Sum of two odd powers: The sum of two odd powers is
always divisible by the sum of the numbers.
Example.
as
:-t
+b
5
= (a
+ b)(a
4 -
a 3b
+a b
2 2 -
ab 4
+ b5).
,1\"'
Note the form of the second factor. The coefficients
of the various terms are alternately + 1 and -1; the
exponent of a decreases by 1 from term to term, while
the exponent of b increases by 1 from term to term, b
first appearing in the second term.
Difference of two odd powers: The difference of two odd
powers is always divisible by the difference of the numbers.
I
Example.
as - b5 = (a - b)(a 4
+ a b + a b + ab + b
2 2
3
3
4
).
The difference of even powers should be factored as the
difference of squares.
Example.
a6
-
b6 = (a 3 )2 - (b 3 )2 = (a 3 + b3 )(a 3 - b3 )
= (a + 6)(a 2 - ab + b2 )(a - b)(a 2 + ab
+
b2 ).
Each factor should be refactored if possible until the
expression is reduced to prime factors.
The sum of two even powers cannot be factored as
such. Note, however, that
a6
+
b6
= (a 2)3
=
2
(a
= (a 2
+ (b 2)3
+ b2)[(a 2)2
+ b )(a
2
4
-
-
,
a1.b 2
2 2
ab
+ (b
+ b ).
4
2 )2]
OTHER TYPES
§3.61
59
3.6. Other types.
Expression reducible to the difference of squares.
Example 1.
X2 - y2 - Z2
+ 2yz
= x 2 - (y2 - 2yz
x2
= [x
= (x
=
+ Z2)
(y - Z)2
-
+
+
(y - z)][x - (y - z)]
y - z)(x - y
z).
+
Example 2.
:,.,
!
Add and subtract 4x2y2:
x.
+ 4x2y2 + 4y4 Example 3.
4x2y2 = (x 2
= (x 2
+ 2y 2)2 - (2xy)2
+ 2y2 + 2xy)(x 2 + 2y2 -
9x 4 - 6X 2 y 2
2xy).
+ 25y4.
This would be a perfect square if the middle term were
If we add and subtract 36x 2y2 we get
±30x2y2.
9X4 - 6X 2y 2
+ 25y4 =
=
=
+
+
9x 4 30x2y2
25y4 - 36x 2y2
2
(3x + 5y2)2 - (6xy)2
(3x 2 5y2
6xy) (3x 2 5y2 - 6xy).
+
+
+
Polynomial perfect square:
a"
+ b' + c' + 2ab + 2ac + 2bc
= (a
+ b + c)".
Example.
4x 2 - 4xy
+ y2 + 12xz -
6yz
+ 9z 2 =
(2x - y
+ 3Z)2.
Factoring by trial division: Sometimes a polynomial
such as
x 3 - 2x 2 - 2x - 3,
in which the coefficients are all whole numbers, that of the
term of highest degree being 1, can be factored by inspection. Possible factors are x ± k, where k is a factor of
the constant term. In this example the possible factors
[Ch. III
FACTORING
60
are x + 1, x - 1, x + 3, x - 3, the last of which is found,
by division, to be a factor. This method will be discussed
further in Chapter XIII.
3.7.
Suggestions for factoring.
As was stated earlier, the secret of factoring is the
ability to recognize type forms. The following sug--~
gestions may prove helpful:
First remove monomial factors, or other obvious factors l
if any exist. It may then be possible to classify the
expression under one of the following heads:
j.,'
.'
/.,
Birwmial:
Difference of squa.res.
Difference of cubes.
Sum of cubes.
Difference of odd powers.
Sum of odd powers.
~:J \ ,
Reducible to difference of squa~ by adding and
subtracting a perfect square.
Trinomial:
Square of binomial.
Type x 2 + (a + b)x + abo
Type acx 2 + (ad + bc)x + bd.
Reducible to difference of squares by adding and
subtracting a perfect square.
Polynomial of four or more terms:
Capable of being grouped.
Reducible to difference of squares by rearrangement.
Cube of hinomial.
Square of polynomial.
Divisible by x ± k, where k is a factor of the constant
term.
EXERCISES
61
Always see whether any factor can be factored further.
A check is afforded by multiplying together the factors
obtained. The product must equal the original expression
if the work is correct.
EXERCISES III. B
',1 ,,'.
Factor completely:
1.
3.
5.
6.
7.
S.
9.
11.
13.
15.
16.
17.
18.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
42.
43.
45.
47.
49.
51.
2. 26x 3 y2 + 52x 2y 3 - 39x 2yz.
6xyz + 9abx.
51x 6y2 - 119x 2y 6.
4. 42abc 2 - 49a 2 - 56ac 3•
7(a - b) + 3x(a - b).
4x(3m - 4n) - 6y(3m - 4n).
(p + 3q)5x - (p + 3q)8y.
(a + 2b)(x - 3y) + (3c + 4d)(x - 3y).
36p2 - 49q2.
10. 25a 2b4 - 64c 6 •
27x 2 - 12y2.
12. a 3b - ab 3 •
(m + n)2 - 25.
14. 36p2 - 49(r + 8)2.
(9x - y)2 - (4u + 3V)2.
(a + b - 2c + 3d)2 - (3d + e + 2f - g)2.
98a 3 - 72a.
L
51ab 2 - 119a 2b2c - 34ab 4c2.
fT
x 2 - 4xy + 4y2.
20. 36x 2 + 60xy + 25y2.
9m 2 + 48mn + 64n 2.
22. 8a 4 - 24a 2b + 18b2. nu[u
x 2 - lOx + 24.
24. x 2 - 5x - 24.
a 2 + 16ab - 36b 2.
26. x 2 + 13xy + 40y2.
I I
X4 - 13x 2 + 36.
28. 2x 2 - 12xy - 54y2.
6x 2 - llx - 10.
30. 6a 2 - 92ab -- 98b 2.
.;}6p2 - 9pq - 81 q2.
32. 6x 4 + bx 2 - 15b 2.
5x 2 + 26xy + 5y2.
34. 18x 2 + 19xy - 12y2.
b(ax 2 - c) + x(b 2 - ac).
36. ab(x2 + 1) + (a 2 + b2 )x. I'
3
x + 2X2 + X + 2.
38. 6ax + 4ay - 9bx - Gby . .t
ax 2 - 3b - a + 3bx 2.
40. a 2 - &2 + a-b.
6xy - 4ay + 9ax - 6a 2.
12abm + 27bm - 8acn - 18cn.
8x 3 + y3.
44. a 6 - b3 •
343a 3 - 125b 3•
46. 192x 3 + 81 y 3.
(a + b)3 - c3 •
48. x 3 - (2y - 3Z)3.
3
x + y3 + 3x + 3y.
50. a 3 - 6a 2b + 12ab 2 - 8bl •
5
y
x + 32 6.
52. x 6 - 64 y 6.
)l"~,,\
62
[Ch. III
FACTORING
53. x 8 - y8.
55. x 7 + y7.
67. x 6 + 27 y 6.
59. a 2 + b2 - (:2
61. a 2
62. a 2
-
2
':'if
: :iiX
x8
-
16y 8.
x7
-
y7.
x6
+
y12.
2ab.
4xy + gz2 - x 2 - 4y2.
2
2
c - 4d - 2ab - 4cd.
4c 2
d 2 - 4ab
4ac - 2ad - 8bc
4bd
+b
+ 4b +
2
' ;
54.
66.
58.
60.
-
+
+
64.
63. 9x 4 + 6X 2 y 2 + y4.
65. 25x 4 - 24x2y2 + 4y4. .t 66.
J, 68.
67. x 8 - 5x + 4.
70.
69. x 3 + 2x 2 + 9.
72.
71. x 3 + 4x 2 + 5x + 6.
73. X4 - 5x 3 + 3x 2 + 2x + 8. 74.
75. a 9 + b9 •
76. x 2z - x2y + y2x - y2z + Z2y -
+
-~,~
9x 4 + 11x2y2 + 4y4.
'
81x 4 + 63x 2y 2 + 16y4.
x 8 - 2x - 4.
x3
x2
7x
18.
3
x - 8x 2 + 4x + 48.
X4 - 2x 3
3x 2 - 3x
+ + +
+
.l',
':!: .
.ll
.0,
+ 1.
z2x.
3.8. Highest common Factor and lowest common multiple •.
The highest common factor (H. C. F.) of two or more
expressions is the expression of highest degree that can
be divided into all of them exactly.
The lowest common multiple (L. C. M.) of two or
more expressions is the expression of lowest degree that
is exactly divisible by all of them.
To obtain the H. C. F. and the L. C. M. of two or more
expressions, separate each expression into its prime
factors. The H. C. F. is the product formed by taking
each factor which appears in all of the expressions to the
lowest power to which it appears in any of the expressions.
The L. C. M. is the product formed by taking each factor
to the highest power to which it appears in any expression.
Example 1.
Find the H. C. F. and the L. C. M. of 2X 2 y 3, 4xy fi, 18x7 y 2z.
SOLUTION.
Rewrite in the respective forms
2X 2y 3, 2 2xy., 2 . 3 2x 7y 2Z.
H. C. F. = 2xy2.
L. C. M. = 2232~7y5z = 36x 7y 5Z.
r
I
i .
I
EXERCISES
63
Example 2.
Find the H. C. F. and the L. C. M. of
x 2 + 2xy + y2, x 2 - y2, x 2 - Z'IJ - 2y2.
SOLUTION.
x 2 + 2xy + y2 = (x
x 2 - y2 = (x
x 2 - xy - 2y2 = (x
H. C. F.
L. C. M.
+ y)2,
+
y) (x - y),
2y).
+ y)(x = x + y.
= (x + y)2(X -
y)(x - 2y).
EXERCISES III. C
1. Prove that the product of the highest common factor and
the lowest common multiple of two expressions is equal to
the product of the two expressions.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Find the highest common factor and the lowest common
"1,1
multiple of the following expressions:
26a 2b3c4, 39a 4b3c2 •
261x 5 y 2z 1o, 116x 2y 7z 8w 3.
a 3 - b3 , a 2 - b2 •
H'HroX oJ 'nrri";:
x 2 - 4x - 12, x 2 - 9x + 18.
6x 2 + X - 12, 8x 2 + 30x + 27.
.,,:n
; •• ) ,1:1
a 2 - ab + b2, a 2 + 2ab + b2 , a 3 + b3 •
2x 2 - 11x + 12, 3x 2 - lOx - 8, x 2 - 11x + 28.
36x 2 - 84xy + 49y2, 42x2 - 25xy - 28y2.
8x 2 + 6xy - 9y2, 12x2 - 17xy + 6y2, 9x 2 + 3xy - 6y2.
8a 3 - 12a 2b + 6ab 2 - b3, 4a 3 - ab 2 •
a 8 - b8 , a 6 - b6 •
a 8 - b8 , a 6 + b6 •
a 2 + b2 - 4c 2 - 2ab, a 2 + b2 + 4c 2 - 2ab - 4ac - 4bc.
x 3 - 5x - 12, x 3 - 5x 2 + 18.
16x4 - 81y4, 4x 2 - 9y2.
16x 4 + 81y\ 4x 2 + 9y2.
x 3 - 3x 2 - 6x + 8, x 3 + 3x 2 - 4, x 3 + 3x 2 + 2x.
CHAPTER IV
4.1. Fundamental principle$.
The value of a fraction is unchanged if the numerator
and the denominator are both multiplied or both divided
by the same number, zero excepted.
NOTE.
2x 2
Expressions such as 3' :3 x, and 2x/3 are entirely
equivalent.
This principle is used in reducing a fraction to its lowest
terms. *
Example.
Reduce to lowest terms:
4X 2y 7z
fu5
y
3'
~,~
Divide numerator and ·4i~r\ ey'ltheir
H. C. F., 2X 2 y 3, getting
v::' t, 'J;; , I
. ,,; ,
SOLUTION.
2y4Z.
3X 3
A fraction may be regarded as having three signs
associated with it: the sign of the numerator, the sign of
the denominator, and the sign preceding the fraction.
Some or all of these may not actually be written, in which
case they are, of course, considered to be +. Any two
of these signs may be changed without changing the value
of the fraction. Thus,
n
-n
n-n
d = - d = - -d - -d
* A fraction is in its lowest terms when its numerator and its denomina.tor
have no common factor other than 1 or -1.
64
~
EXERCISES
65
This is a direct consequence of the laws for the division
of signed numbers. (See § 1.5.)
Example.
(a+b)(c-d)
Reduce to lowest terms: (x + y)(d _ c)'
" .b,
SOLUTION. The factor (c -- d) in the numerator and the factor
(d - c) in the denominator are equal except for sign. If we
change (d - c) in the denominator to (c - d) we have changed
the sign of the denominator, since changing the sign of a single
factor changes the sign of a product. We can compensate for
this change by changing the sign preceding the fraction. Thus,
the given fraction is equal to
(a+b)(c-d)
(x
+ y)(c
- d)
,
wddv8
__
a_+_b.
x+y
This reduces to
i
10
'wt.BDIIIWW;,h (1001
. to:) ,B ~i1i.·llld
EXERCISES IV. A
rIOI
Reduce to lowest terms:
1.
3.
5.
7.
9.
+
x
+
+
III
(x - y)(u - v).
(x
y)(v - u)
X4
4
13.
X4
2x 3 2X2
6
6
15. a-4_ - b-4
a - b
x 3 - 3x 2 - 50
17. 3
x - 3x - 110
11.
'gl
+
+
+
+
+
,dr.:;
fa',
153x 7 yz 2
2. 187(xyz) 6'
4.
x2
-
•
+
",id.T
.l.t
lOx
21
x2 - 9
.
+ 10ab + b
6a2 + ab - b2
4a
12ab + 9b 2
6a2 + tab - 24b2
2
8.
-
x8 _ y8
10. x6 _ y6
b2
a3
+ b3
+ b)3
+ 8b 3
4ab 2
3x 2 - 16
+ 4x 2 - 32x
a2 b3 3
14. a
(a
a3
16. 3
a x3 18. 3
x
12.
.j
9t!j ,i::
6 16a 2
i
(:l(!(lii
'JILl dRU (,'
f
91a 3b2c5
441a 5bc 2
x 2 - 7x
6
2 2x - 24
a 2 - ab - 12b 2
a2 _ 2ab - 8b2
3a 2 6ab - a - 2b .:')
9a2 - 6a
1
'.
X4 __]f_4.
x3 _ y3
bi!': fiT
2
•
.~
'to
66
fRACTIONS
19•
21
x3
X4
+
a2
• a
2
y3
+ x2y2 + y4.
-+- b
2
-
b2
-
20.
c2
+ c + 2ab
2
+ 2bc
- 2ac - 2bc
x
[Ch. IV
3 3
X
4x
2 2
+ 6x
11x + 30
- x- 30 ·l:\).
~
io
.
4.2. Addition and subtraction of fractions.
To add or subtract fractions having a common de~t"na- ,.
tor, add or subtract the numerators and place the renilt over--:
the common denominator.
1.1
Example 1.
Example 2.
2x
5y
2x - 5y.
tI
= 3z 2
To add or subtract fractions which do not have a common denominator it is necessary to reduce them to fractions having a common denominator. It is ordinarily
best to use the lowest common denominator (L. C. D.),
that is, the lowest common multiple of their denominators. This is done by using the fundamental principle
of § 4.1.
3z 2
Example 3.
Add the fractions
-
l~x and
3z 2
1;x 2 '
SOLUTION. The L. C. D. is 30x 2 • (See § 3.8.) To reduce the
first denominator to the L. C. D. it is necessary to multiply it by
the factor 3x (found by dividing the L. C. D., 30x 2 , by the
denominator lOx). We multiply both numerator and denominator of the first fraction by 3x, which, by the fundamental
principle of § 4.1, does not change its value.
Similarly, we mUltiply both numerator and denominator of
the second fraction by 2 (since 30x 2 + 15x 2 = 2).
The required sum is
_2_+~=
lOx
15x2
3x·7
3x . 10x2
+
2·2 =21x+4.
2 . 15x2
30x~
MIXED EXPRESSIONS
§4.3]
67
The dividing line of a fraction performs the same service
as signs of grouping placed around the numerator.
Example 4.
5
x - 2
6x - ~
2x . 5 - 3(x - 2)
=
l2x 2
=
lOx - 3x
+6
12x2
7x
+6
=
Example 5.
Combine into a single fraction:
+
3a
a2
-
b2
b
a2
+ 2ab + b
2
__
2_.
a +b
SOLUTION. Write the fractions with their denominators in
factored 1orm:
' .
+ b _ _2_.
3a
(a + b)(a - b)
(a + b)2
a +b
L. C. D. = (a + b)2(a - b).
The required fraction is
3a(a
+ b) + b(a (a
b) - 2(a
- b)
+ b)2(a
=
+ b)(a
- b)
+ 3ab + ab
- b2 - 2a 2
(a 4- b)2(a - b)
a 2 + 4ab + b2
3a 2
= (a + b)2(a
+ 2b
2
- b)'
4.3. Mixed expressions.
A mixed expression is the sum of a polynomial (or
monomial) and a fraction. It corresponds to a mixed
number, such as 2!, in arithmetic. The following are
mixed expressions:
3a
2b
+ 5c'
x
2
+ 2x -
5
4 - x _ 3'
To reduce a mixed expression to a fraction, write the
integral part (i.e.) the nonfractional part) as a fraction
with the denominator 1 and proceed as in adding fractions.
FRACTIONS
68
EXilmpJe
[eh.IV
1.
2b
+ _.
5c
Reduce to a fraction: 3a
SOLUTION.
3a
+ 2b
= 3a
+ 2b
5c
1
5c
= 3a . 5c
/.
+ 2b =
l5ae
5c
Example 2.
Reduce to a fraction: 3x _ 2
+
+ 2b:
5c
i'\
". if) ,
"I>~.':i
~-
_6_. tEl i: oJt1t l3nidmo:,)
x+45){;
SOLUTION.
;'3x _ 2
+ _6_
= 3x -
x+4
I
=
+ _6_ = (ax - 2)(x + 4) + 6 .
x+4
x+4
•
+ lOx - 8 + 6 3x + lOx - 2
x +4
-----'x-+--,--4-o--
3x 2
2
2
=<:
An improper fraction in algebra is one in which the
denominator is a polynomial whose degree is less than or
equal to the degree of the numerator. To reduce such a fraction to a mixed expression is merely a Droblem in division.
Example.
. d
.
Red uce to a mlxe expressIOn:
++X 3-
6X2
5
2x
.
SOLUTION.
3x - 4
2x
+ 3)6x2 + X 6x + 9x
.5
2
- 8x - .5
- 8x - 12
(
+X +3
6x 2
2x
5
=
7
3x - 4
+ 2:1: + 3'
EXERCISES IV. B
Combine into a single fraction:
J.. -2
3x
3
+_.
4x
a
2. 4b2
e
+ 6b'
a
b
3. be
-+-.
ae
EXERCISES
5
1
4. 36a2 + 24a3 : - l:H ;08
abc
6. b + C+ (1"
8. _3_, _ _
2_.
x +y
x - Y
-:c8
10.
4
_ _3_.
x - 2y
2x + Y ~- '
69
6 ~+....!!_.
3
· 16x y 2
:8
7 _9_ - ~
4xy
.
9.
3
+
6
x 2 - y2
,,$
11
6yz
+ 2...
8xz
5
(x _ y)2
,H
•
5b
lOa
· 2a - 3b
12._1_+_2_+ 2 3
x + y
x - Y
x - y2
1
13. x2 - y 2
14
2x
• x 2 - 7x
18 xy 4
3a
+ 4b'
os - "
'-r.-
1
x2 - x2 y + Y 2
_
5
.
+ 12 x 2 - 2x - 8
, J:&
,0&
OS -- ..•~
'--_
:s;
~ ~-
-,.-.
2
16. a - b - b - a'
SOLUTION. Change the sign of the denominator of the
second fraction and change the sign preceding it. Then we
have
325
--+--=-_.
a-b
a-b
a-b
3
16. _x_ _ _ y_.
17. 5a - 4b
x-y
'II-x
18 3a + 2b _ _
5_ + _4_.
• a 2 - b2
a - b
b - a
7x - 3
2x - 5
20 5x + 6y
19. x + 3 - 2x + 5'
· 6x + 5y
21
· 6x 2
x + 2
+ 19x + 15
3
22. (a -- b)(b - c)
a-c
23. (a + b)(b + c)
z
24. 5x + 6y'
27. 3x
+ 5 + 3x 2~
9x 2
+
+
2x - 5
+ 9x -
(b - c)(c - a)
b-a
(b + c)(c + a)
2c
5'
+ 3x
2x
2
- 5a'
- 4y.
+ 3y
10
4
26. 4a - 3b'
+ 4b
5
(c - a)(a - b)
c-b
+
(c
+ a)(a + b)
26. y
28. 2a - 7b _
+ x- z- _yz·
4(~: -=- ~~b2).
FRAalONS
70
29.
5
x_ 2 - x
a
(Ch.IV
4x - ay
+ 7 + 2.
30. lIx - 2y - 2x _ ay'
?
Reduce to mixed expressions:
31 x + 5.
. x+2
2x - 13.
33
. ax + 8
12x 2 - 7x - 20
36.
ax - 4 •
""
, i,
6x - 7
32. 2x - 11'
%2 + 5x + 4.
34
•
x+3
36 &:2 + 7x
•
2
37. _15-:;-x-::-+_6x_--::2_O.
5x 2 _ X 2X' - 70x 2
x2
39.
38.
2
+ lOx + ax - 4
+ 6.
!
2
x' + x + X + 1
x+2
.
2x 2 + 5
40 8x' - 6x 2
3
•
• 4x 8
+ X2 -
-
7x + 8.
2x - 18
4.4. Multiplication of Fradions.
To multiply fractions, multiply their numerators to obtain
the numerator of the product and multiply their denominators
to obtain the denominator of the product. Factors common
to numerator and denominator should, of course, be
divided out.
. Examples.
a c
b.d =
:1: 2 -
xy
y2
1)t
ac
a' bx! ax!
b a 4y = b2y'
bi
X2
. X2
+ "y2
+ 2xy + y2
(x
z y
y
x'
Z = l'
+ y)(x -
=
y)
z = z'
x2
. (x
xy
(x - y)(x 2 y2)
= xy(x + y) •
+
+ ~8
xy
+ y2
;'
;;,
".IV
+ y)2
4.5. Powers of Fradions.
The power of a fraction is equal to the power of its
numerator divided by the power of its denominator.
Thus, for example,
n)3
(d
=
n n n
d.d.d =
n3
d 3'
EXERCISES
71
4.6. Division of fractions.
Since, by the fundamental principle of § 4.1,..
a
,{]
d
a
b b'c a d
C = Cd = b' c'
d d' c
we have the familiar rule:
To divide one fraction by another, invert the divisor
multiply.
aM
'
.":"."
Examples.
i + a i· ~
=
~~, ~ +
=
+ 8)2 _,_ ~ _
(r
r-
8
•
r2 -
x =
~ +y ~ ~. ~ =
+ 8)2 • (r + 8)(r -
(r
r-
82 -
r8
8
If)
=
(r
;,
+ 8)3
r8
.
When mixed expressions are involved in division it is'
often convenient to reduce them to fractions before proceeding with the division, as in the following example,
which, however, can also be solved by the method of §4.7.
Example.
x
(4_~)+(4+~)=4X-5+4X+5=4X-5.
x
x
x
x
x
4x+5
4x - 5
=
4x
+ 5'
EXERCISES IV. C
Perform the indicated operations and simplify:
3
4 2 3
1 24a b c . 15x 2y 23z 23 .
• 25 xy 2z 5 16a b c
+
p2 - q2 m2
2
2 ·
2
• m - n
+
+
+
3 x 2 2x - 48. x 2 6x
9.
·
x3
27
x 2 - 8x
12
.
4 8a 2 2ab - 15b 2 .
a 2 - b2
2
3
3
b
4a - ab - 5b 2
•
a
+
+
+
3p2
+ 2mn + n2.
+ 2pq
_ q2
72
FRAalONS
[Ch.IV
+
+
[) a + 4 . a + 4a
4.
• a 2 - 4 a2
2a
2
6 6r2 + 59r8 - 108 2 • 98 2 - 4r2.
• 8r2 - 6r8 - 98 2 8 2 - 36r2
7 2x 2 + 5x - 3
10x 2 + 13x + 4 35x 2 - 38x + 8
2
• 16x + 26x + 9 5x 2 + llx - 12
x2 + X - 2
2
2
8 3x + llxy + lOy2 . 6x + 29xy - 5y2 . 3x 2 - 31xy + 56y2.
· 3x 2 + 8xy - 35y2 3x 2 - 19xy - 40y2
X2 - 4y2
a 4 - b4
a 2 - b2
9. x 3 _ ya -;- x2 _ y2'
4
2
+
lOx + 24 _,_ X2 + 4x - 32.
· x 2 + lOx + 24 . x 2 + 4x - 12
11 x 2 - 9xy - 36y2 _,_ x 2 + 9xy + lSy2.
• x 2 - txy - lSy2 . x 2 - 6xy - 16y2
' " \'
~
"~ qme.n
2
2
12. 30x + 71x + 42 -;- 18x + 33x + 14. b!)
~,'I.'l
8X2 + 65x + 8
24x2 + 35x + 4
',...., r f.- ~
2
lOx 2 + 61xy + 33y2
,~:''l -+- ~)
13. 15x2 - llxy - 12y2
4x + 4xy - 35y2 . 8X2 + 46xy + 63y2
",
10 x 2
-!-
14.
•
.
(a - ~) + (c - ~}
"~,
f,.,
I
4.7. Complex Fractions.
A complex fraction is a fraction which has one or more
fractions in its numerator or in its denominator or in both.
Methods of simplifying complex fractions are illustrated
by the following examples. One very useful method is
that of multiplying numerator and denominator of the
complex fraction by the lowest common denominator of
the simple fractions occurring in the complex fraction.
§4.7]
COMPLEX FRACTIONS
Example 1.
73
~+~
Simplify:
3
6
1 -
~
4
SOLUTION. The L. C. D. of the fractions occurring in numer·
ator and denominator is 12. Multiply numerator and denoml
nator by 12:
2
1
12 . 3" + 12 . 6"
8 +2
10
3 = 12 - 9 = 3'
12· 1 - 12· =1
"HIt"
Example 2.
c:rr!.
,~jl) 11
Simplify:
SOLUTION. Multiply numerator and denominator by the
L. C. D. of the fractioJ¥5 occurring in numerator and denomi',\
nator, viz., ab:
ab .
E+ ab· 1
b
a
ab·--ab·a
b
a2
b2
Example 3.
+ ab
a(a
a 2 = (b
-
x
+ b)
+ a)(b
a
- a) = b - a·
1
'(d
'!().
X2=-I- X+1
Simplify:
_x_+
x-I
_l_
x+I
SOLUTION. Multiply numerator and denominator by the
L. C. D. of the fractions in numerator and denominator, viz.
(x
(x
+
+
1)(x - 1):
1)(x - 1)
x
X2=-I x
(x
+
1)(x - 1)
1
XTI
(x + 1)(x - 1) x _ 1 + (x + 1)(x - 1) x
= (x
1
+1
x - (x - 1)
x - x +1
I)x + (x - 1) = x 2 + X + x - I = x 2
+
1
+ 2x -
l'
74
FRACTIONS
[Ch.IV
Example 4.
Simplify:
1
1
a+1
2a--a
2a ".
SOLUTION.
Oonsider the fraction
1
2a _ a
(2)
+ l'
a
which is found in the denominator. The L. O. D. of the fractions occurring in the numerator and denominator of the fraction (2) is a (since there is only one such fraction and its denominator is a), Multiply numerator and denominator of (2) by a.
This reduces (2) to
a
(3)
2a 2 - a - 1"
"'-;:J'
lj'
Substitute (3) in the proper place in (1),
1
2a _
2a 2
-
a
a-I
which h)~mes
:' r\·r~
<
'\
~
(4)
The L. O. D. of the fractions occurrin8 in numerator and
denominator of (4) is 2a 2 - a - 1. Multiplying numerator
and denominator by this gives
/:l.\(
2a 2 - a-I
4a 3 - 2a 2 - 3a'
(5)
A result in fractional form should always be reduced to its
lowest terms. Oonsequently every such result should be
investigated to see whether it can be reduced. If numerator
and denominator of the fraction (5) are factored, (5) takes the
form
(2a
+
a(4a 2
l)(a - 1)
-
2a - 3)'
(6)
and since there is no factor common to' numerator and d.enominator, the fraction is in its lowest terms and cannot be reduced.
Either (5) or (6) is an acceptable form for the result.
75
EXERCISES
EXERCISES IV. D
~ '/~. f ~ J.
Simplify:
3 5
"
a+c
3 · - -b·
,i
':~L'
c+-a
8-"2
_1_ +_1_
"x+ y x-y
1
b+~a_
b- a
Ii. - - - - __a_ _ b
a-b
1
+y
x 2 - y2 - X
6.
b
3
2
4 - {3
1-. 11'
,i':
a-6+~
9
a--
5
a
x
a
x
8.
2
,'/;'
-2-
'"
4-x
x-2
----=--=----
11.
x - 2 _ _ _x __
x - 1
x--x-2
1
1-
1 - x
x+2- _ _x_ _
3x + x-I
x
x - 2y
-----~--~~1-----
2y -
1
x
a+b
a 2 + b2
14. a + b
a 2 + b2
x+l
16.
+ 2y + x
1
a
1
b
x+3
a-b
+ b2
'a - b
a 2 + b2
(l2
1
~---:--
1+_1_
!+3
x -
3'
;
12. ____~x~______
13.
3x
-
2- - - 5
'$,
r· ...
2
5
9.
1+
-3-x
10.
4
xn-x=a
7.
:t _ 27
"
- 2y
1
-
a
l'
+b
x-I
x-3
X+2 - X+4
X+2 - X'+4
x+5
x+7
x+6-x+8
x-5
x-7
x+6-x+o
i, ; "
76
FRACTIONS
[(h. IV
4.8. Extraneous roots.
In § 2.3 it was stated that an equivalent equation is
obtained if the sides of a given equation are increased or
diminished by the same number or expression, or if they
are multiplied or divided by the same number or expression, provided, in the case of multiplication or division,
that this number or expression is not zero and does not
contain an unknown.
Certain operations, when performed on an equation,
may lead to an equation containing extraneous roots,
that is, roots which are not roots of the original equation.
In particular, the following operations may lead to
extraneous roots:
Multiplying both sides by an expression containing an
unknown.
Raising both sides to the same power.
It is therefore absolutely necessary to check the values
obtained in solving an equation when either of these
operations has been performed. For, even if the work
has been correct, some of the roots may be extraneous,
in which case they will have to be discarded.
Example 7.
The equation x + 2 = 5 has only root, x = 3. Multiply
both sides by x - 2, obtaining x 2 - 4 = 5x - 10. The new
equation has the roots x = 3 and x = 2, as can be verified by
substitution. Thus, the extraneous root x = 2 has been
introduced.
Example 2.
The equation x + 2 = 5 has only one root, x = 3. Square
both sides, obtaining x 2 + 4x + 4 = 25. The new equation
has the roots x = 3 and x = -7, the latter being an extraneous
root introduced by squaring.
If both sides of an equation are divided by an expression
containing the unknown, a root may be lost.
14.9]
EQUATIONS INVOLVING FRAOIONS
77
Example 3.
The equation x 2 - 4 = 5x - 10 has the roots x = 3 and
x = 2, as can be verified by substitution. Divide both sides
by x - 2, obtaining x
2 = 5. This equation has the single
root x = 3, the root x = 2 having been lost.
+
Example 4.
The equation x 3 = 25x has the roots x = 0, x = 5, and
x = - 5. Divide both sides by x, obtaining x 2 = 25. This
equation has the roots x = 5 and x = - 5 only; the root x = 0
has been lost.
4.9. Equations involving fractions.
In solving an equation involving fractions, W~ usually
clear of fractions by multiplying both sides of the equation
by the lowest common denominator. If a common
denominator other than the lowest is used as a multiplier,
extraneous roots are likely to be introduced. if both
sides of an equation are multiplied by the lowest common
denominator of the fractions involved, any root of the
resulting equation is also a root of the original equation,
provided it does not, when substituted, make any denominator of the original equation equal to zero .
.,." :f:t.
{; '_;j;
(;,
x
1
+3
+
1
_
2
x - 3 - x 2 - 9'
\i --
-t" 3;
Example 1.
d
C'I
f,
...: ..-- 8 .~":::".:.....,.... _._.:_.Solve the equation ._ t:t;'
A!!."."'- I~
f. - ;r;!::
<:J;
+- 8--':+..._._
.\'
:s.S!
('
.!.._ -+- _~_..
- \i . r· ,
.~, ,
e
SOL:rTION. Clear the equation of fractionS~fi~lUltip~ing
both sIdes by the L. C. D., (x
3) (x - 3):
'.
+
+
x - 3
x + 3 = 2,
2x = 2, x = 1.
CHECK. Substitute x = 1 in the original equation:
1
1
2
4 - 2 = - 8'
1
- 4= -
r1
.,\
.1 ...
FRACTIONS
78
[Ch.IV
Example 2.
Solve the equation
1
x
+3 +x
1
- 3 ...
6
X2 -
9'
SOLUTION. Clear of fractions by multiplying by (x
(x - 3):
+ 8)
x - 3 +x + 3 = 6,
2x = 6, x = 3.
When we attempt to substitute the value x = 3 in the original
equation we get
116
-6
+= _.
00
But division by zero is meaningless; hence x = 3 is not a. root .
of the original equation. In fa.ct the equation has no root.
EXERCISES IV. E
Solve:
3
5
1. x _ 7 - x + 7 == O.
3x
4
3. x _ 4 - x + 5 = 3.
14
7
1
5. x2 - 9 - x + 3 == x - 3'
7
•
9.
x
5
1
2'3x+7+x+5=3'
x-4
2
4· x + 4 - x _ 4 =1.
x
2
1
6. 3x+4 +5x+7=r
5
3
2
_2_+_3_= 12 .
2
8.
x2
_
9
+
x
+
3
=
x
3'
2x + 3 2x - 3 9 - 4x
1
1
3
7
x + 2 + 1J _ 4 ... 0,
10. 3x + 7 - 5y _ 6 == 0,
1
4
2
3
+ y + 4 == O.
2x - 2 + 3y + 10 = O.
x
y
~6 + _L3
12. x _ 2 + y + 1 = 2,
xy- == 2,
2x
y
1
2x + 3 _ 3y + 3 ... -1.
3x + 1 - 2y - 1 == tf
x
y
j
/
x
11.
13. Solve for z:
!z = !x +!.
y
14. Solve for r: 1 == R
2E
+ 2r'
/
EXERCISES
16. Solve for T: H
= 8;5 (T _
16. Solve for P: T
=
W;2}
~ (p +~) (v -
b).
17. Find a number which, when subtracted from the numerator
and added to the denominator, changes the fraction H into
a fraction whose value is i.
18. If 5 is added to the numerator and the denominator of a
fraction the value of the result is t. If 5 is subtracted
from the numerator and the denominator of the fraction
the value of the result is t. What is the fraction?
19. A motorist set out from A at a certain rate and would have
reached B in 4 hours and 48 minutes if he had continued
at this rate. However, when he had gone three-fourths
of the way he had motor trouble which delayed him for an
hour. During the remainder of the journey he incrmsed
his speed by 10 miles an hour and reached B 48 minutes
behind schedule. Find the distance from A to B and the
rate at which he started out.
.
/
CHAPTER V
t) t;8 .~. ':\
Exponents and Radicals
,\'.1 ._ ';1/
t ...· +
"\;"I.I
I
H ftO!J::Hl11 ~~I{,t e'~~rus£h i'fOOaffir
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5.1.
Laws of exponents.
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ei C U ,81:
';"', ,,·,1
The subject of exponents was treated briefly in Chapter
I (§§ 1.7, 1.9, and 1.10). We summarize here the laws
of exponents. The numbers m and n represent positive
integers; a and b are any algebraic expressions, except
that if either appears in a denominator it cannot be zero.
1J
I'd ~)!i
aman = am+n,
(1)
tlS-h} ihLL! 8f. H I-.;j
(am)n = amn •
:);
(2)
m - n if m > n ();,
I', t:
91H hr.1R n 00 A r
am
'
.1 j
(3)
Ii=
1.
a
a n- m If m < n.
I
{a
•
The symbol > means "is greater than," the symbol <
means "is less than." *
(ab)n = anb n,
(4)
'.
(~r =
(6)
::.
In this chaRter we shall extend the definition of exponents to cover cases in which they are not positive
integers,
5.2. Roots.
If xn = a, where n is a positive integer and a is a real
number (i.. e., any positive or negative number or zero),
* Other symbols sometimes used are:
__ ~/
;?; meaning "is greater than or equal to,"
~
meaning "is less than or equal to,"
::I> meaning "is not greater than,"
<I: Meaning "is not less than."
80
~
I
§5.3]
RADICALS
81
then x is said to be an nth root of a. In particular, if
= 2, x is a square root of aj if n = 3, x is a cube root
of a.
n
Examples.
Since 2 3 = 8, 2 is a cube root of 8.
Since 3 2 = 9, 3 is a square root of 9; since (-3)2 = 9, -3 is
also a square root of 9.
Since (- 5) a = -125, - 5 is a cube root of -125.
It will be shown in Chapter XII that every number
except zero has exactly n distinct nth roots if we include
imaginary numbers. (See § 5.16.) Some numbers do
not have any real nth roots for certain values of n. For
example, - 4 has no real square root, since the square of
any real number (except zero) is positive. For the
present we shall consider only values of n and a for which
a real root exists.
When a has an nth root which is positive or zero, that
root is called the principal nth root of a. If a does not
have a positive nth root but does have a negative nth
root, the negative nth root is the principal nth root of a.
Thus, the principal square root of 9 is 3, although both 3
and -3 are square roots of 9; the principal cube root of
8 is 2, the principal cube root of -8 is -2.
According to the foregoing definitions, the principal
nth root of a is the positive nth rootof a when a is positive
and the negative nth root of a when a is negative and n
is odd. 'Ve do not define the principal nth root for the
case in which a is negative and n is even.
5.3. Radicals.
The principal nth root of a is denoted by the symbol
This symbol is called a radical. The integer n
is the index, or order, of the radical. The number a,
under the radical sign, is the radicand. If no index is
{Ya.
EXPONENTS AND RADICALS
82
[Ch. V
written, it is understood to be 2; that is, the symbol Va
means the principal square root of a.
The symbol v'9 is equal to 3 and is not equal to - 3, !
although - 3 is one of the square roots of 9. However, !.
it is permissible to say that the squate roots of 9 are ± 3,
the symbol ± being read" plus and minus." (In certain
situations the symbol is more appropriately read "plus
or minus.")
NOTE. It is assumed for the present that if the index of a
radical is even, all literal factors in the radicand are positive.
EXERCISES V. A
Find the following roots:
1.~.
5. 42fX27.
8. ~O.OOlx9.
11. ~6.
14. ~.
17.
2. ~27a6.
3. ~.
4. .y4(i4.
6. ~-100Oxs
7. v'if.OIX4.
9. ~.
10.~6.
12. ~.,
13. {I64x6.
16. ~. ".;, +; 16. ~ -216a 16 •
3
216x 6
19.
18.
x 8•
- I25Yl2'
~ 1~99
20.
22.
-¢' V15625a 12 •
24. 06a 16•
5.4. Fractional exponents.
In this section and the next secti<ms, we shall extend
the exponent notation by assigning, so far as possible, a
meaning to am, where m is any rational number. The
extension is guided throughout by the assumption that
the exponent laws of § 5.1 still hold, even when m and n
are not positive integers.
First we investigate what meaning should be assigned
to the symbol a 1/ 2• Assuming that (2) of § 5.1 holds, we
obtain
ZERO EXPONENTS
§5.5]
83
Thus a l/2 could denote either square root of a.. For the
sake of simplicity, we define
a 1/ 2
= Va,
a ~
O.
That is, a l / 2 will mean the principal square root of a.
Similarly, since the assumption that (2) still holds leads
to the equation
we define
alln =
Va.
That is, a l/n is the principal nth root of a. When a iF
negative and n is even, no meaning is assigned to a lln •
Again, let m and n be positive integers with no common.
factor other than 1. Assuming that (2) of § 5.1 may be
applied, we have
so that we are led to define
am/n = (va)m.
Here too, if a is negative and n is even we
ing to the symbol.
Examples.
_gn
no mean·
(25) 1/2 = 05 = 5.
(-27)1/3 = ~ -27 = -3.
82/3 = (~)2 = 22 = 4.
5.5. Zero exponent.
If we assume that the relation am I an = a-- n still
holds when m = n, we are led to the result
But
if
a
¢
O.
EXPONENTS AND RADICALS
84
[eh. V
Consequently we define
aO = 1,
a
~
o.
We assign no meaning to the symbol 0° ..
Examples.
(2x)O
= 1 (x
~
0),
5.6. Negative exponents.
Finally, let us assume that the relation am/an = am-n
holds even when m < n. Then if, for example, we set.
m = 4, n = 7, we get
a4
a7
-
=
a 4-
7
=
a- 3•
But
and if the two results are to be consi~
a- 3 = 1/a3• In general we define
a
Examples.
~
O.
we ~,"'v.e .
EXERCISES
85
EXERCISES V. B
Perform the indicated operations and simplify the results if
possible:
' -,
,d
C
"
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
33.
34.
35.
36.
37.
38.
(49)112.
(16x lS )1I'.
(64x S) 1I 2 .__
(64x S)1!6.
(23x) 0.
13x- l •
(2X)-3.
<
2.
4.
6.
8.
10.
12.
14.
16.
18.
',i,;"
tx 1l2 . tX1!4.
(tx 11 3)3.
(_27a 27 )-liS.
(4a 4 ) 3/2.
(64 p 3q-S)-2/S.
125a-3bSc9 )2/3
( 216x3y Sz 9
256X 64 )-3 / 4.
( 81 y 8
2- 4 - 3- 3
4
2
(-,125) 113.
(36xI2) 112.
(64x 6 )1/3.
(36 x 36) 1/2.
23x o.
(13x)-1.
2x- 3•
txl/2 + tX1!4.
(tx1!2)-2.
20. (9a 4)-1/2.
' ",' "
22. (- 8a 9 ) 2/3.
"
24. (-8m- 9n O)-213. "'.;:
8-3a 3/2b-3/2)-2/1'. '
'
", 26. ( 343c s
f',
"
28.
(a-b- 3)2/3.
30.
~. 8 2/ 3
•
'
2
-
~ • 8- 2/.,
a- 2 - b- 2
32. a-1b-2 _ a- 2b- l '
(3X 3/2 - 4xl/ 2 + 5X-l!2) C2X 3/2 - 3Xl!2).
" :
(4X7!3y-1/3 + 7X4!3y2/ 3 - 8Xl!3y5/3)(6x2l3y1!3 - 5x-1!'JI'~;!)'~'1
(3x 3 - x 2 - 18x + 16) + (Xl!2 - 2X-l!2).
(14x 512 - 25x 312 + 9X- 112) + (2;r1l2 - 3X- 112).
(5X 5/3 - 2X 2/3 - X-l!3)(X1!3 + 4X- 2/3 ).
(6X2 - 19x - 15 + 28x- 1) + (2X 2(3 - 7X- 1/3 ).
Any number may be written as the product of a number
having a single nonzero digit to the left of the decimal point,
and a positive or negative power of 10. A number so writttn is said to be written in scientific notation. W rite each
of the following numbers in scientific notation:
EXPONENTS AND RADICALS
86
[Ch. V
, I
39. (a) 14,700,000; (b) 0.0000328.
SOLUTION.
(a) 14,700,000 = 1.47 X 10,000,000
= 1.47 X 10 7 ;
(b) 0.0000328 = 3.28 + 100,000 = 3.28 X 10-6 •
I
I
Note that the power of 10 is the number of places which
the decimal point would have to be moved, from a position
just after the first nonzero digit, to its actual position. It
is plus if the move is to the right, minus if the move is to ,
the left.
i'
40.
42.
43.
44.
36,000,000.
41. 0.00000017.
The number of seconds in a year.
0.000058 (wave length, in centimeters, of yellow light).
The speed of light is 186,000 miles per second. A light-year
is the distance that light travels in one year. Find, to the
three significant figures (see § 14.7), the number of miles
to the North Star, which is 1160 light-years away, expressing
the result in scientific notation.
45. An automobile wheel is 2.5 feet in diameter. Find, to four
significant figures, the number of revolutions it makes in a
trip of 5000 miles. Express the result in scientific notation.
Write the following in ordinary decimal notation:
46. 1.2 X 10-8 (closest approach, in centimeters, of atoms in
a hydrogen molecule).
47. 2.15 X 10-8 (closest approach, in centimeters, of atoms in
an iodine molecule).
Perform the following indicated operations and write the
results in scientific notation. It is suggested that each number
be written in scientific notation before the operations are
performed.
48.
49.
60.
51.
52.
63.
650,000,000 X 12,000,000.
/
/
0.00000082 X 0.00001l.
'.
38,000,000 X 0.00000061. "
750,000 + 0.0000015.
0.000072 + 240,000,000.
A fibrinogen molecule is a rod-shaped molecule 7 X 10-6
centimeters long and 3.5 X 10-7 centimeters in diameter.
I
\ '
LA WS OF RADICALS
§S.8)
87
Find the volume of this molecule. (A combination of these
protein molecules in blood creates clots, which stop bleeding.)
5.7. Rational and irrational numbers.
A rational number is a number which can be expressed
as the quotient of two integers. For example, the
numbers 3, -4, i, 0.3, 5.23 are rational numbers, since
they can be written, respectively, as the following ratios
of integers:
2
523
3
-4
3
100'
10'
3'
I'
l'
An irrational number is a real number which is not
rational. The numbers V3, ~, 7r are irrational numbers. * They cannot be expressed, except approximately,
as the ratio of two integers.
A number is a perfect nth power if it is the nth power
of a rational number. A perfect second power is called a
perfect square, a perfect third power is called a perfect cube.
If a number a is not the nth power of a rational number,
and {Ya is real, then {Ya is irrational * and is called a surd
of order n. A surd of order two is called a quadratic
surd, a surd of order three is called a cubic surd. Thus,
V3 is a quadratic surd, ~ is a cubic surd. Although
V25 is written with a radical sign, it is not a surd, since
V25 = 5.
5.S. Laws of radicals.
The following are the fundamental laws of radicals:
(va)n
van
a.
= a.
=
-\iva = Va.
nr= _n/L
vavb
=
_n / -
vab.
~=~.
• This can be proved, but a proof is not ghten in this book.
"
(1)
(2)
(3)
(4)
(5)
88
EXPONENTS AND RADICALS
[Ch. V
These formulas follow from the definition of an nth
root and from the laws of exponents. Some of them do
not hold without the restrictions assumed in the Note of
§ 5.3. For example, (2) does not hold if a is negative
and n is even.
5.9. Removing factors from and introducing them into
radicals.
A factor which is a perfect power may be removed from
the radicand as in the following illustrations:
e,
;
v'8
v'I6X3
=
v'4-2
= v16x 2
=
•
X
v4v2 = 2V2,
= vf42X2Vx = 4xVx,
v75a 2 b3 = v25a 2b'.l . 3b = v'5 2 a2 b2 v'3b
= 5abV3b,
~54,x4y5z6 = ~27x3y3Z6. 2xy2 = ~ 33x3y3z6~
t"
:
3xyz2~,
+ 2ab + b2)
= v2(a + b)2 = (a + b)v'2,
=
.;"dJ),
v{-+ 4ab + 2b
b'H},:}
2
= v2(a 2
under the assumption of the Note of § 5.3. (But see
§ 5.17.)
Conversely, a factor may be introduced into the
radicand, thus
5avbc =
V'25a2v'bC = v'25a 2bc.
5.10. Reducing the index of a radical.
The following examples illustrate how the index of a
radical can sometimes be reduced:
~ =
-\I(7a)2 = (7a)2/4 = (7a)1I2 =
{125 = {I5 2 = 5 2/ 6 = 5 113 = ~.
..../fa,
One advantage of fractional exponents is evident
from these reductions.
RATIONALIZING THE DENOMINATOR
§ 5.11]
89
EXERCISES V. C
Remove all perfect powers from the radicands:
1.
3.
6.
7.
9.
11.
13.
vma.
"''.~
.y242x 3y 2Z.
-y'9i19.
V64x·y6 Z 7.
v'<f09U9.
,,
-<I-24x- 24 .
v4a z - 8b z•
2. V§8b3.
4.
6. 06a16 .
~\
8. ~250XlO • "<.;,,
-\yO.016
J6.
p
10.
12. ~.
14. ~4b3 - 64a3b4.
0W.
Introduce all coefficients under the radical signs:
16.
17.
16. 5ab 2V21ib.
4Vab.
5a~Z V2abC.
18. lOy 3/ X •
3x "J 2y
c
19. (a - b)
20. (a
•
','\
'"
.~ ,"'
21. xny'X.
23. xVX2.
22. x n
24.
+ 3b) ~a2a+b
_ 9b 2'
0.
X2YX.
Reduce the indexes of the following radicals:
25.
V8fX6.
27.~.
26. ~.
28. ~.
10'
29.
30.
31. .zyaw.
33.~.
5.11.
. '"
32.~.
34.~.
Rationalizing the denominator.
When a radical occurs in the denominator, the denominator can be rationalized (that is, freed from radicals) as
shown in the examples below:
Example 1.
0 =0,
-1= _1 . _
v'3
0,0
3
or
1
3' 0.
EXPONENTS AND RADICALS
90
[Ch. V
The principal advantage of rationalizing the denominator is in finding the numerical value of the given expression. Thus V3 = 1.7321, and in example 1 it is easier
to evaluate 1.7321/3 than 1/1.7321. There is no advantage, however, if one is using a calculating machine or
logarithms.
;
I
Example 2.
We rationalize in a similar way when a fraction occurs
under a radical sign.
I'
Example 3.
J=~= f'
or
~0.·
When the denominator is aYb + f:Vd, we multiply
numerator and denominator of the fraction by aYb cYd; if the denominator is aYb - cVd we multiply by
aYb + cVd. This will eliminate the radicals in the
denominator, since the product of the sum and difference
of two numbers is equal to the difference of their squares.
Example 4.
1+0
'\1'3+V2
1+0 .'\1'3-0
'\1'3+0 '\1'3-0
_'\1'3-0+00-00
('\1'3)2 - (V2)2
'\1'3-0+0-2
=
=
3-2
v'3 - v'2 + V6 -
2.
EXERCISES V. 0
Rationalize the denominators of the following fractions or
remove the denominators from the radical signs:
/
EXERCISES
1
:(H:Yi,:/':;).;I:'
1. - .
o10
4.
7.
10.
13.
_31'<li:'
v 25 '
4r
,,~.
.i
9
v 81a B
12
V6 a
22.
'\It;,
,»
0
J.
6.
~~.
8.
~::.
9.
~.
50
J,
. r"
17.
~ 20.
+ v'6.
3+ V3
a + bye.
1
12. ~ -16a'
a
,qg
15. Vb + yc'
'\Iii'
i
a
Vb + c·
Vf3 - VIT.
V 13 + VII
23.
-dye
+ V2
(
V2 + 0' . •
1
24. 1 -
1
26.
10
.'
11. ~.
1
14. v'I5 +
yb +c
a
3
6.
'(in"
_5/-'
19. 6
-3 .
V6
"
r
16.
2.
91
;>'
25.
18.
0 + 2.
y7 - 2
21.
3V5 - 2V13.
+ aV13
7y5
avx
+ bVY.
Cvx +dyy
1
.
3+V5+0
, I
V2 - 0 - V5'
~i)
.,<1.'
Simplify:
2x...jXi'TI _ x(x 2
+ 2)
...jXi'TI.
27 .
3x(5x + 4) _ 5y3x2
y3x2 + 7
28.
(5x + 4)2
x2 + 3
29. Show that either of the expressions
1
+ Xl/B,
is rationalized if multiplied by the other, similarly for
Rationalize the denominators of the following fractions:
1
30. X 118 + Y113'
32.
149
.
3~-2~
l+a
31. 1
33.
+ a2/3 + a4!3'
1
a
-¢b
~.
b+ c
+7
•
EXPONENTS AND RADICALS
[Ch. V
Rationalize the numerators of the following expressions:
34.
V~-Vx.
36.
(x
+ h) 1/3
h
-
X 1/ 3
.
I
I
r' !
Find the decimal values of the following expressions,
to thousandths, first rationalizing the denominators:
1
36. _ d
v 5
39.
v'i7 - va
vT7 - 3 .
2
Jl 37.
40.
38.
y'6'
05 - ViO
V5 + V2 .
co~
Ii
'
V5 + 2.
V5-2
41. 2 v'IT - 3 V7.
5v'IT - 20
5.12. Reduction of radical expressions to simplest form.
An expression containing radicals is said to be in its
simplest form when
(i) no factor can be removed from the radicand,
(ii) no index can be reduced,
(iii) there are no fractions under a radical sign,
(iv) there are no radicals in a denominator.
,,hI):
:r
The radical VB violates (i). It can be reduced to
simplest form by the method of § 5.9.
The radical {125 violates (ii). It can be reduced to
simplest form by the method of § 5.10.
The expressions
1
-,
1
VB
violate (iv). .They can be reduced to simplest form by
rationalizing the denominators. (See § 5.11, examples 1
and 4 respectively.)
In many cases the so-called simplest form of a radical
expression is the easiest form to evaluate numerically.
However. this is not always the case.
,
/
§5.13]
ADDITION AND SUBTRACTION OF RADICALS
93
EXERCISES V. E
Reduce to simplest form:
C',
~
10.
~~~.
11.
13.
3 .
__
14.
16.
(36)-1/3.
~9a2b
19.
(~y/6 .
.~~;,
::S;
f
t.
~-a4
5b •
12.
4
.
0 - 0 ..
15.
2
4y
VIQX3Y'
30.
30 + 40
20 -
17. (54x)-2 / 3.
18. (8x- 7)-2!3.
20.
21.
(_D-2!s.
24.
a- 1/2
aI/2
G)-S!4.
~_
2a - b
22. 2al/2 _ (2b)1Ii'
1
25.
3. ~40a4b5c(
6. Y208x I2 y I6·.
9. {/729a 2bs.
2. ~54p3q5 .
5. V"121m 2n 6•
2 2
. 1 8 . {/361a b .
.1. V5Op3q6.
____:_4. V"289x2y4.
7. {/343a 3b s.
vx-vY
23. a 1/2
1
26.
VX+Vy
1
+b
11("
1
v'3 + 0.
27.
1
1
v3- 0
28. Ya n +2bn+3.
31. Ya3n+2b6n+3.
29. Ya 2n b3n .
32. ~.
30.
+ b-1/2
+ b1l2
~5 .
~~-~~
tya Snb 6n .
J.:
1.
t
33.~
5.13. Additio~ and subtraction of radicals.
Radicals c'an be combined into a single term by addition
or subtraction if and only if they have the same index
and the same radicand.
Examples.
30 +
(3 +
= (5 -
0' =
1)0
=
40,
5va - 2va
2)va 3va,
avx + bvx - cvx = (a + b- c) 0,
70 + 0 -
~0 =
(7 + 1 -
=
2)~5 =
60.
We cannot combine into a, single term radicals such
and V3, ~r VZ and ~Z, or v'5 and ~7.
v'2
8.8
EXPONENTS AND RADICALS
94
[Ch. V
However, radicals can sometimes be reduced to forms
which can be added.
I
t! i
EXERCISES V. F
Simplify, and combine like terms:
1. 73V17 - 84V17
3. 27 V3 - 307.
+ 2v'f7.
""-
2. 18ya + 240 - 70.
4. 50+40-90.
8. ~/3000 - ~81 o. v'27 + V75 - -02.
4
!I _1M
{2a
{7a
7. V3 - '\j3 + v 27.
8. '\j 7b - '\j 2b'
9. 6{/125 + 5025 + 4V'25 - 2VI25.
10. y'64 - 04 + .,y64 + ~ - {/64.
11. (0.6)0.6 + {"O.216 + {I164025.
{2
2
3/2 2
12. '\js - VIO - '\}5 + -00'
13. V50 + v'<f.5 + _~ + {IS + {12500 ..
v O.5
14. v'729 + ~729 + .,y729 + {I729 + {lffi}.lk'o~
U
~.
.
f'
~
5.14. Multiplication of radicals.
Radicals having the same index can be 1ll'U1iiplied hI
§ 5.8, formula (4).
Example 1.
Example- 2.
20· 50 =
2.5.
0 .0
== 10y2I.
Note that this method cannot be used if the radicands
are negative.
95
EXERCISES
Radicals having different indices but the same radicand
can be multiplied by using fractional exponents.
Example 3.
.y'5{/5
= 5112 .5114 =
51n-H/4
= 53/ 4
=
{I5i
{/125.
=
Radicals having different indices and different radicands
can be multiplied by first reducing them to radicals having
the same index.
J
Example 4.
00 =
51/2. 41/3
=
/j'/6 . 42/6
= {l5 3 {142
,
,;:: {l5 3 . 4 2
=
{12000.
EXERCISES V. G
Perform the indicated operations and simplify results:
1.
4.
6.
8.
10.
12.
14.
16.
18.
'y'21.
3. ~. ~.
6. y3a' ~3a.
7. 5v'7x· 7 V5x.
j~. ~.
9. ~8a2. ~4a3.
3v'3a' 3a~.
11. v'3.~.
(i0)2.
13. 2~ . 4~2a2b2.
xv'yz· YVxz . zv'XY.
16. 0· V'3 . {/4 . V'5.
(0 + -0)(0 - ~). 17. (0 + ~4) (V2 - '\Y4)
(0 - 0)2. 19. (i) 3/2(-t) 4/3.
20. (i)_2/3(!)_3/\
21.
22.
(0 -
23.
24.
26.
V7'
v'6'
v'19.
~.
~. y'2X2y.
v'I4'
(3V6 - 20)(4V6 + 30).
~6)2.
(x _3 + ;0)(x _3 - ;~.
(x - 7+2Y!)(x - 7-2V~'
(x - 4+;Y!)(x _4-~.
-b
26.
2.
+ v'b
2a
2
-
4ac. -b - v'b 2 - 4ac.
2a
EXPONENTS AND RADICALS
96
27. Find the value of x 2
[Ch. V
+ 2 when x = 3 + 0.
5 - V6
20x + 19 when x = --2---
6x
-
28. Find the value of 4x 2
-
+ 12x + 12 when x = 0 - 2-0.
-b + v'b 2 - 4ac
Find the value ofax 2 + bx + c when x =
2a
;
29. Find the value of x 3
,~O.
';1
..
-b - v'b 2
2a
when x =
-
4ac
i
I. :-;-:---
•
5.15. Division of radicals.
Division of radicals and of radical expressions is Jsually
best effected by expressing the problem in fractio1l.i.l form
and rationalizing the denominator (if necessary).
Example 1.
v'i4
-
o
=
ff4-=-0 .,
7
',i:'.'
Example 2.
.8
1
Example 3.
(20 + 30) -;- (0 - 0)
~t
20 + 30 0 + 0
t
v'5 - v'2 v'5 + v'2
2·5 + 2Vlo + 3v'IO + 3 ·2
5 - 2
=
=
16
+ 35v'IO.
When the indices are different but the radicands are
the same, fractional exponents can be used to advantage.
Example 4.
112
V5
-0 = 55
1/;
= 5112-113 =
51/6
=
.6r,;5
vo.
IMAGINARY AND COMPLEX NUMBERS
§ 5.16]
97
The following example illustrates the case in which dividend and divisor have both different indices and different
radicands.
,\ Example 5.
v'6 _ 6 112
o -
_
6 3/ 6
_
51/ 8 -
52/ 6
-
6[63 _
"J 52 -
~6a. 54 _ {!135000
52 • 54 5
.
EXERCISES V. H
Perform the indicated operations and simplify results:
1. vID -74. y84 -77. ~17 -710. VI5 -7-
3[2
13.
16.
17.
19.
20.
21.
22.
24.
26.
26.
27.
VI3.
0.
VI7.
0.
13
"J"3 -7- V4'
2. V92 -7- Y23.
6. V"18 -7- 0.
8. yB -7- VITI.
11.0-7- 0.
14.
~~ -7- ~~.
3. ~198 -7- V"33.
6. V23 -7- V"23.
9. V"7 + 0.
12. V"12 -7- 0.
16.
ar -;- ~~.r
~E
10x2y~ -7- (5x2y~).
(~ -
VI4x) -7- y'2:j;. 18.
(y6 + yS) -7- (V6 - YS).
(3
+ 0)
-7- (2 -
0).
,y;
;~i!
(yB - 0) -;- (yll + Y7). ~klml;)!J ~.
~
(6y2 - y3) -;- (y'2 + 50).
--. ~:' Y
;:m:
(a + b.yx) -7- (c + dYx). 23. (Va + Vb) -7- (c + ~~
(Va + Vb) -7- (Va - Yb).
\:1
(a.yx + bvy) -7- (cVx + bYY).
H
(1 + 0) + G1 - 0).
,ni
(1 + 0) + fj1 + 0 + ~X2).
5.16. Imaginary and complex numbers. *
A negative number has no real square root, for no positive or negative number, when squared, will yield a negative number. Consequently, in order to solve certain
equations, for example, X2 + 4 = 0, it is necessary to
introduce a new kind of number, called an imaginary
number.
• For a more complete discussion of this topic see Chapter XII.
EXPONENTS ANP. RADICALS
98
[Ch. V
The imaginary unit is the number i having the property
,"" =
-1.
It is assumed that this new number obeys all of the laws
of addition and multiplication assumed for real numbers.
From the definition of i it is seen that we may write
v=I
:lj\
=
i,
v-3 =
v=4 = v'4v'-=1
vavCI = iva.
=
2i,
A number of the form a + bi in which a and b are real
numbers is a complex number. The number a is called
the real component, or real part, of the complex number;
the number b is called the imaginary component, or
imaginary part.
Two complex numbers a + hi and c + di are equal if
and only if a = c and b = d, that is, if and only if their
real components are equal and their imaginary components are equal.
If b ~ 0, the complex number is called an imaginary
number. Thus, 2 - 3i, 5i, and iv'3 are imaginary numbers. If b ~ and a = 0, the complex number reduces
to the form bi, which is called a pure imaginary number.
If neither a nor b is zero, a + hi may be called a mixed
imaginary number.
If b = 0, the complex number reduces to the real number a, for example, 7, - v'5, 0.
Thus, the system of complex numbers includes the real
numbers and the pure imaginary numbers as special cases.
The numbers a + bi and a - bi are called conjugate
complex numbers, either being the conjugate of the oth.r;r.
°
5.1 7.
Operations with complex numbers.
We add or subtract complex numbers by adding or
subtracting their real parts and imaginary parts separately.
EXERCISES
99
Example 1.
(2
(2
+ 5i)
+ 5i)
+ (6 - 4'£") = (2 + 6) + (5 - 4)i = 8 + i,
- (6 - 4i) = (2 - 6) + (5 + 4)i = -4 + 9i.
Pure imaginary numbers are multiplied as are any
literal numbers and the product is simplified by making
use of the relation i2 = -1. A number such as V -5
must always be reduced to the form iV5 before multiplication, otherwise incorrect results may be obtained.
Thus, while vC5V'=6 = iv's . iv'6 = i 2 V30 = - v30,
the following is incorrect: -v=5V'=6 = v' -5( -6)
v'3O.
Example 2.
(3
+ 5'£") (6 -
7'£") = 18
= 18
+ 9i - 35i
+ 9i + 35 =
2
53
+ 9i.
To divide one complex number by another, write the
quotient as a fraction and then multiply numerator and
denominator by the conjugate of the denominator.
(3
Example 3.
+ (6 - 7i)
+ 5i)
'1
= 3
::
6 - 7'£
I
(3 + 5i)(6 + 7'£")
18 + 51i + 351'1
= (6 - 7i)(6 + 7i) =
36 - 49i 2 (J.
lS+51i-35
-17+51i
cU
= 36 + 49 = --:S;:;-:5~EXERCISES V. I f-
/'
Reduce to the form a
+ -v'=36.
+ v=a.
+ .y:::y.
~+
.y=s + ~ -S.
1. 5
4. 3
7. x
10.
R·
13 .
16. y-=2
+ {I- 5.
+ 5~
+ bi:
2. 8 - v'"=289'"
6. 7 - vi -S. 1)
8. p + yCq2.'
11. S -
g. ::.
14.
v=m.
3. 11 +
6. 18 + v'=98.
9. a - ...;=-ti2C.
12.
~-
v'=4 + ~.
R·
[Ch. V
EXPONENTS AND RADICALS
100
Find the sums and differences of the following numbers:
16. 16
+ 5i,
18 - 3i.
17. 71 - lli, 10 - Z4i.
yC3,3 - Y-6.
v=zg. 19.6va + V -12.
13 - v=t7, 14 + yCTS.
(20 + 3y2) + (5V6 + 6V5)i,
18.5+ Y-16, 3 +
20. V2 + v=-3,
21.
22.
(70 - 3V2)
+
(2V6 - 3V5)i.
Find the following, -products and -powers:
23.
26.
27.
29.
31.
33.
35.
36.
3'7.
39.
v=-av-s.
24.
26.
28.
30.
32.
34.
V"=2Y-6Y-15.
(V-TI)s.
~Y-7)4.
'-,
(2 + 3~)(7 + 6i).
(13 - 2i)(2 - 9i).
(4 - yC3)(5 + .y=3).
(9 - 2v=:tO)(1l - 6v=--rs).
(7 - 3i)'2.
3S.
(2 + 3i)3.
40.
41. ( -
~+f
.iy'
. to
,fl·
v!=14V=2I.!! i;
,/
y-13Vl3.
(- Y -3)3.
(yC5)s.y=15.
(4 - 5i)(8 - 9i).
(4 - iy5)(7 - 2i.y5).
(2 - iV3)'2.
(1 + i)4.
42. (-
~-
f .i)'.
Find the following quotients:
43.
45.
l7.
49.
01.
52.
53.
64.
55.
67.
58.
v'-TOO -;- v'=25.
44. V-3O -:- -v=6.
46 . .y17 -:- -yCI7.
48. (8 - 3i) -:- (6 + IH).
60. (9 - i) -:- (13 + 2i).
v=--rs -:- y'I5.
(6 + 5i) + (2 + 7'/,).
(3 - 8i) + (10 - 9t).
(5 - i.y3) -:- (2 + h/i).
(40 + 3V7 . i) -:- (70 + 40· i).
(y6 + iVU) -:- ( 0 + iv'2i).
(4 + 3iy'I5) + (8 + 7iv'fO).
2 -:- (9 + 7i).
56. 5i -:- ~6 - 1h).
Find the value of ~2 - 4x + 13 when x = 2 + 3i.
Find the value of x 2 - lOx + 32 when x = 5 - iV7.
.
59. Fmd the value of 2x 2
60. Find the value of x 2
-
-
18x
+ 43 when x
2V3 . x + 5 when x
=
=
9
+2iV5
V3 - iV2.
,
I
!
CHAPTER VI
Quadratic Equations
~,!, •• "
6.1. Quadratic equations. ,,,,,, , \;"
('I " "
A quadratic equation (in one unknown) is one in
which the highest power of the unknown occurring is
two. The first-degree term and the constant term may
or may not be present. The following are quadratic
equations:
3x 2
-
2x
+ 4 = 0,
X2
+5
= 0, 2x 2 + 3x = 0,
2X2
= 0.
6.2. Solution by Factoring.
"
Frequently a quadratic :equation)cal{1be solved by
factoring.
Example 1.
Solve x 2
-
5x
+6 =
O.
SOLUTION. In factored form, the equation is
(x - 2) (x - 3)
=
o.
The product is zero if and only if one of the factors is zero.
Therefore, all the roots can be found by equating each factor in
turn to zero. Thus,
x - 2 = 0,
x - 3 = 0,
CHECK.
22 - 5 . 2
32 - 5 . 3
+6 = 4 +6 = 9 101
x = 2.
x = 3.
10
15
+ 6 = O.
+ 6 = O.
~
I
[Ch. VI
QUADRATIC EQUATIONS
102
Example 2.
+ 2x
x(3x + 2)
Solve
3X2
SOLUTION.
=
o.
=
O.
+2 =
0,
:t
3x
= O.
x
= -
2
3'
The roots are 0 and -i, as it can be readily verified W~
numbers satisfy the equation.
6.3. First-degree term missing.
If the first-degree term is missing the equation can easily
be solved as in the following examples. . ~,
Example 1.
Solve
3 = O.
4X2 -
SOLUTION.
4X2 =
3.
x2 = 3
_.
4
X=
(The sign
(or_
+ 1v- /93)
- y'3
2
-+ '\j@.=
4 +
2 .:>.
± is read
Ii
plus or minus."
See § 5.3.) .. d,
Example 2.
Solve
4x 2
+3
=
O.
-3.
3
SOLUTION.
-4'
X=
+
F3
- '\jT
=
± vI-3
2
=
+
-
iV3.
2
6.4. Completing the square.
A quadratic equation can always be solved by completing the square.
SOLUTION BY FORMULA
§6.S]
103
Example.
3x2
Solve
2x - 4 = O.
-
SOLUTION.
_- -Transpose the constant term:
3x 2
2x = 4.
-
Divide both sides by the coefficient of x 2 :
x2
-
2
3
4
3
·l
- X = -.
To complete the square of the left side, add the square of half the
coefficient of x, namely, ( -
~y =~.
be added to the right side.
We get
x2
~'.t
2
-
-
3
X
+ -1
9
This must of course also
1
= -
9
+ 4-,
3
(x _~y = I;.
flBH!i-:
Take the square root of both sides:
1
or
X -
or
x
Vi3 :).,\,
3 = - ---3'
1
Vi3
=:3 -
-3-'
!3 -+ v'I3
3
== 1 ±
We usually write
X -
1
03
3- = -+ - 3- ,-
6.5. Solution
x =
03.
3
by Formula.
The quadratic equation
+ bx + c =
(a ~ 0)
(1)
0
is representative of all quadratic equations. For by giving the proper values to a, b, c we can reproduce any
quadratic equation whatever. Thus, (1) becomes 3x 2 2x - 4 = 0 if we set a = 3, b = .-2, c = -4. We can,
ax"
104
[Ch. VI
OUADRA TIC EOUA TlONS
however, solve (1) by completing the square, and obtain
a formula by which any quadratic equation may be solved
by mere substitution.
+ bx
X2 + ~x
a
Transpose c:
ax 2
Divide by a:
=
-c.
=
c
a
-:
".,
'n;'.'~ ~l'J~.,'\, .
L
,t',
1
Add to both sides the square of half the toe.ftMiea$- 01 ~
b
namely, (~~y =
= 4
(:aY
;2:
b2
2
+ -ba X + -4ab
X2
(""l'i
2
.!)2
+ 2a
X
(
C
= -4a 2 - -.
a
b
=
2
4ac.
4a 2
Take the squar~ root of both sides:
x
Subtract
+
b
2a
+vb 2
=
4ac
-
2a
-
.
t ;':',
:a from both sides:
b
x = - 2a
-b
or
::
JLn f1
,
'" i: ."~'
±
vb 2
+ vb2
x = - - - 2a
4ac
-
2a
I
<,-
't-. '.
4ac
-
(2)
That the two numbers
-b
+ vb? 2a
4ac
XI =
-b -
vb2
2a
-
4ac
(3)
are roots of the quadratic equation (1) can be verified by
actual substitution.
EXERCISE
Verify, by actual substitution, that
are roots of (1).
Xl
and
X2,
given by (3),
EXERCISES
Example 1.
Solve the equation 3x 'l.
SOLUTION. a = 3,
-
2x - 4
= O.
b = -2,
c = -4.
± V(_2)2
- 4·3· (-4)
SUbstitute in (2):
_ -(-2)
x -
2. 3
2
=
==
±
V4
6
+ 48
=
2
± v'52
6
2 ± 2v'f3
1 ± vIf3
6
=
3
.
Example 2.
Solve the equation kx 2
+ 3kx
- 2x
+4=
k.
SOLUTION. Rearrange the equation in the form
kx 2
+ (3k
a == k,
Then
- 2)x
= 3k -
b
+ (4 -
k)
= O.
c = 4 - k.
2,
Substitute in formula (2):.
10 m.!;
-3k
+ 2 ± V(3k
-3k
+2±
x =
- 2)2 - 4k(4 - k)
2k
v'-::-:13=k-;:-2---2""'So-:"k-+--'-4
2k
EXERCISES VI. A
Solve:
1. x 2
3. x 2
5. x 2
7.
9.
11.
13.
15.
17.
-
12x
+ 35
+ x-56
-
=
= O.
O.
121 = O.
+ 289 = O.
x 2 + 4x + 1 = O.
x2
+ 2x - 7 = O.
x 2 - 3x + 10 = O.
2x 2 - x-I = O.
15x2 - 2x - S = O.
x2
2. x 2
4. x 2
6.
8.
10.
12.
14.
16.
18.
3x - 40 = O.
13x + 36 = o.
x 2 - 121x = O.
4X2 + 7 = O.
x 2 - 6x - 8 = O.
x 2 - 3x - 10 = G.
x 2 + x + 1 = O.
3x 2 - llx + 10 = O.
12x 2 - 73x + 6 = O.
-
-
QUADRATIC EQUATIONS
19. 9x 2 + 42x + 49 = O.
21. x 2 - (r + 8)X + rs = O.
23. 3x 2 + 4x + 5 = O.
26. 4x 2 + 4x + 1 = O.
.~;
21. 2x 2 + 5x + 2 = O.
29. x 2 - 0.7x + 0.12 = O.
31. 300x 2 - 750x + 625 = O;:~
33. 2x 2 + 3x + 4 = kx.
35. x - 16 = 6.
x
37
1 + 1
x +2
x - 2 =
x - 2
3x - 4
39. 2x - 3 = 4x - 5'
.
2
3'
20.
22.
24.
26.
28.
30.
32.
34.
+
[Ch. VI
16x 2
88x + 121 = O.
x 2 + px
q = O.
6x 2 - lOx - 7 = O.
5x 2 - X
2 = O.
21x 2 - 58x + 21 = O.
x 2 0.5x - 0.84 - O.
2X2
3x + 4 = k.
2X2
3x + 4 = kx 2•
+
+
!
I
I:
+
+
+
36. :: - ~ = x-a.
a
x
38 _1_ _ _
1_ = !.
.x- 3 x+5 6
40 x + 4 _ 5x - 4.
"-"
x - 2 - 3x - 8
+:
41. x + 11 + x - 22 = 3.
42. 4 5x 3 - 3x
= 6. ,i
xx+
xx43. The length of a rectangle is 3 inches greater than its width.
Its area is 70 square inches. What are its dimensions?
44. The diagonal of a rectangle is 2 inches greater than its
length and 9 inches greater than its width. Find the
dimensions of the rectangle.
•
46. Find two consecutive positive .integers whose product is
1332.
~, /
46. Find two consecutive positive even integers the sum of
whose squares is 580.
47. Find two consecutive positive odd integers whose product
is 783.
48. One number is 10 more than another. Their product is 299.
What are the numbers?
49. The outside dimensions of a picture frame are 24 inches
and 36 inches. The area of the frame is three-fifths of the
area of the picture which it encloses. Find the width of
the frame.
60. A tray with square base and a volume of 648 cubic inches
is to be constructed from a square sheet of tin by cutting
2-inch squares from the corners and bending up the sides.
What size of sheet of tin should be used?
61. A rectangular sheet of cardboard is twice as long as it is
wide. An open box having a capacity of 168 cubic inches
EXERCISES
G2.
1)3.
G4.
GG.
GS.
G7.
liS.
G9.
60.
107
is made by cutting 3-inch squares from the corners and
bending up the sides and ends. Find the dimensions of
the cardboard.
A party of fishermen rented a boat for $240. Two of the
men had to withdraw from the party and, as a result, the
share of each of the others was increased by $10. How
many were in the original party?
A man drove 156 miles at a constant rate. If he had driven
5 miles per hour faster he would have made the trip in
12 minutes less time. How fast did he drive?
An airplane flying north at the rate of 240 miles per hour
passed over a flying field at noon. A second plane flying
east at 200 miles per hour passed over the same field 5
minutes later. When were they 250 miles apart?
A grocer paid $360 for some crates of fruit. Four crates
spoiled and had to be thrown away, but he sold the rest
at an increase in price of $2 a crate, gaining $88. How
many crates did he buy?
A speculator sold some shares of stock for $2880. Several
days later, the stock having dropped in price $2 per share,
he repurchased, for the same amount of money, 6 more
shares than he sold. How many shares did he sell?
A tank can be filled in 12 minutes by two pipes running
simultaneously. By the larger pipe it can be filled in 10
minutes less time than by the smaller. Find the time
required for each to fill it.
A boy is at a point A on a straight river. He can row downstream at a rate which is 2 miles an hour faster than he can
run. If he rows downstream to a point B, which is 4 miles
below A, and runs, in a direction perpendicular to the river,
to a point C, 3 miles from B, it takes him 10 minutes longer
than it does to run directly from A to C. How fast can he
run?
A circular flower bed is bordered by a 3-foot walk. The
area of the walk is t of the area of the bed. What is the
radius of the bed?
A rod which weighs 2 pounds per foot can be made to
balance at a point 2 feet from one end by attaching a
weight of 6 pounds to that end. . How long is the rod?
[Ch. VI
QUADRATIC EQUATIONS
108
61. A motorcycle messenger left the rear of a motorized troop
7 miles long and rode to the front of the troop, returning at
once to the rear. How far did he ride if the troop traveled
24 miles during this time and each traveled at a constant
rate?
In analytic geometry the standard form of the equation of
a circle is
(x - h)2 + (y - k)2 = r2,
in which hand k are the coordinates of the center and r is the i
radius. The quantities hand k may be positive or negative, !
but r is restricted to positive values. By completing
squares, reduce each of the following equations of circles to
the foregoing form and give the coordinates of the center,
also the radius:
U. x 2
+ y2 -
6x
SOLUTION.
+ 4y -
3 = O.
X2 - 6x
(x - 3)2
+ 9 + y2 + 4y + 4 = 3 + 9 + 4,
+
(y
+ 2)2
= 16.
Coordinates of center are (3,-2), radius =
v'I6 = 4.
+ y2 - lOx - 6y + IB = O.
+ y2 + 4x - 12y + 15 = O.
x + y2 + 3x + 5y + 8 = O.
x + y2 - 14x + 2y + 1 = o.
2X2 + 2y2 - 2x + 2y - 17 = O.
2
63. x
x2
64.
66.
66.
67.
2
2
1\
In integral calculus it is frequently desirable to change
an expression of the form vax 2 + bx + c, a > 0, into the
form vaV(x - h)2 + k or an expression of the form
VC + bx - ax 2, a > 0, into the form -vaVk - (x - h)2.
Transform each of the following expressions into the appropriate one of these forms:
68.
70.
72.
74.
76.
78.
V2X2
V2X2
V6x 2
+ 4x + 10.
+ 2x + 7.
+ 36x.
y2 - 6x - x 2 •
y6 - 3x - 3x 2•
y7 - 3x - 5x 2•
69. y3x 2 - 18x + 12.
71. V5x 2 + 40x - 10.
73. y7x 2 - 42x + 77.
76. y6 + 8x - 4x 2 •
77. ylOx - 2x2 •
79. y-8x - 2x 2•
I
EXERCISES
109
6.6. Equations in quadratic form.
An equation is in quadratic form if it is a quadratic in
some function of the original unknown. Thus,
+ 5x 36 =
2x + 3Vx - 5 =
0 is a quadratic in x2,
0 is a quadratic in Vx.
2 -
X4
Example 1.
Solve
+x
X4
20 =
o.
+ 5)
=
2 -
SOLUTION. Factor:
(x 2
4)(x 2
-
o.
Set each factor separately equal to zero, getting
±2.
x =
x = ± v'=5
= ±iV5.
r
Example 2.
x2
Solve
x2
SOLUTION.
-
x2
5
+3 +x
+3
2 -
5 = -2.
,(
~
: ,,_ 'f
x2
".', + :t.~
5
x 2 + 3 = y.
Then
.1(:
! 'I
.~'
1/1
-
\ .lIt{
.as
-
Y
+!y =
-2 ,
y2
+1=
-2y,
0,
+ 2y + 1 =
+
y = -1, or
X2 = 1,
x = ±1.
Xl X2
5
+3 =
-1.
Note that the two fractions in the equation are reciprocals,
and that we could equally well have represented the second,
instead of the first, by y.
EXERCISES VI. B
Solve the following equations. If an equation involves
radicals or fractional exponents all answers must be tested by
substitution in the original equation.
1.
3.
6.
7.
9.
11.
13.
15.
17.
19.
20
X4 - 11x2 + 18 = O.
X4 + 19x 2 + 48 = O.
12x4 - 31x 2 + 9 = O.
x + 2VX - 15 = O.
X2/3 - 3 X1!3 - 18 = O.
X2/3 - 2Xl/3 - 6 = O.
6X 2/ 3 + ll x 1!3 - 10 = O.
x4!8 - 13x2/8 + 36 = O.
6Xl/4 - Xl/2 - 9 = O.
5)
•
5- 18
(X ~ 2 + X~
X2 + 2 + x - 6 +
• x - 6
X2 + 2
21. X2 _ 2
22
[Ch. VI
QUADRATIC EQUATIONS
110
~X
+
2.
4.
6.
8.
10.
12.
14.
16.
18.
X4 - 2x 2 - 35 = O.
x 6 - 7x 3 - 8 = O.
6X4 - 23x2 - 225 = O.
(
x - 10y'X + 21 = O.
X2/3 + 12x1!3 + 35 = O.
X2/3 - Xl/3
4 = O.
9X2/3 - 33 x 1!3 + 28 = O. ---' i
X4/3 - 5X 2/ 3 - 6 = O.
1;1/3 + X1!6 - 6 = O.
+ 10 + '-J["'X=6
_
X+I6
+
: :~:
= O.
13 = 0 ;"
x2 + 2
6
.
X2 - 2
170
X2 + 2 - 77 = O.
x - 6
J,
J
,(,;
.;';L'J-;",,'·j;';;
',
L \.
I
34 = 0
15
.
_ ~X - 5 _ ~ = O.
x-5
x+7
2
24. (X2 + X - 2)2 - 5(X2 + X - 2) - 50 = O.
26. (X2 - 2x + 3)2 - 5(x 2 - 2x + 3) + 6 = O.
23.
~X + 7
6.7. Equations involving radicals.
An equation involving radicals can sometimes be solved
by the method of the preceding section. Another method
of solution is illustrated in the following examples.
Example 1.
Solve
x
+ Vi -
6 = O.
;SOLUTION.
,t,
Vi:
6 = -vx.~"
X2 - 12x + 36 = x,
' .,
x 2 - 13x + 36 = 0,
(x - 4) (x - 9) = 0,
X = 4,9.
The value 4 checks, since
. --~ ..,
Transpose
Square:
4
X -
+ y'4 -
6 = 4' + 2 ~ 6 = O.
/"
,,;,/..;
bJ)' •
EQUATIONS INVOLVING RADICALS
§ 6.7]
111
The value 9 however does not check, sinc{l
9
+ v'9 -
6 = 9
+3 -
(3 ~ 0.
This is not because we have made an error, but because by squaring both sides of the equation, we have introduced an extraneous
root. (See § 4.8.) The value 9 must be discarded as a possible
root, since it does not satisfy the original equation. Thus, the
only root of the equation is x = 4.
In dealing with equations involving rCldicals, all solutions
obtained must be tested, as some may be extraneous. In this
connection it should be recalled that ...IX, when x is positive, means the positive square root of x only.
Example 1 can be solved by the method of the preceding sec\)'\)"Il, 'il.'O ~\)\\\)'W'i'I'.
x
+0 -
(0 + 3)(0 Since
0
cann<1-, be negative,
0-2
=
0,
6 = 0,
2) = 0.
0 + 3 cl1nnot be zero; hence
0=2, x = 4.
This method is preferable here.
Example 2.
.y3x
Solve
+7+
VX+1 - 2 = 0.
SOLUTION. Transpose, keeping one radical alone on one side
of the equation. (If there is any difference in the complexity of
the radicals always isolate the most complicated.)
Square:
.y3x
3x
+7
+7
= 2 - VX+1.
= 4 - 4.yx + 1 -t- x
+ 1.
Collect terms, keeping the radical alone oIl one side: \ to J§:!loS
Divide by 2:
Square again:
4VX+1
2VX+1
4x
4
x 2 - 2x - 3
(x - 3) (x
1)
x
+
+
= -2x - 2.
=
=
=
=
=
-x - 1.
x 2 + 2x + 1,
0,
0,
3, -1.
----~
-
QUADRATIC EQUATIONS
112
[Ch. VI
The value a does not satisfy the equation and must be discarded
as a root.
Example 3.
Vx+I +3
Solve
=0.
DISCUSSION. We could proceed as above, obtaining values for
x. This is useless, however, for both terms are positive, namely, _
the positive root of x + 1 and the positive number 3, and it is
impossible for their sum to equal zero.
Example 4.
Solve
6x 2
+
lOx
+
v'3x 2
+ 5x + 1 = -
1.
nJ;ftJf] 1
SOLUTION BY REDUCING TO QUADRATIC FORM. Add,2 to both
gjdes;
f~l~{tr;i.-.r~"
6x 2 + lOx + 2 + v'3x 2 + 5x + 1
2(3x 2 + 5x + 1) + v'3x 2 + 5x + 1
Let .y3x 2 + 5x + 1 = y:
2y2 + y - 1
(2y - 1)(y + 1)
= 1,
= 1.
= 0,
= 0,
y =
1
"2'
-1.
Since a positive radical cann,ot equal -1, we need only to consider the equation in which the right side is t. Squaring thi~,
we get
1
3x 2 + 5x + 1 = 4'
12x 2
+ 20x + 3
x
= 0,
1
=
3
-"6' - '2'
Both of these values satisfy the original equation.
j
EXERCISES VI. C
Solve:
1. v'X=2 - 3 = O.
3. v'4x + 15 - 2x = O.
2. v'5x + 4 - 13 = O.
4. v'5x _:_ 4 + 2x - 12 = O.
EXERCISES
6. y5 - x =' 2x.
6.
7. y'2X-g + Vx - 6 = O.
8. y2x + 4 - ~ - 1 = O.
9. y2x - vX+l - 1 = O.
- 10.
11.
12.
113
vx=3 + 2v'X=5 =
Y x + 2 y'2x + 3 - 3 = O.
Yx + y;; + 2 - -v.x=-1 = 1.
Y x + 4x + 6 + Yx 4x + 6
2
2
13. Y2x - y'x - 4 -
-
VXTI
= 5.
= O.
14. YX+2 ~ y'x - 3 - ~ = O.
15. y3x + 4 - vx=--3 - V2X+1 = O.
16.
Y x 2 + 2x
-
17. 2.q37 - x 3
O.
2yx2 - 2x + 10 + 30 = O.
x
=
'I
'I
O. '-
18. One side of a rectangle is 12 inches. If the other side is
decreased by 4 inches the diagonal will be decreased by 2
inches. Find the length of the other side.
19. One ladder is 7 fe~t longer than another. Each is placed
on level ground with its lower end 8 feet from the foot of a
vertical wall. The longer ladder reaches 9 feet farther up
the wall. How long is it?
20. Two ladders are placed against a straight vertical wall.
The upper end of each is 24 feet above the level ground.
One ladder is 1 foot longer than the other. The lower end
of the longer ladder is 3 feet farther from the foot of the
wall than is the lower end of the other ladder. Find the
lengths of the ladders.
21. Two straight wires are attached at a point 24 feet above the
base of a vertical pole standing on level ground. One wire
is 5 feet longer than the other and reaches a point on the
ground 11 feet farther from the base. Find the length of
each.
22. The radius of the base of each of two right circular cones is
8 inches. The altitude of the first cone is 9 inches greater
than that of the second cone and its slant height is 7 inehes
greater than that of the second. Find the altitude of the
taller cone.
QUADRATIC EQUATIONS
114
[Ch. VI
23. A man is at a certain point on the east bank of a straight
river which flows south. Some distance below him is a
dock, and 8 miles directly east of the dock is a farmhouse.
He can row 6 miles an hour downstream and he can walk
4 miles an hour. If he goes by boat to the dock and then
walks to the farmhouse it will take him 15 minutes longer
than if walks directly to the farmhouse. How far is he
from the dock?
6.8. Character of the roots.
It will be recalled that the two roots of the quadratic
equation ax2 + bx + c = 0 are
Xl
=
+ v'b
-b
2
-
4ac
2a
-b - v'b 2
2a
,
-
4ac
(1)
The eXJ)ression
b" - 4ac,
appearing under the radical sign, is called the discriminant of the equation l}X 2 + bx + c = O.
If a, b, c are real numbers, then the discriminant gives
us the following information regarding the character of
the roots:
If b 2
-
4ac is
{i;}-
the roots are
{~:~~ ~:~ ~;:~~al.
'tmagmary.
If abc are rational, * the roots are rational when b 2 4ac is a perfect square (including zero) i otherwise they are
irrational.
If the roots are imaginary they are of the form p
qi and p - qi, that is, they are conjugate imaginary
quantities.
Two eqTIal roots are often called a double root.
The examples tabulated on page 115 illustrate the use
of the discriminant in determining, without solving the
quadratic equation, the character of its roots. The final
+
* See
§ 5.7.
EXERCISES
115
column is unnecessary, but is included here for purposes of
verification.
If the constant term in a quadratic equation is missing,
one root of the equation is zero. Thus, the equation
2X2 - 3x = 0 has one root o.
(The other root is 3/2.)
If both the constant term and the first-degree term are
missing, both roots of the quadratic equation are zero; this
is, of course, a trivial equation. Thus, the roots of 2X2 =
o are 0 and o.
~-----------.------------~----------~--------
Discriminant
b2 - 4ac
Equation
4x - 7 = 0
Roots are
real and unequal
(-4)2 - 4·3· (-7) (also rational,
= 16 + 84 = 100 since 100 is a
perfect square)
3x 2
-
2Xl
+ 2x + 5 =
0 22 - 4 . 2 . 5
imaginary
= 4 - 40 = -36
4x 2
+ 12x + \} =
0 (12)2 - 4 . 4 . 9
= 144 - 144
=
real and equal
(also rational,
0 of course)
Values of
roots
t,
-1
-t + ti,
-~ - ti
-1-, -t
If the first-degree term is missing, but the constant term
is present, one root is the negative of the other. Thus,
the roots of 2x 2 - 3 = 0 are x = V372 and x = - V372;
the roots of x 2 + 4- = 0 are x = 2i and x = -2i.
EXERCISES VI. 0
Determine the character of the roots of the following equations without solving:
1. x 2 - 7x - 12 = o.
2. x 2 - 7x + 12 = O.
3. 9x 2 + 42x + 49 = o.
4. 2X2 - 2x + 3 = o.
6. 14x2 - 22x + 13 = o.
6. 7x 2 - 8x - 10 = o.
7. 6x 2 + 19x + 15 = o.
8. 4x 2 + 20x + 25 = o.
9. 8x 2 - 7x - 9 = o.
10. 12x2 - 19x - 18 = o.
11. 289x 2 + 306x + 81 = o.
12. 2166x2 - 1923x + 2087 = o.
QUADRATIC EQUATIONS
116
[Ch. VI
Determine the values of k which will make the roots of the
following equations equal:
13. x 2 + kx
+k
SOLUTION.
= O.
b2
-
:;
4ac = k 2
k
-
ti
= 0,
= 0,4.
If k = 0 the equation reduces io x 2 = 0, which has both
roots zero. If k = 4 the equation reduces to x 2 + 4x + 4 = 0,
both of whose roots are x = - 2.
14.
16.
18.
20.
21.
22.
23.
24.
+
kx 2 + 2x - 3 = O.
15. 16x 2 - 8x
3 = k.
2
2x - 7x + 8 = kx.
17. 16x 2 - 8x + 3 = kx 2•
kx 2 + 3x + k = O.
19. kx 2 + kx + 7 = O.
'!
2
2
(k + 3)x + (x - 3)k = O.
(x - 1)2k - (k - 1)2x = O.
(x - 3)(3 - x)
(x - 3)(3 - k)
(k - 3)(3 - k) = O.
(x + l)(x + 2) - (k + 2)(x + 2) = O.
(x + l)(x - 2) + (k - 2)(x + 1) + (k - l)(x + 2) = o.
+
+
Determine k in each of the following equations so that one
root of the equation will be zero.
3x 2 - 7x + 11 = k.
26. (x + 2)2 - (k
x2
k 2 - 2(x 2 - 1)
3(x - k) = O.
(kx - 2)2
3(x - 2)2 - (k - 2)2 = O.
(kx - 3)2 - (k - 3)(x - 3)2 = O.
x-I
k+2
30· k _ 1 x + 2 =1.
25.
27.
28.
29.
+
+
+
+ 3)2
= O.
+
Determine k so that one root of each of the following equations
will be the negative of the other.
31. x 2 + k 2 + X + k + kx = O.
32. (x
k)2 + 3(x + k) + 7 = O.
+ 5x _ kx _ ~ ~ 0
33. kX2
5
k
5
k
.
2
x - 4
x + k
x - k
34. k2 _ 4 = k - 2 + k + 2'
+
35. (kx
36. (7x
+ 3)2 + (x + 2k)2 + 3x + 2k =
+ k)2 - 7(7k + x) + tkx = O.
-
01
. ),
SUM AND PRODUCT OF THE ROOTS
§6.91
117
37. Given the equation
(x - 1)2
+ 2(k
l)(x
-
+ 1) + (k + 1)2 = O.
Determine k so that (a) it will have equal roots, (b) one
root will be zero, (c) one root will be the negative of the other.
6.9. Sum and product of the roots.
If we write r for the radical vb 2
(1) of § 6.8 reduce to
Xl =
4ac, then equations
-
-b + r
2a '
-b - r
2a
Then,
Xl
+
X2
X1 X 2
-b -t. r
-b - r
2b
b
- = - -,
2a
+ 2a = _2a
a
b2 - r2
b 2 - (b 2 - 4ac)
4ac
c
=
- a
2
4a
4a 2
4a 2
That is, in the quadratic equation ax2
sum uf the roots is -
~
+ bx + c =
and the product of the roots is
0, the
E.
a
a
It is perhaps simpler to divide the equation through by
the coefficient of x 2, reducing it to the form x 2 + px + q
= O. Then we can state that in the quadratic equation
+ px + q = 0, the sum of the roots is -p and the
product of the roots is q.
X2
Example.
Find the sum and the product of the roots of the equation
3x 2
4x - 7 =
-
o.
SOLUTION. Divide by 3:
x2
Sum of roots = - ( Product of roots = -
4
-
~)
~.
3'x -
=
~.
7
3'
=
0
[Ch. VI
QUADRATIC EQUATIONS
118
These results can easily be checked in the present example by
solving. The roots are i and -1.
6.10. Formation of an equation with given roots.
If Xl and X2 are the roots of a quadratic equation, then'
the equation can be written (x - Xl)(X - X2) = o. _\~::)_
Example 1.
Find an equation whose roots are 2 and -3.
SOLUTION.
Method 1. (x - 2) (x
x2
+X
+ 3)
= 0,
6
-
=
0.
, ::,
Method 2.
Sum of roots = 2 - 3 = -1 = -p, p = 1.
Product of roots = 2( -3) = -6 = q.
x2
+ px + q == x +
2
6
X -
=
Example 2.
I,
Find an equation whose roots are
~t.
SOLUTION.
Multiply by 3 . 2 to clear of fractions:
3
(x - ~) 2 (x +~) = 0,
(3x - 2)(2x
6x 2
-
+ 1)
X -
= 0,
2 = 0.
Example 3.
Find an equation whose roots are 2, -3, 7.
SOLUTION.
(x - 2) (x
x3
-
6x 2
+ 3) (x
-
13x
- 7) = 0,
= 0.-
+ 42
Note that this is not a quadratic equation.
0,.,
EXERCISES
119
Example 4.
Find an equation whose roots are
(X -
SOLUTION.
x2 -
(Xl
xI)(x -
and
X2.
X2) =
0,
Xl
+ X2)X + XIX2
=
o.
,.---
f'
,.'.,\")r
Compare this result with § 6.9.
Example 5.
,~I
Find an equation whose roots are 1
+ i and 1 -
i
"
, 1 · ',
i.
SOLUTION.
[X - (1 + i)J[x - (1 - i)] = 0,
[(x - 1) - tl[(x - 1) + i] = 0,
(x - 1)2 - i2 = 0,
x 2 - 2x + 2 == O.
(
Or, since (1
'tJ = 2, q
(1-
+ i) + (1
=
2.
+ tJ
- i) = 2, p = -2; and since (1
Thus Method 2 is simpler in this case.
EXERCISES VI. E
Find, without solving, the sum and the product of the roots
of following equations:
1.
3.
6.
7.
9.
11.
x 2 + 7x + 12 = O.
2. x 2 - 3x - 4 = O.
5x 2 - 22x - 17 = O.
4. 8x 2 - X + 22 = O.
7 - 8x - 9x 2 = O.
6. (5x + 4)2 = 18.
tx2 + tx + t = O.
8. iX2 + Ix - t = O.
32x 2 - 5x = O.
10. 32x 2 - 5 = O.
(x - 1)(x - 2) + (x - 2)(x - 3) + (x - 3)(x - 4) =
o.
Determine the value of k in the following equations:
12. x 2
+ kx +
18 = 0, given that one r09t is twice the other.
SOLUTION.
r
+ 2r
r'2r
Let r = one root; then 2r is the other.
= - k }
= 18;'
or
{3r = - k
r2 = 9, 'r
k = -3r =
13. x
2
= ±3.
+9.
+ kx + 5 = 0, given that one root is 4.
J
QUADRATIC EQUATIONS
120
I
[Ch. VI
f
SUGGESTION FOR ALTERNATIVE METHOD OF SOLUTION.
Since 4 is a root it may be substituted for x and the resulting equation solved for k.
14. 2x 2 - 9x + k = 0, given that one i'Oot is twice the other.
15. 3x 2 - lOx + k = 0, given that one root is 2 more than the
other.
16. 2x 2 - 9x + k = 0, given that one root is'3 more than twice
the other.
17. 4x 2 - 15x + k = 0, given that one root is the square of
the other.
18. 3x 2 - kx + 6 = 0, given that one root is 3.
19. 4x 2 + kx + 7 = 0, given that the difference between the
roots is 3.
20. 2x 2 + kx + 9 = 0, given that one root is twice the other.
21. 7x 2 + kx - 4 = 0, given that the quotient of the roots is -/>i.
22. ax 2 + bx + k = 0, given that the roots are reciprocals of
each other.
23. ax 2 + bx + k = 0, given that either root is the negative reciprocal pf the other.
Find the equations, quadratic if possible, and with integral
coefficients, whose roots are
24.
27.
30.
33.
36.
38.
40.
42.
44.
46.
7, 9.
0,35..
25. -4, 8.
28. t, -to
± 07.
31. ± 13i.
±3V5' i.
34. 6 ± 7i.
i(7 ± 30· 1j. .
37.
y'3 ± iV6.
39.
4, t, -j.
41.
't_.
3, 1 ± i.
43.
1,8 ± i-v3.
!('i;(
45.
0,5 ± 3.yf.
47.
26. 6, 6.
29. ±t.
32. ± iv15.·' :.'
35. 5 ± iva.,., ~.
{-(17 ± 20· i)
.. ".,
2, -5,6.
6, ± 5. .
5, 3 ± 0.
;,
2,3 ± 20 ,·i.:J 1
-2,5 ± 30
·t,,;,:rl
6.11. Factoring by solving a quadratic.
A quadratic expression with rational coefficients can be
factored into rational factors if and only if its discriminant
is a perfect square.
- !i
;
;.
121
EXERCISES
Example.
3x 2
Factor
SOLUTION. b2
X
-
-
-
4
4ac
-
4x - 7.
= 100. Solve the equation
3x 2 - 4x - 7 = o.
± y'ufT84 _
6
4
-
± 10
6
_ ~, -1
-3
.
I:.
The factors of the original expression are
3 (x -
D+
(x
1),
or
In general, the factors. ofax 2
(3x - 7)(x
+ bx + care ff .
a(x - Xl) (X -
where
Xl
and
X2
+ 1).
X2),
are given by (1) of § 6.8.
EXERCISES VI. F
Factor the following expressions by first solving the corr«*;
sponding equations:
1. 168x 2
3. 500x 2
+ 1l0x - 7623.
+ 4140x + 3577.
2. 128x 2
4. 646x 2
1896x
-
+ 9711x
+ 6075.
- 256.
Find the irrational or the imaginary factors of the "following
expressions:
6. x 2
-
2x - 1.
SOLUTION. Set x 2
Xl
= 1+
-
2x - 1 = 0 and solve, getting
y2,
X2 =
0.
1 -
The factors are
+
[x - (1
0)l(x - (1 - 0)],
(X - 1 - 0)(x - 1 + 0).
6.
8.
10.
12.
14.
16.
x2
x2
+ 36.
+ 6.
2x 2 - 3x + 7.
2:r: 2 - 3x + 5.
x 2 + 2Y3· x-I.
3x 2 - 40 . X + 2.
7. x 2
9. x 2
6.
-
-
11. 5x 2
13. 3x 2
15. x 2 -
17. 4x 2
or
X
-
-
-
+ 1.
3x - 4.
8x + 6.
20' x - 4.
60· x - 3.
[Ch. VI
QUADRATIC EQUATIONS
122
6.12. Graphic representation of a quadratic function.
If we have a quadratic function such as x 2 - 2x - 3, we
can set y = X2 - 2x - 3, find values of y corresponding
to assigned values of x (see accompanying table), and plot.
y = xl _ 2x _ 3 The graph of this function is shown in
Fig. 6.1. The curve is called a parabola.
x
Y
(The graph of any quadratic function is a
parabola.) The points at which the curve
-2
5
the x-axis, namely, x = -1 and
crosses
-1
0
x = 3, are the real zeros of the quadratic
-3
0
1
-4
function, or the real roots of the corre2
-3
sponding
equation, x 2 - 2x - 3 = O.
0
3
The graph of the function y = x 2 - 2x
5
4
+ 1 is shown in Fig. 6.2. The equation found by setting the function equal to zero has equal
roots (Le., a double root) x = 1, and it will be observed that
the curve merely touches the x-axis at the point x = 1.
y
y
y
"
I
I
I
\
\
X'
(
x
1
l\
~
y
I
FIG. 6.1.
Graph of quadratic
function x' - 2x - 3.
Roots of corresponding equation real and
unequal.
I
1\
.\
\
I
I
1/
1\
r
I'-
y
.
I
i\
_L
I
,/
\
1
I
1\
X r
FIG. 6.2.
Graph of quadratic
function x' - 2x + 1.
Roots of corresponding equation real and
equal.
X
y
I
FIG. 6.3.
Graph of quadratic
function x' - 2x + 2.
Roots of corresponding equation imaginary.
Fig. 6.3 shows the graph of y = x 2 - ~x + 2. The corresponding equation has imaginary roots x = 1 ± i, and
the curve does not meet the x-axis.
In general, the graph of a quadratic function crosses the
x-axis in two distinct points if the roots oj the equation
--c...
MAXIMUM OR MINIMUM VALUE
obtained by setting the function equal to zero are real and
unequal, touches the x-axis if they are real and equal, and does
not meet the x-axis at all if they are imaginary.
6.13. Maximum or minimum value of a quadratic Function.
The lowest point on the curve representing the quadratic function corresponds to the minimum, or least, value
of the function. If the curve is inverted from the positions shown in Figs. 6.1, 6.2, 6.3, as will be the case if the
coefficient of x 2 is negative, the highest point on the curve
corresponds to the maximum, or greatest, value of the
function. A method of finding the maximum or minimum value of a quadratic function by completing the
square is illustrated in the examples below.
The determination of maxima and minima is of great
practical importance.
Example 1.
Find the minimum value of 3x 2
SOLUTION.
Y = 3 (X2
- ~ x) + 4
2
'-:),{, = 3 ( x
=
+ 4.
2x
-
2
-
1)
~
3' x + 9 +
~)2 +
3(x _
1;.
" .'
1 ,.. ',,' 4
4 - ! ,+ -i
-.
-~,
I"
'~:~~r
JJJ
,,i,l1:;
,'i'}1
AUl
The term 3(x - i)2 is always positive or zero; consequently y
can never be less than ¥. It is equal to this minimum value
of ¥ when x = i.
Example 2.
Find the maximum value of - 5x 2
SOLUTION.
Y = -5x 2 - 4x + 7
-
4x
+ 7.
x) + 7
- 5 (X2 + ~ X+ _!) + 7 + ~
5
25
5
- 5 (X2
+~
39
5 + 5"
( + '2)2
-5 x
QUADRATIC EQUAliONS
124
tCh. VI
The term -5(x + i)2 is always negative or zero, and the maximum value of y is ~¥; y is equal to llr! when x = -to
EXERCISES VI. G
f
I
Draw the graphs of the following functions:
1. x 2 - 16.
3. x 2 - 5x.
6. x 2 + 4x.
7. x 2 - 2x - 8.
9. x 2 - X + 3.
11. 2X2 - 3x - 4.
13. 12 + 5x - 2x 2•
2. 4 - x 2 •
.4
4. 7x - x 2 •
I
6. x 2 + X - 6.
8. 5 - 4x - x 2•
10. iX2 - 5x - 7.
12. tx2 + x + 1.
14. 9 + 2x - tx 2.
1
'J
. .
- ~ 1.
Ai:
-~:__~
Find the maximum or the minimum value of each of the
following functions, and the corresponding value of x. State
whether maximum or minimum.
16.
17.
19.
21.
23.
26.
27.
29.
x 2 - 8x + 2.
x 2 + 4x + 11.
10 + 6x - x 2 •
2x 2 - 8x + 7.
9x - x 2•
tx2 + 2x - 3.
tx2 + tx + i.
x2
-
0 . x + 2.
1
16.
18.
20.
22.
'~24.
26.
28.
6x + 12.
2x - x 2 •
5x + 3.
4x - 12.
7x.
- 2x + 4.
x2 13 x2 3x 2 x2 +
tx2
t-
30. 6 -
1;X - iX2.
V3 . x -
20 . x 2•
31. A stone wall of indefinite length is available for one side of
a rectangular enclosure which is to be fenced in on the other
three sides. What is the maximum area that can be
enclosed by 400 feet of fence?
32. A man wishes to fence in a rectangular plot of ground, one
side of which is on the dividing line of his property. He
has $600 to spend for the fence, which costs $2 per foot.
He is to pay for the three sides of the enclosure which are
entirely on his own land and for half of that part which is
on the dividing line. What is the maximum area that he
can enclose?
33. A motion picture theater which had a daily average of 1500
paid admissions, decreased the admission price from 80 cents
to 75 cents. This resulted in an increase of 100 in the
/
EXERCISES
34.
36.
36.
37.
125
daily average of paid admissions. Assuming that for each
5-cent decrease in admission price the number of daily paid
admissions will increase by 100, find the price which will
afford maximum rereipts.
A manufacturer can ship now a load of 200 tons at a profit
of $15 a ton. He estimates, however, that by waiting he
can add 10 tons per week to the shipment, but that the
profit on all that he ships will be reduced by 50 cents per
ton per week. How long will it be to his advantage to wait?
A merchant can sell 100 articles per day at a price of $5
apiece. He can sell 5 less articles per day for each 25 cents
that he adds to the price. If he pays $3 apiece wholesale
for the articles, what retail price will yield him the greatest
profit?
Prove that the rectangle having a maximum area with
a fixed perimeter is a square.
Prove that if a positive number is separated into two factors
whose product is a maximum, the factors must be equal.
"
'\
128
SYSTEMS OF QUADRATICS
[Ch. VII
------------------------------------~-2
+ y2 + 6x + 4y - 12 = 0,
7x - y - 6 = O.
12. x 2 - y2 = 9,
x~ + y2 = 9,
x - y = 9.
x
y = 9.
14. x 2 - 2y - 3 = 0,
2X2 - 3y - 6 = 0,
x - y - 1 = O.
5x - 4y + 1 = o.
y2 - 12x - 8y - 8 = 0,
3x - y + 11 = o.
9x 2 + 16y2 + 18x - 32y - 119 = 0,
x - y - 3 = O.
x 2 + y2 + 6x - 2y - 15 = 0,
x - 7y + 35 = O.
3x 2 + 2y2 - 563 = 0,
10. x
11.
13.
15.
16.
17.
18.
+
2:t - 3y - 8 = O.
"
16y2
18x
32y - Hi - 0,
3x - 4y - 13 = O.
20. x 2
y2 - 14y - 120 = 0,
5x - 12y + 84 = O.
21. 7x 2
5xy
2x - 3y
2 = 0,
4x - 3y
14 = O.
22. 3x 2 2xy
y2
X
5y - 6 = 0,
x - y - 3 = O.
19. 9x 2
+
+
+
l:
I
+
+
+
+
+
+ + + +
23. The hypotenuse of a right triangle is 13 inches. Its perimeter is 30 inches. Find the length of its shortest side.
24. The area of a right triangle is 210 square inches. Its
perimeter is 70 inches. Find the length of its hypotenuse.
25. The total area of the surfaces of two cubes is 312 square
inches, the total length of their edges is 120 inches. Find
the length of the edge of each.
26. The difference of the squares of the digits of a two-place
positive number is 27. If the digits are reversed in order
and the resulting number subtracted from the original
number, the difference is also 27. What is the original
number?
7.3. Equations linear in the squares of the unknowns.
\Vhen the unknowns occur only as squares, simultaneous
quadratic equations can be solved by the methods used in
solving simultaneous linear equations.
EXERCISES
119
Example.
Solve the simultaneous equations
4X2 + 3y2 = 127,
5x 2 - 4y2 = 89.
(1)
(2)
SOLUTION. The operations employed in effecting the solution
are indicated at the left.
16x 2 + 12y2 = 508.
,I,
15x 2 - 12y2 = 267.
31x2
= 775.
X2 = 25,
x = ±5.
3v2 = 127 - 4x 2 = 127 - 100 ::::: 27,
y = ±3.
4· (1)
3· (2)
(3)
+
(4)
(5) -+ 31
(1)
(3)
(4)
(5)
(6)
The results should be paired as follows:
(x = 5, y = 3),
(x = -5, y = 3),
(x
(x
= 5, y = -3),
= - 5, Y = - 3).
All four pairs of values satisfy both equations.
EXERCISES VII. B
Solve for x and y:
1.
xi!
+ 2y2
= 17,
2. 3x 2
2
3.
5.
7.
9.
11.
x - y2 = 5.
2X2
3y2 = 20,
3x 2 + 2y2 = 25.
2X2 - 9y2 = 170,
5x 2 - 10y2 = 525.
4X2 - 5y2 = -47,
3x 2
2y2 = -18.
lOx 2 7y2 = 8,
l1x 2 5y2 = 4.
13x 2 + 27y2 = 295,
22x 2 - 23y2 = -119.
+
+
+
+
4.
6.
8.
10.
12.
5y2 = -5,
4x 2
y2 = 116.
'-J,
7X2 - 2?/2 = 27,
3x 2 + 4112 = 31.
3x 2 - 8y2 = 80,
5x 2 + 6y2 = 56.
5x 2 - 26y2 = 111,
7x 2
8y2 = O.
7x 2
8y2 = 17,
8x 2 + 9y2 = 20.
95x 2 + 18y2 = 23,
85x 2 + 42y2 = -1.
+
+
+
13. The medians to the two legs of a right triangle are 45 and
60. Find the lengths of the legs.
SYSTEMS OF QUADRATICS
130
[Ch. VII
7.4. All terms involving the unknowns of second degree. *
If all terms involving x and yare of the second degree
we proceed as in the following example.
Example.
Solve:
+
2x 2 5xy - 10y2 = 8,
x 2 - 2xy + 3y2 = 3.
r-,
(1)
(2)
• f SOLUTION. Multiply (1) by 3 and (2) by 8, so as to make the
.:constant terms the same:
6x 2 + 15xy - 30y2 = 24,
(3)
8x 2 - 16xy + 24y2 = 24.
(4)
Subtract (3) from (4):
2x 2 - 31xy
Solve for x in terms of y.
+ 54y2 = O.
(5)
This can be done here by factoring.
(x - 2y)(2x - 27y)
x = 2y,
=
27
0,
(6)
Y'
(7)
3y2 = 3,
y = ± 1.
x = 2y = ±2.
(9)
=2
x
Substituting X !- 2y in (2) ,: we get
Substitute x
=
(8)
2; y in (2) and solve for y, geiting
2
;"'
y
= +-=.
- y'211
27
x =-y = +
2
-
27
-.
v'2TI
The values must be paired as follows:
(x = 2, y
(x
= 1),
27
= V2IT'
y
2 ),
= y'2TI
(x
=
-2, Y = -1),
(x = -
27
2 )
V2TI'
y = - V2TI .
* Note that simultaneous quadratics of the type considered in the preceding section (viz., linear in the squares of the unknowns) are also of this
type.
SYMMETRIC EQUATIONS
§ 7.5]
131
The values containing radicals in the denominator can be rationalized if desired, e.g.,
27
V2TI
=
27yill
211 '
2
y211
2V2TI
211 .
NOTE. Equations of this type may also be solved by the
following method: Set y = tx in both equations. Solve each
equation for x 2 as a function of t. Equate these two functions
and solve for t. Then find x. Finally find y, which is equal
to tx.
EXERCISES VII. C
1. Solve the above illustrative example by the method suggested in the note.
Solve for x and y:
2. x 2 + 2xy = 16,
3. 3x 2 - 2xy =- 7,
xy = 6.
3xy + y2 = - 2.
4. x 2 + xy = - 2,
5. X2 + 5xy = - 36,
2xy + y2 = 3.
xy - y2 = -24.
6. x 2 + 2xy = 16,
7. 8x 2 + 3y2 = 140,
xy - 2y2 = -12.
8xy + 5y2 = 84.
8. x 2 - 4xy = -20,
9. x 2 - 3xy + 2y2 = 20,
x 2 - xy + 3y2 = 25.
2x 2 - xy - y2 = - 35.
2 + xy = 6,
11.
8x
10. 4x 2 + 3y2 = 48,
.,' :j
2xy + y2 = 16.
2xy + y2 = 7.
12. 2x 2 + 3xy + y2 = 3,
13. 3x 2 - xy - y2 = 9,
x 2 + 3xy - y2 = -3.
x 2 - y2 = -27.
14. x 2 - xy + y2 = 3,
15. x 2 - 2xy + y2 = 16,
x 2 + 2xy = 5.
2x 2 + xy + y2 = 88.
16. The area of a right triangle is 84 square inches; its hypotenuse is 25 inches. Find the lengths of its two sides.
17. The area of a right triangle is 108 square inches; the
median to one of its sides is 15 inches. Find the lengths
of its two sides.
7.5. Symmetric equations.
An equation is symmetric in two unknowns if the equation is unaltered when the two unknowns are inter-
leh. VII
SYSTEMS OF aUADRATICS
132
changed;* for example,
+ 3y2 + X + y
3x 2 - 2xy
- 7 =
o.
Simultaneous quadratic equations both of which are symmetric can be solved by replacing one unknown by u + v,
the other by u - v.
"'i" ' ,,~-q
,. <
. .',-/
Example.
:.~, , :1 .~ "
r
i
,:,r
Solve:
3x 2 - 2xy + 3y2 + X + y - 50 = 0,
2X2 + 2y2 - 3x - 3y - 29 = o.
+ 1,', y = U - t'.
+ 8v + 2u - 50 = O.
+ 4v + u - 25 = O.
+ 4v 6u - 29 = O.
Solution. Let z = u
From
(3) +
From
(5) -
(1),
2
(2),
(4)
4u 2
2u 2
4u 2
(1)
(2)
1.1:
2
j,
2
2
2u
2
7u - 4 =
-
(4)
(5)
(6)
o.
u = 4, -
(3)
1
.
"
2
Substitute u ... 4 in (4): '
4v 2 = -2u 2
v= ±
-
U
+ 25
= -11,
~ VIT.
x=u+v=4+
- i2 v-1[f
~~,
y = u - v = 4
{j
+ ~ VIT.
* Note that some symmetric simultaneous quadratics may also be other- _
wise classified. Thus, the equations
x2
+ y2
= 25,
xy = 12
are of the type considered in the previous section and can be solved by the
methods of that section or by the methods of the present section. They
can probably be solved most easily by substituting y = 12/x from the
second equation into the first.
EXERCISES
Substitute u = 4v 2
(4):
5
v = ± 2'
1 5
y = - 2 - 2 = -3.
= 25,
}J I
x
~ in
1
5
= - 2 + 2 = 2,
x = -
133
!2 - 2~ =
1
5
-3, (-", y = - 2 + 2 = 2.
The anSW(lrs must be paired as follows:
(, = 4
+ ~ VTI,
(x = 4 - ~ vrr,
I
(x = 2, y
4 - ~ vrr),
y =
Y=4+~vrr),
= -3),
(x
= -3, y
=
2).
It will be noted that the solutions are symmetric, as is necessarily the case.
It will be helpful to note that in symmetric quadratic
equations the unknowns can always be grouped into the
forms x 2 + y2, xy, X + y, and that for the substitution
x = u + v, y = u - v, we have
x
+y
= 2u.
EXERCISES VII. D
Solve for x and y:
1. x 2 - 3xy
3.
6.
7.
8.
xy xy X2
xy
X2
X2
X2 x2 -
+
+ 11
+
+
+ +
+
4 O.
+ +
xy
+
0, 2.
y2
=
x - y - 5 = O.
x - y
7 = 0,
y2 - 13 = O.
x
y - 2 = 0,
y2 =
xy
y2 - 3x - 3y
3xy + y2 + 1 =
3xy + y2 - 2x - 2y 2x + 2y - 1 =
o.
o.
+
+ +
20
0,
x2
y2
X
y =
xy - 5 = 0 ..
4. 2xy - 3x - 3y - 27 = 0,
x2
y2 - 13 = O.
6. x 2 y2 - X - Y = 12,
+- 2
+
+
xy
=
+x+y
0,
15 = 0,
=
9.
134
[Ch. VII
SYSTEMS OF QUADRATICS
+ y)2 - 3(x - y)2 + X + y - 52
(x - y)(y - x) + xy - 5 = O.
10. x(y - 2)
y(x - 2)
2 = 0,
x 2 + y2 - (x + 2) (y + 2) + 1 = O.
9. 2(x
+
= 0,
+
11. The area of a rectangle is 192 square inches, its diagonal is
20 inches. Find its dimensions.
7.6. Miscellaneous methods and types.
Sometimes special methods are shorter than those given
above.
Example 1.
Solve:
.z' +
y2 = 25,
xy = 12.
;)
(1)
(2)
SOLUTION. These equations are symmetric, and they also
come under the case in which all terms involving the unknowns
are of second degree. They can therefore be solved as either of
these cases. See footnote of § 7.5. However, the following
method is perhaps neater:
From (2),
(1) + (3)
2xy = 24..
(3)
x 2 + 2xy + y2 = 49, . . x + y = :1:7.
(4)
(1) - (3) x 2 - 2xy + y2 = 1,
x - y = ± 1.
(5)
x
y = 7,
x
y = 7,
x
y = -7,
x + y = -7,
x-y=l;
x-y=-l;
x-Y=l;
x-y=-l;
(x = 4, y = 3),
(x = 3, y = 4),
(x = -3, y = -4),
(x = -4, Y = -3).
+
+
+
The methods that have been developed for systems of
equations involving quadratics may sometimes be applied
to systems involving equations of higher degree, as IS
shown by the next example.
Example 2.
Solve:
x 3 + yS = 35,
x + y = 5.
Note that one equation is a cubic, the other linear.
I
I
i
MISCELLANEOUS METHODS AND TYPES
,7.6]
135
SOLUTION. Divide the first equation by the second:
x2
xy
-
+ y2
= 7.
Substitute y = 5 - x in this last equation and reduce the result:
x2
-
5x
+ 6 = 0,
x = 2,3.
y = 3,2.
(x = 3, y = 2).
(x = 2, y = 3),
r
Example 3.
Solve:
6x 2 - txy
+ 2y2
Y
= 0,
= x2
(6)
-
4.
(7)
SOLUTION. In equation (6) each term is of degree two. Such
an equation is called homogeneous of degree two. (Note that
in homogeneous equations there can be no constant term, for
such a term is of degree zero.)
Solve (6) for y in terms of x, getting
y = 2x,
Substitute y
y =
tx.
= 2x in (7):
x 2 - 2x - 4 = 0,
x = 1 ± 0.
y = 2x = 2(1
± 0).
Substitute y = tx in (7) and reduce:
2x 2 - 3x - 8 = 0,
x
=
3
± 403.
3 9 ± 303
Y="2 x =
8
.
The complete solution is
[x = 1 +
[x = 1 -
(x
(x
= 3
+
= 3 -
0, y =
v5,
;n,
;n,
2(1 + 0)],
y = 2(1 - 0)],
y = 9
+~~)
y = 9 -
~03)-
SYSTEMS OF QUADRATICS
136
[Ch. VII
EXERCISES VII. E .
Solve for x and y:
1. x 2 + y2 = 61,
,;:
2. x 2 + 4y2 = 85,
xy = 21.
4. x 2 + xy = 8,
xy + y2 = 8.
6. x 3 - 27 y 3 = -91,
x - 3y = -1.
8. 8x 3 - y3 = 152,
i •:
4x 2 + 2xy + y2 = 76.
10. X4 - y4 = 240,
x 2 + y2 = 20.
12. X4 - 16y4 = 1040,
x 2 - 4y2 = 20.
14. x 2 - 6xy + 9y2 = 0, .
x 2 - 2x - 3y - 18 = O.
xy = 30.
3. x 2 + 4y2 = 37,
xy + x + 2y = 10.
6. x 3 + y3 = 91,
x + y = 7.
7. x 3 + y3 = 280,
x 2 - xy + y2 = 28.
9. x' - y4 = 544,
x 2 - y2 = 16.
11. x' + 4y4 = 20,
x 2 + 2xy + 2y2 = 10.
13. 3x 2 + 4xy - 4y2 = 0,
x 2 - 2y - 4 = O.
+ 2V6.
V5 + 2V6.
5 + 2V6.
16. Find the positive square root of 5
SOLUTION.
Square:
Let Vx +
x + 2vXY
VY
+
=
y =
-",
;
,."1'
,
,'j.;
+ y = 5,
2.yxy = 2V6.
It follows that
x
. These equations have the solutions x = 3, y = 2, and x = 1
2, y = 3. Thus, V 5 + 2V6 = V3 + 0. This may
be checked by squaring V3 + 0.
Find the positive square roots 6f the following quantities:
16. 11
+ 2V30.
18. 13 - 202.
20. 21 + 600.
22. 35 - 14V6.
7.7.
,
-
1 .'.'
17.
19.
21.
23.
9
+ 60.
25 - 10V6.
55 - 22V6.
34 + 2v'I73.
Graphic representation.
Quadratic equations as well as linear equations can be
represented graphically, although the plotting of the
former is usually more difficult. Graphic methods are of
137
GRAPHIC REPRESENTATION
§l.7]
little use in the actual solution of equations but often
give a clearer meaning to results. Points of intersection
of curves correspond to pairs of real values of both variables
satisfying the eq'uations; imagirtary values do not appear
on the graph.
Points of tangency of the curves correspond to double
solutions.
Example 1.
Draw the curves represented by, and solve, the following
equations:
x2
+
= 25,
y = x 2 - 5.
(1)
(2)
y2
SOLUTION. The first equation is that of a circle with its center
at the origin and having a radius of 5 units. For the (J(JOldinates x and y of any point on such a
y
y = x2 _ I)
circle are (since x 2 + y2
= 52) the sides of a
x
y
right triangle whose
hypotenuse is 5. (See X'
x
-5
0
Fig. 7.1.)
-4
±1
The other equation
-1
;1:2
y'
be plotted by givcan
4
±3
ing values to x and
11
±4
FIG. 7.1.
finding the corresponding values of y. A number of such values are shown in the
accompanying table. The curve obtained by plotting them
y
is a parabola. (See § 6.12.)
Both curves are shown in
ill
\
Fig. 7.2.
),
(-=~. p)
Following is the algebraic
1/
,..
~
J
solution:
,
"
I
\
x,
x
0
T
7
I
\
Vi
"(.
y'
FIG.
7.'2.
;)
From (2),
x2 = y
+ 5.
(3)
Substitute (3) in (1) and combine terms:
y2
+y -
20 = 0,
y =
-5,4.
[Ch. VII
SYSTEMS OF aUADRATICS
138
Substitute y = - 5 in (3):
x=o
x 2 = 0,
I
";1
(double root).
,
,!
Substitute y = 4 in (3):
±3.
x =
,
The values are paired as follows:
------'-~-
(x = 0, y = -5) (double solution),
(x = 3, y = 4), (x = -3, y = 4),
,.ii
Note that at the point corresponding to the double solution
y
the curves are tangent.
Example 2.
L
/(4
"1.--7 ~-)
Plot and solve:
t5.)
x,
\
I'
'"
x 2 + y2 = 25,
4y2 = 9x.
X
If
7
VC
:""'-..!.. -j).
I:
I"
i'
(4)
(5)
SOLUTION. Equation (4) is
the circle of example 1. Equa,
Y
tion (5) is that of another
FIG. 7.3.
parabola. Both curves are
shown in Fig. 7.3. Following is the algebraic solution:
v
I
(6)
From (5),
Substitute in (4), clear of fractions, and transpose the constant
term:
4x 2 + 9x - 100 = 0,
25
x = 4 , - _.
4
Substitute x
= 4 in (6):
y2 = 9,
Substitute x = -
~ in
9 25 ,
y2 = ____
4 4
y =
±3.
(6):
y
=
+
~~~-i
- 2 2
=
± 145 i.
I
I'
t
EXERCISES
139
The values are paired as follows:
= 3),
(x = 4, y
(X =
-
25, y = 1:
4
i}
(x
= 4,
(X =
y = -3),
25, y
4
-
= - ~1
i}
All four pairs of values satisfy both equations, but the last two
pairs, in which y is imaginary, do not appear on the graph.
The subject of graphic representation of equations
belongs more properly to the field of analytic geometry, in
which simpler and more systematic methods than pointby-point plotting are developed. Consequently we shall
not go further into the discussion here.
EXERCISES VII. F
The instructor may require the student to draw graphs for
certain of the preceding exercises of the chapter.
For what values of k will the line and the curve, or the
two curves, be tangent?
1.
X2
+ y2
= 20,
y = 2x
+ k.
,
(7)
(8)
:t",;
SOLUTION. Equation (7) is represented by a circle with
center at the origin and with radius V20 = 4.47. Equation (8) is a series of parallel
Y
r
V I I
lines, one line correspond1
Ii
ing to each value of k.
OS' ~
1'::'"1
/ iVl
Substitute the value of y
1{T
,(i
from (8) in (7):
~I
~~I
X2 + 4X2 + 4kx + k 2 = 20,
5x 2 + 4kx + k 2 - 20 = O.
(9)
t-V
1/
x
,
t1
lY
~h
If'
,
"
7
Ordinarily there would be
two distinct roots of this
equation, Xl and X2, which,
if real, would be the abscissas of the points in
which the line cuts the circle. 1£ the
.......
."11\
c'l.
~~f-
I
f7
IJ
x
j
r-,...
rt
y'
FIG.7.4.
line were tangent
140
SYSTEMS OF QUADRATICS
[Ch. VII
to the circle these values of x would be the same, that
is, equation (9) would have equal roots. Making use
of the fact that if a quadratic equation has equal roots, its
discriminant, b2 - 4ac, must be zero, we find that for
tangency we must have
or
4.
6.
+ y2 =
- _'---
thereforebetangentt6:~e~~14!1.;
. .• '.,',',.,,'"
3. x 2 + y2 = 17,
. ~:"",I"
Y = 4x + k.
y = kx + 17. I I'
x 2 + y2 = k,
6. x 2 + y2 = 25,
3x - 4y = k.
x + y = 6.
yZ = 4x,
7. y2 = 6x,
y = kx + 3.
y = 2x + k.
"1
'
9. y = 6X2,
y = 4x 2,
y = 3x + k.
Y = kx - 6.
y2 - 8x + 2y + 17 = 0,
11. y2 + 12x - 4y + 16 = 0,
y = kx + 4.
y = 2x + k.
13. 9x 2 - 8y2 = 144,
x 2 + 2y2 = 8,
y = 2x + k.
y = -2x + k.
16. 16x 2 + 9y2 = 144, ,'.:
8X2 - 3y2 = 96,
y = 2x + k.
y = x + k.
x 2 + y2 - 4x - 6y - 81 = 0,
5x + 12y = k.
y2 = 2px,
18. x 2 + y2 = r2,
y = mx + k.
y = mx + k.
x2
y2
x2
y2
20. a 2 - b2 = 1,
a 2 + b2 = 1,
2. x!
17,
"
8.
10.
12.
14.
16.
17.
19.
y
= mx
+ k.
y = mx
") ~'
+ k.
MISCELLANEOUS EXERCISES VII. G
Solve:
1. x 2 + y2 + lOx - 14y + 49
7x - y + 17 = O.
2. 2X2 + 3xy + y2 + 2 = 0,
3x 2 + 2xy + 3 = O.
=
i
'"
-
Thelinesy = 2x ± 10 will
(See Fig. 7 . 4 . ) . .
f
I
I;
4· 5(k 2 - 20) = 0,
k 2 = 100,
k = ±10.
16k 2
0,
3. 3x 2 - 7y2
+ 48 = 0,
8x 2 -- 3y2 - 13 = 0.
./
I
EXERCISES
141
4. x 2 + y2 - 2x - 2y - 27 = 0,
xy - x - y - 9 = O.
O. x 2 - 2xy + y2 + 16 = 0,
x 2 + xy + 2y2 + 88 = O.
'.'
6. 2x - 3y - 1 = 0,
4x 2 + 9y2 - 16x + 18y - 11 = 0.
7. xy - 4x - 4y + 20 = 0,
x 2 + y2 - 11x - 11y + 48 = 0.
8. 6X2 + 55y2 - 447 = 0,
9x 2 + 22y2 - 126 = O.
9. x 2 - y2 - 6x + 5y + 3 = 0,
3x + 5y - 24 = O.
10. 5x 2 + 3xy - y2 + 80 = 0,
2X2 + 5xy + 2y2 = O.
11. x 2 - 3xy + y2 + 2x + 2y - 15 = 0,
xy - 2x - 2y - 1 = O.
12. /)x 2 + xy - 4y2 - 6 = 0, 13. 5x 2 - y2 = 35,
6x 2 - xy - 4y2 - 4 = 0.
3x 2 - xy - y2 = 1!: ...
2
14. x - 3xy + y2 - X - Y - 12 = 0,
xy + 4x + 4y + 20 = O.
16. 8x 3 + 27 y 3 - 37 = 0,
2x + 3y - 1 = O.
16. 27x 3 - 64 y 3 - 152 = 0,
9X2 + 12xy + 16y2 - 76 = 0.
17. x 3 - 18xy + y3 = 0,
18. x 3 + y3 = 4xy,
x - xy + y = o.
x + y = xv.
19. 2x2y2 - 25xy + 12 = 0,
20. x 3 - y3 = lOxy,
3x - 2y - 1 = O.
x - y = xv.
21. x 3 - y3 - 2 = 0,
22. x 3 + y3 = 9,
x2y - xy2 + 2 = O.
x2y + xy2 = ti.
23. X4 - y4 = 0,
x 2 + y2 = 98.
24. 8X 3 + 12x2y + 6xy 2 + y3 = 0,
x 2 + 3xy + 2y2 = 27.
26. 4X4 - 25x 2y2 + 36y4 = 0,
26. X4 + 5X 2y 2 - 36y4 = 0,
4x 2 - 2xy + y2 - 28 = O.
5x - 4y - 6 = O.
28. X4 + y4 = 2, (a)
27. X4 + y4 = 32, (a)
x + y = 2. (b)
x + y = 4. (b)
( ,\ ....
,','
r~i\'
SYSTEMS OF QUADRATICS
141
lCh. VII
SUGGESTION for solving exercises 27 and 28. Raise (b)
to fourth power and subtract (a). Divide by 2 and label
resulting equation (c). Square (b), mUltiply by 2xy, and
subtract (c). Solve resulting equation for xy and combine
with (b).
29.
3X2 -
4x 2
2y2 - 7z 2
+ 5y2 + 8z
2
= 41,
= 37,
5x 2 - 6y2 + 6z 2 = -44.
31. xy + x + y = -13,
yz + y + z = -17,
zx + z + x = 11.
33. x 2 + xy + xz = 9,
y2 + yz + xy = 6,
Z2 + XZ + yz = - 6.
30. x 2 + y2 + Z2 = 14,
x + y + z = 4,
x + 2y + 3z = 9.
32. x 2 + y2 + Z2 = 21,
x + y + Z = 3,
Z2
= xy.
SUGGESTION. Add the three equations.
34. The perimeter of a rectangle is 72 inches. I ts area is
288 square inches. Find its dimensions.
35. The perimeter of a right triangle is 20 inches. Its area is
15 square inches. Find its hypotenuse.
36. A man fenced in a rectangular lot of area 31,200 square feet
at a cost of $600. The fence for three sides of the lot cost
$0.75 per foot, that for the front cost $1.00 per foot. Find
the dimensions of the lot.
37. The area of the page of a book is 40 square inches, the area
of the printed part is 2H square inches. The margin at
the top and sides is ! inch wide, the margin at the bottom
is 1 inch wide. Find the dimensions of the page.
38. A speculator sold some shares of stock for $19,800. A week
later, the price of the stock having fallen $2 per share, he
repurchased, for the same amount of money, 5 more shares
than he sold. How many shares did he sell?
39. A group of fishermen chartered a boat for $300. Two
members of the group were unable to go on the trip and,
as a consequence, each of the others had to pay $12.50
more. How many were there in the original group?
40. A sum of money invested for a year brought $81 interest.
If the rate had been i per cent more and the principal
i·
EXERCISES
41.
42.
43.
44.
45.
46.
41.
48.
49.
143
$350 less, the interest would have been the same. Find
the principal and the rate.
Two pipes together can fill a tank in 5 hours. The smaller
pipe alone requires 3t hours longer than the larger to fill it.
How long does it take each pipe alone to fill it?
A typist would require 45 hours to type a certain manuscript. A second typist, who takes 2 minutes more to each
page, would require 50 hours. How many pages of manuscript are there and how long per page does it take each
typist?
On a certain day two inspectors, A and B, employed by a
factory, were inspecting a lot of manufactured articles.
In the morning A had inspected 1500 articles before B
started in. During the remainder of the morning they
inspected 4000 more, A working 4 hours altogether. In
the afternoon A worked 4 hours and B 2 hours, in which
time they inspected 5200 articles. Find the average number of articles that each inspected per hour and the number
of hours that B worked in the morning.
The sides of a triangle are 5, 6, and 8. Find the medians.
The medians of a triangle are 6, 8, and 10. Find the sides.
A beam rests on a fulcrum 2t feet from One end. A weight
of 30 pounds suspended from this end of the beam causes .
the beam to balance. If the weight is suspended from the
other end of the beam it is necessary to suspend a 95-pound
weight from the first end in order to effect a balance. _.
Find the length and the weight of the bf~am.
In a two-place number the square of the digit in the tens'
place is 2 less than the sum of the digits. If the digits are
transposed the number is increased by 45. What is the
number?
The sum of the squares of the digits in a three-place number
is 84. The square of the middle digit is equal to the product
of the other two. If the digits are reversed in order the
number is increased by 594. What is the number?
If 5 is added to the numerator and to the denominator of a
fraction and the result added to the original fraction the
sum is 1~. If 5 is subtracted from the numerator and the
denominator of the fraction and the result subtracted from
SYSTEMS OF QUADRATICS
60.
"l
-I"
J~;l
a.
rl;),
'i';
II.
H
,;
e.
:'),
64.
[Ch. VII
the original fraction the remainder is -h. What is the
fraction?
A man drove a certain distance at a uniform rate. If his
rate had been 3 miles per hour slower the trip would have
taken 24 minutes longer. If his rate had been 6 miles
per hour faster he would have saved a quarter of an hour.
Find the distance and his rate.
A motorboat can go downstream 15 miles and back in 4
hours. Traveling at half speed it would require 10 hours
to make the same trip. Find the rate of the boat at full
speed and the rate of the current.
A boat made a trip of 9 miles with the tide and back against '
the tide in 3 hours. If the tide had been half as strong
the trip would have been made in 36 minutes less time.
Find the rate of the boat in still water and the rate of the
stronger tide.
Two trains started at the same time from cities A and B
respectively, the first going from A to B, the second from _
B to A. They traveled at uniform, but different, rates.
The first train reached B four hours after they passed each
other, the second reached A two hours and 15 minutes
after they passed. Find the time that each train required
to make the trip.
I
A man is at a point A on the bank of a straight river. He
can row in still water at a rate which is 20 per cent faster
than he can walk. If he rows downstream to a point B,
which is four miles below A and on the same side of the
river, and then walks, in a direction which is perpendicular
to the river, to a point C, three miles from B, it takes him
the same amount of time that it does to walk straight from
A to C, If he walks back from C to B and then rows
upstream to A, the return trip takes him an hour and 36
minutes longer. Find his rate of walking, his rate of rowing in still water, and the rate of of the current of the river.
CHAPTER VIII
Inequalities
".,
,
8.1. Inequalities.
The statement "a is greater than b" (a > b), a and b
being real numbers, means that a - b is a positive number. If we have an axis (Fig. 8.1) on which the positive
---~6---~4---~3---~2---~1--~O--Tl--~2--~3--4~-6~+
FIG.
8.1.
direction is to the right, a will be to the right of b. Similarly, "a is less than b" (a < b) means that a - b is a
negative number, and on an axis directed toward the right,
a will be to the left of b. Such expressions are called
inequalities. All inequalities refer to
--1b~--a+----++
real numbers.
a>b
Two inequalities a > b, c > d in
FIG. 8.2.
which the signs point in the same
direction are said to be alike in sense (or order); two
inequalities a > b, c < d in which the signs point in opposite directions are said to be opposite in sense (or order).
The statement a G; b is read" a is greater than or equal
~-+-_ _--I-_-++ to b"; the statement a ~ b is read
aa <b b
"a is less than or equal to b."
Some inequalities are satisfied by all
FIG. 8.3.
real numbers, for example, X2 + 3 > O.
Such inequalities are called absolute inequalities. Other
inequalities, called conditional inequalities, are satisfied
only by certain numbers. Thus, the inequality x - 3 > 0
is satisfied only by numbers greater than 3.
145
146
lCh. ViiI
INEQUALITIES
8.2. Fundamental properties.
The following properties are useful in dealing with
inequalities. Their proofs are simple, following almost
immediately from the definition of an inequality. The
proof of Property I only will be given. Proofs of the
others can be effected in a similar manner.
I. An inequality is unchanged in sense if the same number
is added to or subtracted from both sides. That is, if a > b,
then a + c > b + c and a - c > b - c.
Proof. By hypothesis, a > b.
Let a - b = p, which will be positive, since a > b. Then
(a
That is,
+ c)
- (b
+ c)
I>
= a -
b = p. ;
,
,
,
,
+ c > b + c.
a
Similarly we can prove that a - c
> b - c.
Example 1.
10> 8,
10
+ 3> 8 + 3.
Example 2.
10
10 - 3
> 8,
>8-
3.
II. An inequality is unchanged in sense if both side8 are
multiplied or both divided by the same posi#ve number.
Example 3.
2 < 3,
5·2<5·3.
Example 4.
3x
~
x ~
15,
5.
III. An inequality is changed in sense if both sides are
multiplied or both divided by the same negative number.
EXERCISES
147
Example 5.
2 < 3,
-2> -3.
(Each side has been multiplied by -1.)
Example 6.
8> 6,
8 -+- (-2) < 6 -+- (-2),
-4 < -3
IV. An inequal-ity of positive quantities is unchanged in
sense if the same positive power or same positive root of each
side is taken.
Example 7.
Example 8.
49> 25,
Y25.
V49 >
EXERCISES VIII. A
Prove the following:
1. If the first of three quantities is greater than the second, and the
second greater than the third, then the first is greater than the
third; that is, if a > band b > c, then a > c.
2. If un equals are added to unequals in the same order, the sums
are tmequal in the same order; that is, if a > band c > d, then
a
c > b d.
3. If unequals are subtracted from equals, the results are unequal
in the opposite order; that is, if a > b, then c - a < c - b.
(Show, by means of an example, that if a > band c > d, then
it does not necessarily follow that a - c > b - d.)
4. If a > b ~ 0 and c > d ~ 0, then ac > bd.
+
+
INEQUALITIES
148
o.
If a and b are real, then a 2
SUGGESTION.
(a
± b)2
~
j
leh. VIII
I
!
+ b2 ~ 2labl.
O.
,
6. If a, b, and c are positive, then a 2
I
+ b + c > ab + ac
+ be. '
,
2
t
,t~,
8.3. Solution of inequalities.
Many inequalities can be solved, by applying these
fundamental properties, very much as equations are solved
by the use of axioms concerning equals. Graphs aiso are
often ·of assistance, especiaUy in the case of quadratic
inequalities, in determining when an inequality is satisfied.
I
!
Example 1.
y
Find the values of x for which
1/
il
X'
7
1/3
II
1/
!I
y'
FIG. 8.4.
3x - 4>
x
7x
2 +
2.
SOLUTION.
Multiply by 2:
6x - 8
Subtract 7x and add 8: -x
Multiply by -1:
x
>
>
7x + 4.
12.
<
-12.
Example 2.
Give a graphic interpretation of the
inequality 3x - 7 > O.
SOLLTION. Set y = 3x - 7 and draw the graph. It is seen
that the graph crosses the x-axis between 2 and 3. Solving
algebraically, we find
7
x> _.
3x > 7,
3
The graph makes the inequality visual.
Example 3.
Find the values of x for which
x2
+ 3x -
4> O.
I
§ 8.4]
THE GENERAL. aUADRA TIC FUNCTION
149
Draw the graph (Fig. 8.5) of the function
SOLUTION.
y = x2
+ 3x
- 4.
It is seen that the graph crosses the x-axis at -4 and 1 and that
the function is positive (i.e., > 0) for values of x outside these
values, namely,
y
x < -4 or x > 1.
II
.Algebraically we can solve b:p<completing the square:
II
,
X
+ 3x > 4,
+ 3x + 4~ > 4 + ~,
4
x2
X2
:J
f4
x
II
1\
\
25
(x +23)2 > 4"'
Ii-
"y
FIG. 8.5.
This last ineq uality will hold if
x
+ 2> 2
3
I)
or if
x>
1
or if
that is, if
x
<
-4.
A satisfactory way of solving a quadratic inequality is
to replace the inequality sign by an equality sign, solve the
resulting equation, and note from the graph, or by testing
with numbers, whether the inequality is satisfied between
the roots or outside of the roots.
8.4. The general quadratic function.
Consider the general quadratic function,
ax 2
+ bx + c,
(1)
which may also be written in the form
a(x - Xl)(X - X2),
(2)
and X2 being the roots of the equation obtained by setting the function equal to zero.
If these roots are imaginary, the function cannot change
sign. -For we recall (see § 6.12) that when a quadratic
Xl
(Ch. VIII
INEQUALITIES
150
equation has imaginary roots, the curve representing the
corresponding quadratic function is wholly above or
wholly below the x-axis (because if it crossed, the equation
would have a real root). It must therefore have the sign
of c, since when x = 0 the function is equal to c. But
if the roots are imaginary we must have b2 - 4ac < O.
This requires that a and c have the same sign, since they
are real numbers. Therefore we can state that in this
case ax 2 + bx + c has the same sign as a and c for all
values of x.
If the roots are equal (Xl = X2), we may write the function in the form
,
I
i
(3)
This will be positive when and only when a is positive and
x
~
Xl.
If the roots are real and different and a is positive, sup- _
pose Xl < X2. From (2) we see that the factors X - Xl and
X - X2 will both be negative when X < Xl, and that consequently for such values of X the function will be positive.
When X is between Xl and Xi, x - Xl will be positive and
X - X2 will be negative. The expression (2) will then be i
the product of three factors having the signs +, -, +,
respectively, and the function will be negative. When
X > X2 all the factors will be positive and the function will
be positive.
If a is negative the situation is reversed, and the function a(x - Xl) (X - X2) will be positive only when X is
between Xl and X2. Results are summarized in the table
below:
b2
-
If
4ac is
+
Then ax! + bx
c will have
same sign as a for
0
+
X
All real values of x.
x ;t. Xl.
< Xl, X > X2 (if Xl <
X2).
EXERCISES
151
EXERCISES VIII. B
Solve the following inequalities:
< 9.
3. 9x + 3 !?; 2x - 4.
6. lOx - 3 < 3x + 4.
7. 6 - 7x ~ 3 + 2xa .
9 5x + 11 < 3x - 5. .
1. 5x
.
2
""
. (,,?,
J
4
4x - 3 < 8.
3x
5 ~ 7x - 3.
2.
4.
6.
8.
10
+
8x - \} > 11x + 6.
2x - 19 ~ 5x - 1.
3 - 4x
.
7>
2 - x.
8
6x - 16 > O.
::)~ 12. x - 9x + 20 < O.
13. x 2 - 8x + 7 ~ O.
': .
14. 8 + 4x - x 2 !?; O.
15. 2X2 - 5x - 12 > O.
16. .6x 2 + 19x - 7 > O.
2
17. 25 - 20x - 12x < O.
18. x 2 + 3x + 2 > x 2 - 2x 4- 3.
19. 4x 2 + 3x + 1 > O.
20. 2x 2 - 5x + 10 < O.
2
21. x + 2x + 7 < 3x + 4.
22. 6x 2 + 5x - 7 > 4x 2 - 3x + 17.
23. x 2 + 2x - 5 > O.
24. x 2 - 7x + 1 > O.
26. (x + 3)(x - 2)(x - 5) > O. 26. (x + l)(x - 2)2 > O.
11. x
2
2
SUGGESTION. § 13.11 will be helpful in solving exercises
25-32.
27. (x + l)(x - 2)2(X - 4)3 !?; O.
28. x(x - l)(x - 5) > o.
29. (x + 1)2(x 2 + 1) > O.
30. x 2 (x - l)(x - 5)
31. (x + 1)(2x - 3)(3x + 2)2 > O.
;
32. (5x - 4)2(2x + 1)2 > 0 ..
3
33. x
+2>x
> O.
2
- l'
.0
SOLUTION. Subtract 2/(x - 1) from both sides:
3
2
x + 2 - x - 1 > O.
Clear of fractions by multiplying both sides by the expression (x + 2)2(X - 1)2, which is positive unless x is equal tc
lor -2:
+
+
3(x
2)(x - 1)2 - 2(x
2)2(X - 1)
(x + 2)(x - 1)[3(x - 1) - 2(x + 2)J
(x
2) (x - t) (x - 7)
+
•
>0
> 0,
> O.
1
152
[Ch.
INEQUALITIES
vm
By studying the signs of the three factors when x is in
various intervals, we see that the inequality is satisfied for
-2
3
<x <
38.
x>
and
2
5
+ 3 < x + 2'
2
3
9. + --4 > O.
x' xx - I > x + 5.
34. x
36.
1
x+3
35. 2x
+
7.
I·.
3
7 > 3x - 2
37. x + 1 > x + 3. ~'_-"'_-.
x+2
x+4
39. x - 2 > x - 5.
1x-5
x+l
I
x-7
I
Find the values of k for which the roots of the foll~
quadratic equations are real and different. (It is assumed1~~
k is real.)
: '.
"
SUGGESTION.
40.
42.
43.
44.
46.
j
The discriminant must be positive.
2kx + k = O.
41. &:2 - 6x + 3 =
+ k 2 - X - k = O.
(k + 2)x 2 + (x + 2)k 2 = O.
k(x + 3)2 + x(k - 3)2 = O.
(x - k)(x - 2) + (x + k)(x + 3) = o.
x2
-
k.
x2
Find the values of k for which the roots of the following equations are imaginary. (It is assumed that k is real.)
+ 4x + 5 = O.
47. 36x
+ k + 4x + kx + k + 4 = O.
46. kx 2
2
2
48. x
x+k
x-2
49. x _ 2k - 2x _ k = O.
2
6x
//
+5 =
kx.
-<
Find, in the following exercises, the values of k for which the
line intersects the curve in two distinct points.
SUGGESTION. The roots of the quadratic ,equation
obtained by substituting from the linear equation into the
quadratic must be real and unequal.
&0. x 2 + y2 = 10,
y = 3x + k.
12. x 2 + y2 = k,
x + y = 5.
61. x 2
+ y2
= 1,
= kx + 2.
153. y2 = 12x,
y
y = 3x
+ k.
,r-
I
EXERCISES
2
64. x = 12y,
y = 3x + k.
66. x 2 - y2 = 25,
y = kx - 5.
58. 9x 2 + 4y2 = 36,
y = 2x + k.
:...:-
"'.:
66. y2 = 12x,
y = kx + 3.
67. x 2 + 6y2 = 42,
Y = x + k.
69. 2x 2 - 5y2 = 20,
y = -2x + k.
."
153
,
CHAPTER IX
I
i
!
Variation
",
:;,
!
9.1. Ratio and proportion.
"The ratio of a to b is the quotient a + b, that is, the
fraction a/b. The ratio may also be written a: b.
A proportion is a statement that two ratios are equal,
for example,
The quantities band c are called the means of the proportion, a and d are called the extremes.
The quantity d is called the fourth proportioc:lal to a, b,
and c.
If ~ =
~,
If ~
~, x is called a mean proportional between a and b.
=
x is called the third proportional to a and b.
Problems involving proportions can always be handled
by the methods used in dealing with fractions.
9.2. Direct variation.
Each of the statements "y varies directly as x," "y
varies as x," "y is directly proportional to x," "y is proportional to x," in a mathematical sense means
y
=
kx,
(1)
in which k is a constant called the constant of variation or
constant of proportionality. (The statement "y varies
as x" is sometimes written y a: x, but the form (1) is
154
i
~
EXERCISES
155
decidedly preferable.) The constant k in (1) can be
determined if any pair of corresponding values of x and
y is known, provided x t6- O.
The statement "y varies as the square of x" means
y = kx 2•
9.3. Inverse variation.
The statements "y varies inversely as x," and ely
inversely proportional to x" mean
y
IS
k
=-.
x
9.4. Joint variation.
The statement "z varies jointly as x and y" means
,:P
\
z = kxy.
The statement" w varies directly as x and the cube of y
and inversely as the square root of z" means
w =
k xy 3
v-z-.
EXERCISES IX. A
1. If y va riCH as x and is equal to 11 when x is equal to 7, find
the v&lue of y when x is equal to 5.
SOLUTION.
Y
11
=
kx.
= 7k,
k
11
= -,;;-.
{
Having determined the constant k, we can state the law of
variation:
y =
11
'lx.
This law is now ready for use.
we have
For example, when x = 5
11
55
y = --·5 = -.
7
. 7
VARIATION
156
[Ch.IX
2. The horsepower that can be safely transmitted by a shaft
varies jointly as the number of revolutions it makes per
minute and the cube of its diameter. If a 3-inch shaft
making 250 revolutions per minute can safely transmit 180
horsepower, what horsepower can be safely transmitted
by a 2-inch shaft of the same material making 300 revolutions per minute?
SOLUTION. Let horsepower be denoted by P, number of
revolutions per minute by n, and diameter by d. Then,
P = knd 3 •
=
Substitute P
180, n = 250, d = 3, and solve for k:
= k . 250 . 27,
k = -.fr;.
= T\nd 3 = -h . 300 . 8 = 64.
180
P
3.
4.
6.
6.
7.
e,
8.
9.
10.
That is, a 2-inch shaft making 300 revolutions per minute
can safely transmit 64 horsepower.
If y varies as x and is equal to 24 when x is equal to 15,
what is the value of y when x is equal to 10? 3H 6t?
If y varies as x and is equal to i when x is equal to t, what
is the value of y when x is equal to H? t? 27?
If y varies as the square of x and is equal to 2 when x is
equal to 6, what is the value of y when x is equal to 9?
-1? H?
If y varies as the square root of x and is equal to 10 when
x is equal to 6, what is the value of y when x is equal to 216?
The quantity z varies jointly as x and y and has the value
15 when x is equal to 6 and y is equal to 10. (a) Find the
value of z when x is equal to 8 and y is equal to 2. (b)
Find the value of y when x is equal to 12 and z is equal to-i.
If y varies inversely as x and has the value l when x is
equal to 10, what is the value of y when x is equal to t?
If z varies directly as x and inversely as y and is equal to 4
when x and y have the values 12 and 8 respectively, what is
the value of z when x is equal to t and y is equal to ?J;\?
The quantity z varies directly as the cube of x and inversely
as the square of y; z has the value 14 when x is equal to
4 and y is equal to 6. Find the value of z when x equals 2
and y equals 3.
i
1/
EXERCISES
157
11. Hooke's law states that the extension of a spring or elastic
string beyond its natural length varies as the force applied.
If a weight of 9 pounds attached to a spring causes it to
stretch from a length of 15 inches to a length of 16i inches,
what weight will be necessary to stretch it to a length of
20 inches?
12. The period of a simple pendulum (the time for it to swing
across and back) varies as the square root of its length.
(a) If a pendulum 8 inches long has a period of 0.45 second,
what is the period of a pendulum 32 inches long? (b)
What is the length of a pendulum whose period is 1 second?
~3. Boyle's law states that at a constant temperature the
pressure of a gas varies inversely as its volume. If a certain amount of gas is under a pressure of 20 pounds per
square inch and has a volume of 250 cubic inches, what will
be the pressure if the volume is decreased to (a) 200 cubic
inches? (b) 150 cubic inches? (c) What will be the
pressure if the volume is increased to 400 cubic inches?
14. If a 2.5-inch shaft making 240 revolutions per minute can
safely transmit 2oo horsepower, what is the maximumnumber of revolutions per minute that is safe for a 1.5-inch shaft
transmitting 100 horsepower? (See exercise 2.)
15. The amount of electric current required to melt a fuse wire
varies as the three-halves power of the diameter. If the
current required to melt a wire of diameter 0.02 inch is
15 amperes, what current will be required to melt a wire
of the same material having a diameter of 0.08 inch?
The electrical resistance of a wire varies directly as its
length and inversely as the square of its diameter. If a
wire 50 feet long and 0.015 inch in diameter has a resistance
of 3.5 ohms, what is the resistance of a wire of the same
material 75 feet long and 0.025 inch in diameter?
17'. The force of a wind blowing perpendicularly onto a flat
surface varies as the area of the surface and the square of
the velocity of the wind. When the wind is blowing 15
miles per hour the force on an area A is 24 pounds. What
is the force on an area 2A when the wind is blowing 20
miles per hour?
18. The power required to propel a ship varies as the cube of the
l6.
158
19.
20.
11.
22.
23.
24.
25.
26.
VARIATION
[Ch.IX
speed. If a speed of 10 knots requires 5600 horsepower,
what horsepower is required for a speed of 15 knots?
The power required to operate a fan varies as the cube of
the speed. If i horsepower will drive a fan at a speed of
100 revolutions per minute, (a) how fast will 4 horsepower
drive it? (b) What horsepower will be required to give it
a speed of 120 revolutions per minute?
If one weight draws up another by means of a string passing
over a fixed pulley, the distance passed over by each weight
in a given time varies directly as the difference between the
weights and inversely as their SUill. If a 5-pound weight
lifts a 3-pound weight through a distance of 16 feet in 2
seconds, how far will a 6-pound weight lift a 2-pound weight
in the same length of time?
The weight of an object above the surface of the earth
varies inversely as the square of its distance from the center
of the earth. If it were possible to project an object weighing 9 pounds at the surface of the earth to a point 2000
miles above the surface, what would it weigh at that height?
(Assume the radius of the earth to be 4000 miles.)
At what height above the surface of the earth would an
object have a weight equal to one-fourth of its weight at
the surface? (See preceding exercise.)
The weight of an object below the surface of the earth varies
as its distance from the center of the earth. What would
a 100-pound object weigh at a distance one-fourth of the
way from the surface to the center?
The critical load of a cylindrical column varies directly
as the fourth power of its diameter and inversely as the
square of its height. What would be the effect on the critical
load if both the diameter and the height were doubled?
The safe load for a beam supported at both ends varies
directly as the breadth and the square of the depth of the
beam and inversely as the distance between supports. If
the safe load for a 4 by 6 (inches) beam whose supports
are 12 feet apart is 1000 pounds, what is the safe load for
a 3 by 10 beam of the same material, the supports being 8
feet apart?
The quantity of water discharged over a weir in a given
EXERCISES
159
interval of time varies as the length of the weir and the
three-halves power of the head. (The head is the difference in height between the level of the crest of water flowing
over the weir ar.d the level of still water above thE: weir.)
If the rate of discharge over a weir 4 feet long and haying
a head of 3 inches is 100 cubic feet per minute, what is the
rate of discharge over a weir 10 feet long and having a
head of 1 foot?
1'1. The number of vibrations made by a stretched string varies
directly as the square root of the stretching force, or tension, and inversely as the product of the length and the
diameter. If a string 2 feet long and 0.025 inch in diameter
vibrates 1440 times per second under 40 pounds tension,
(a) how many vibrations will be made by a string 3 feet
long and 0.03 inch in diameter under a tension of 90 pounds?
(b) To what tension must a string 1 foot 3 inches long and
0.10 inch in diameter be subjected to make it vibrate 720
times per second?
28. The intensity of illumination from a source of light varie~
inversely as the square of the distance from the source.
(a) At what distance from a source of light will the intensity'
of illumination be half as great as it is at a distance of 6
inches? (b) How much closer to a source of light must an
object be moved so that the intensity of illumination which
it receives will be 44 per cent more than that which it
receives at a distance of 3 feet?
29. A piece of cardboard is placed midway between two sources
. of light of equal intensity which are 12 feet apart. Its
plane is perpendicular to the line joining the two sources.
How far must it be moved, being kept always in this line
and perpendicular to it, so that the total illumination which
it receives (i.e., on both sides) will be ten times as great?
CHAPTER X
Mathematical Indudion
and the Binomial Formula
10.1. Mathematical induction.
An extremely important type of reasoning is the process
called mathematical induction. It is very powerful in
effecting certain proofs, especially in establishing the
validity of certain types of formulas. We shall illustrate
the method by proving that the sum of the first n odd
integers is equal to n 2 , that is,
1
+ 3 + 5 + . . . + (2n
- 1) = nt.
(1)
The proof consists of two parts:
Part I. The formula is true for certain particularlvalues
of n, e.g.,
I '
for n = 1,
for n = 2,
for n = 3,
1
1
3
1 = 1"',
= 2 2,
5 = 3 2,
+3
+ +
or
or
or
1 = 1;
4 = 4;
9 = 9.
Part II. Let k denote any positive integer for which
the formula is true, that is, such that
1
+ 3 + 5 + . . . + (2k
- 1) = k 2•
(2)
{So far, in this example, we see that k could be 1, 2, or 3.)
Then we can show that the formula is true for the next
integral value of n, namely, k + 1.
The last odd integer appearing on the left in (2) is
2k - 1. The next odd integer is (2k - 1) + 2 = 2k + 1.
160
/
MATHEMATICAL INDualON
§ 10.1)
161
Adding this to both sides of (2), we obtain
1
+ 3 + 5 + . . . + (2k
- 1)
+ (2k + 1)
= k + 2k + 1
= (k + 1)2,
2
(3)
Since (2) is a true equation and we have added equals to
equals, (3) must be the correct formula for the sum of the
first k + 1 odd integers. However, if we subEtitute k + 1
for n in (1), we also obtain (3). That is, if (1) holds when
n = k, then it also holds for n = k + 1. But the formula
is true for n = 1, 2, 3. If we let k = 3, by Part II the
formula must be true for n = k + 1 = 4. Then, if we
let k = 4, it follows that the formula must be true for
n = k + 1 = 5, and so on, for any positive integer
whatever.
In general, a proof by mathematical induction consists
of two parts:
Part 1. A verification that the proposition or formula is
true for the least integral value of n for which it is to hold.
Part II. A proof that if it is true for any integral value of
n, say n = k, then it is true for the next value, n = k
1.
+
The formula will then be true for all values of n from
the verified value on up.
The two parts of the proof do not have to be established
in the order in which they have been given, as they are
quite independent of each other. Both parts, however,
are absolutely necessary to the proof, as is shown by the
following illustrations:
Consider the formula
1
+ 3 + 5 + . . . + (2n
- 1)
=
n3
-
5n 2
+ Un
We have
for n = 1,
1
=
13
-
5 . 12
+ 11 . 1 -
6,
or
1 = 1;
- 6.
MATHEMATICAL INDUCTION
162
[Ch. X
for n = 2,
1
+3
=
23
-
5 . 22
+ 11 . 2 -
6,
4
or
=
4;
f :
for n = 3,
1
+3 +5
Ii
=
33
-
5 . 32
+ 11 . 3 -
or
6,
9 = 9.
Thus, we might be tempted to assume that the formula
is true for all positive integral values of n. However,
Part II of the mathematical induction proof is impossible
to demonstrate; the formula is not true for any values of n
other than 1, 2, 3. If we substitute n = 4, for example,
we are led to the contradictory result 16 = 22.
Similarly, the formula
1
+ 3 + 5 + . . . + (2n - 1)
= n + (n - 1) (n
2
- 2)
(n - 100)
holds for all positive integral values of n up to and including 100, but fails for n = 10l.
On the other hand, consider the formula
1
+2 +3+ . . . +n
=
1
2 n(n + 1) + 1.
Assume that it is true for n = k.
Add k
1
1
+2+3+ ... +k
+
1 to each side:
Then
1
= 2 k(k
+ 1) + 1.
+ 2 + 3 + . . . + k + (k + 1)
=
,
I
I
~ k(k + 1) + 1 +
(k
+
1) =
~ (k + 1) (k + 2) + 1.
Thus, the formula is true for n = k + 1 if it is true for
n = k, and Part II of the proof has been demonstrated.
However, Part I cannot be proved, since there can be
found no value for n for which the "formula" is true.
EXERCISES
163
EXERCISES X, A
Prove by mathematical induction:
_-.1.1+2+3+' ,_.
+n=-in(n+l),/
+ 22 + 3 + ' , . + n = in(n + 1) (2n. + 1). _ :
3. 13 + 2 + 3 + ' .. + n = in2(n + 1)2,
.
4. 14 + 24 + 3 + .. , + n
= ion(n + 1)(2n + 1)(3n2 + 3n ~ 1). ----G. 1 + 2 + 3 + ' , . + n
= -fin (n + 1)2(2n + 2n - 1).
6. 1 + 3 + 5 + ... + (2n - 1) = n
7. 2 + 4 + 6 + ... + 2n = h(n + 1).'
8. 12 + 3 + 52 + ... + (2n - 1)2 = tn(2n - 1)(2n + 1).
9. 22 + 42 + 6 + ... + (2n)2 = -in(n + 1)(2n + 1).
10. 1 . 2 + 2 . 3 + 3 ' 4 + ... + n(n + 1)
2. 12
3
6
2
2
3
3
4
4
6
6
2
6 •
2
2•
v
2
2
= tn(n +
l)(n
+ 2).
+ 3·4 + 5 . 6 + ... + (2n
= tn(n + 1)(4n - 1).
1 . 3 + 3 . 5 + 5 ' 7 + ... + (2n
= tn(4n 2 + 6n - 1).
11. 1 ·2
- 1) ·2n
12.
- 1)(2n
+ 1)
+ 2 ' 3 2 + 3 ' 42 + ... + n(n + 1)2
= -fin(n + 1)(n + 2)(3n + 5).
1·2·3 + 2 . 3 . 4 + 3 ·4 . 5 + ... + n(n +
= tn(n + 1)(n + 2)(n + 3).
13. 1 ' 22
14.
1)(n
+
2)
+ 3 . 4 . 5 + 5 ' 6 . 7 + ... + (2n - 1)2n(2n + 1)
+ 1)(2n2 + 2n - 1).
1 . n + 2(n - 1) + 3(n - 2) + . . . + n . 1
= tn(n + l)(n + 2),
1
1
1
1
n
1. 2 + 2· 3 + 3~ + ... + n(n + 1) = n + r'
15. 1· 2 . 3
= n(n
16.
17.
1
1
1
1
n
18. 1. 3 + 3· 5 + 5· 7 + ... + (2n - 1)(2n + 1) = 2n +
19. 2 + 22 + 2 3 + ... + 2" = 2(2" - 1).
1
1
1
1
1
20. '2 + 22 + 23 + ... + 2" = 1 - 2'"
21. 3° + 3 1 + 3 2 + ... + 3,,-1 = -i(3" - 1).
22.
1
'13 + 32
+ 331 + . . . + 3"1 = '1:(
2 1i
1)
3" .
f
MATHEMATICAL INDUCTION
164
1
23. 12
+ 221 + 321 + ... + n21
[Ch. X
1
<2-
>
11, for n
1.
24. The sum of the interior angles of a polygon of n sides is
(n - 2) . 180°.
26. The number of diagonals of a polygon of n sides is in(n - 3).
26. xn - yn is divisible by x - y.
SUGGESTION.
xn -
yn = x(xn-l _ yn-l)
+ yn-l(x
_ y).
Prove or disprove that the following are general formulas:
+
27. 1· 3 + 2 . 4 + 3 . 5 + . . . + n(n
2)
= t(n 3 + n 2 + 6n - 2).
28. 1·5 + 5·9
9·13 + ... + (4n - 3)(4n
= tn(16n 2
12n - 13).
+
+
1
29.
M
1
1
+ 5· 9 + 9· 13 + ... + (4n
I
i
1
- 3)(4n
+ 1)
,
'
I
+ 1)
n
+
+
4n
1
30. 1· 4
4 .7
7 . 10
= 4n 3 - 3n 2 9n - 6.
31. 12 .2 + 22 • 3 + 3 2 . 4
= t(n 4 lln 2 - 12n 8).
32.
+ ... + (3n -
+
+
2) (3n
+ ... + n (n,+
+
1)
2
+
1 + ~ + ! + . . . +! < 1 + 10.
2
3
n
n
111
33. 13 + 23 + 33
111
34. 12 + 22 + 3 2
111
36. Iii + 22 + 32
+ 1)
I
1
"1'
1
t
+ ...
::
I
.f
+
1
+ n2
+
1
+ nZ
n . ; . ;.,(
<1+
n
+ 1" '.
1
< 1 + li"
By the method of multiplying polynomials we find
+ b)l
(a + b)2
(a + b)3
(a + b)4
= a
= a2
= a3
= a4
l i.U
'
10.2. The binomial formula.
(a
, ;
+ b,
+ 2ab + b
2
,
+ 3a b + 3ab + b
+ 4a b + 6a b + 4ab + b'..
3
3
2
2
2
2
,
3
EXERCISES
165
If n represents the exponent of the binomial a
any of the above expansions, * we see that
+ b in
I. The first term is an.
II. The second term is nan-lb.
n I. The exponents of a decrease by 1 from term to term;
the exponents of b increase by 1 from term to term.
IV. If in any term the coefficient is multiplied by the
exponent of a and divided by the number of the term,
the result is the coefficient of the next term.
We therefore write
,
(a' +
. b)n = an
' n n
+
-1 a -
+ n(n -
- 1)
+ n(n1·2
an-Ib
l)(n - 2) an- b + ...
1
b
l
3
1· 2·3
3
'
This is the binomial formula. Its validity for the case in
which n is a positive integer, will be proved in § 10.5.
Other properties of the expansion to be noted are:
V. The sum of the exponents of a and b is n.
VI. There are n + 1 terms.
VII. The coefficients are symmetric (e.g., the coefficient
of the third term from the beginning is the same .
as that of the third term from the end).
EXERCISES X. B
Assuming that the binomial formula holds, expand the
following:
1. (x
4. (1
7. (a
+ y)5.
+ x)7.
+ b)9.
2. (x - y)5.
5. (x + 1)8.
8. (a - b)9.
3. (a + b)6
6. (1 - X)8.
9. (1 - X)lO.
Expand and simplify:
10. (2x 3 - .!y)4 .
... It is assumed for the present that n is a positive Integer.
MATHEMATICAL INDUCTION
166
(Ch. X
SOLUTION.
(2x 3 - iy)4 = (2X 3)4
+ 4(2x 3)3( -iy) + 6(2x 3)2( -iy)2
+ 4(2x3)(-iy)8 +
= 16x 12 -
-\~X9y
(_-!y)4
+ iX y
6 2 -
+ rfr y 4.
";"X 3y 3
NOTE. In problems of this type do not attempt to simplify
before completing the expansion. In simplifying, always
divide out factors common to numerator and denominator.
12. (3a 2
11. (3x + 4y)3.
14. (-~X2 - 2 y 3) 1'>.
16. (0.3a 3 + 0.2b 2 )O.
18. (2a 2 + tb1!2)8.
-
5b)3.
13. (2X2 +
ib1!3)8.
15. (t a 1!2
2
17. (a b - 2c 2)1.
19. (lOx - 2y)9.
+
y1!2)4.
Find the first four terms of the following expansions:
20. (a + b)20.
~. Expand (x
21. (x
+ y + Z)8.
+ y)60.
SUGGESTION. First consider (x
22. (1
+ y)
+ X)100.
as a single quantity.
"
24. Find, by using the binomial formula, the value of (0.98)4.
SUGGESTION. 0.98
=
1 - 0.02.
Find, by using the binomial formula, the value of ..
25. (2.2)4.
28. (1.04)6.
31. (1.004)8.
26. (1.3) 6.
29. (99)6.
32. (0.998)3.
27. (0.97)4.
30. (105)4.
33. (1003)'.
Find, by using the binomial formula, values of
34. (1.1)16 correct to three decimal places.
35. (1.02)20 correct to four decimal places.
36. (1.03) 25 correct to three decimal places.
37. (0.99) 10 correct to four decimal places.
38. (0.998) 26 correct to five decimal places.
10.3. Pascal's triangle.
The coefficients of the binomial formula can be obtained
by a clever scheme known as Pascal's triangle, exhibited
below. This triangle of numbers, bordered by l's, can be
y
/
§10.41 THE GENERAL TERM OF THE BINOMIAL FORMULA
167
formed by adding together any two adjacent numbers in a
given row to obtain the number between them in the row
next below. The number 1 at the vertex of the triangle
may be thought of as the expansion of (a + b)O = 1, the
second row gives the coefficients of (a + b)l, the third row
gives the coefficients of (a + b)2 = a 2 + 2ab + b2, and
so on.
,"
1
1
1
2
1
'. .' ~
',.
1
,~~-:,
1
1
1
15
21
7
28
8
10
10
5
6
20
70
56
1
5
1
6
21
1
7
28
56
¥
';.f
1
15
35
35
.i::' f! 'J
1
4
6
4
1
1
1
3
3
8
1
10.4. The general term of the binomial formula.
By examining (1) of § 10.2 we may see that the term of
the binomial formula involving br is
n(n_- _l)(n - 2) ._
. ._to
factors
.....;..
_7_
_ _ an-rb r
,1·2'3"'7
'
~~=---:-
(1)
which may also be written
+ 1)
n(n_- _1) (n_- _2).' .. (n _
- _
7
_.:.,
~
.;......~_.:.__
7
!
where the symbol 7 1, called
positive integer, as
7
!
For example,
=
7
.:......~
an-rb r,
(2)
factorial, is defined, for r a
r(r - 1) (r - 2) . . . 3· 2 . 1.
(3)
4!=4·3·2·1.
Actually, the term involving br is term number r + 1,
since b appears in the second term, b.2 in the third, b3 in the
MATHEMATICAL INDUCTION
168
fourth, and so on.
,
lCh.X
I
The rth term is
n(n -- l)(n - 2) . . . to r - 1 iactors n
a -·,+lb,-l.
1 . 2 . 3 . . . (1 - 1) _ _
....:.--~---,,.....-~--~-~---~
(4)
Either term number r + 1 (formula (1) or (2» or term
number r (4) may be called the general term of the expansion of (a + b)n.
EXERCISES X. C
1. Find the 4th term of (2x - iY)10.
SOLUTION.
The 4th term will involve the 3rd power of i
-iYi it is
lO' 9, 8
1.2. 3 (2Ix)7( -iY)' = -240x7 y'.
Find:
2.
3.
4.
6.
6.
7.
8.
9.
10.
11.
12.
13.
14.
16.
16,
The 5th term of (x + y)18.
The 7th term of (3x - 2y)10.
The 6th term of (2x 2 - 5 y 3)15.
The 8th term of (3 a l/ 2 + 2b 1/ 3) 12.
The 6th term of (1 + X)100.
The middle term of (6x - 7y)10.
The middle term of (2a 2 + ib 3 )18.
The term involving b5 in (a + b)u.
The term involving a 5 in (a - b)15,
The term involving y10 in (2x 3 - 3y2)t2. //
The term involving X 12 in (5x 3 + 2y2)~,
The term involving X 3y 5 in (x - y + 1<) 11,
The term involving a 2b5 in (a - b + c)1°,
i
The term involving y4z12 in (x + y - Z)1S,
. .... f
The middle term in the expansion of (as + 2b 2) .. is Ca18b~l.
Find C and n.
/'
-
I
/
10.5. The binomial theorem for positive integral exponents.
We shall now prove that the binomial formula (1) of
§ 10.2 holds when the exponent n is a positive whole num-
ber. This is called the binomial theorem for positive
integral exponents. The proof wj}} be effected by the
I,
§ 10.5J
THEOREM FOR POSITIVE INTEGRAL EXPONENTS
169
process of mathematical induction, explained earlier in
the chapter.
.
Part I. The formula is true for certain values of n, viz.,
n = 1, 2, 3, 4, as has already been seen.
Part II. If it is true for n = k, it is true for n = k + l.
Assume that it is true for n = k. Then,
(a
+ b)k = ak + kak-1b + ...
+ k(k - 1) . . • (k - r + 2) ak-rtlbr-l
f{) ;
1 . 2 . . . (r - 1)
term number r (= T.)
-+- k(k
,.
- 1) . . . (k - r + 1) ak-rb r
l' 2 . . . r
term number r + 1 (- Tr+l)
Multiply both sides by a
i
+ .. :+ Mr
+ b:
+ b)k+l
= a H1 + kakb + ... + k(k
(a
- 1) ... (k - r
l' 2'" r
+ 1) ak-rt1b
r
+ ... + akb + ... + k(k -
1) . ',' (k - T + 2) ak-rt1b r
1 . 2 ... (r - 1)
(bTrl
+ ... + bH1 .
The coefficients of ak- rt1 b" in the terms aTr+1 and bTr
combine as follows:
k(k - 1) . . . (k - r + 2) (k - r + 1)
l' 2 . . . (T - 1)r
+ _k(,__k---::-:--:::1):--._._-;-(k_--c:-;T,-:+_2-,-)
1 . 2 . . . (r - 1)
= k(k - 1) . . . (k - T + 2) (k _ T + 1
1 . 2 . . . (r - l)r
(k + l)k(k - 1) . . . (k - T + 2)'
=
1· 2 . . . r
.
+ r)
MA THEMATICAL INDUCTION
170
[Ch. X
Therefore,
(a
+ b)k+l = ak+l + (k + l)a kb +
+ (k + l)k . . . (k - r + 2) ak-r+1br + ... + bk+l,
1· 2 . . . r
which is precisely what we should get by substituting
k + 1 for n in the binomial formula.
That is, if the formula is true for any value n = k, it is
true for the next value, n = k + 1. But bY' Part I it is
true for n = 1, 2, 3, 4, and consequently must be true for
n = 5,6, . . . .
Therefore we may write, for any positive whole number
n,
(a + b)" = a" + nan-1b + n(n ~ 1) a n- 2b2 + ...
2
+ n(n - 1) . . . (n - r + 1) an-rbr + ... + b".
r !
'(,il
10.6. The binomial series.
If in the binomial formula (1) of § 10.2 we set a = 1,
b = x (that is, if we make use of I-V of that section, in
the same way that we do when n is a positive integer),
we obtain
(1
+ x)n
=
1
+ nx + n(n2!-
+ n(n -
1)
1) . . . (n -
r!
+ ...
'
r + 1)
XT +
x'"
'.
(1)
If n is a positive whole number, this expansion will terminate with x"; otherwise it will continue indefinitely, in
which case it is called the binomial series. It can be
shown that if x is numerically less than 1 (-1 < x < 1),
formula (1) holds for any real value of n, in the sense that
as we include more and more terms on the right, we get
closer and closer to the value on the left. (See Chapter
XX.)
171
THE BINOMIAL SERIES
§10.6]
Example 1.
(a) Find the first four terms of (1) when n = -2.
Substitute x = t and evaluate. '
(b)
SOLUTION.
(a)
(1
+ ~)-2 =
2~ +
1-
.<"::
•
The true value is ( 1
2x
+ 3x
1·2
+
-2(-3)( -4)
1.2.3
4x 3 + . . . .
1
1
1
5
1 - 2 . 4 + 3 . 16 - 4 . 64 = 8 = 0.625.
= 1-
(b)..
-2( -~_) x 2
+ 41)-2 =
8
x+
2 -
(5)-2
4 = (4)2
5 =
16 = 0.64.
25
f
Example 2.
Find the square root of 1.2 by using the first three terms of the
binomial expansion.
SOLUTION.
v'f.2
= (1
+ 0.2)
~
- 1
+
= 1
+ 0.1
1/2
(0.2) )
- 0.005
~~
P
(0.2)'
+
+ ...
= 1.095.
This is correct to three decimal places.
It is not difficult to show that any binomial (a + b)n,
in which a ;;c b, can be expanded. Suppose, for definiteness, that a is numerically greater than b. Then
where b/a is numerically less than 1.
can be expanded by (1).
The last binomial
MATHEMATICAL INDUCTION
172
[Ch. X
Example 3.
Find the square root of 23 by using the binomial formula.
SOLUTION .
. V23 =
(25 - 2)1/2
(1 - T!2 =
;5)
+~(-0.08) )\~2~)
=5[1
= 5(1
= [ 25
- 0.04 - 0.0008
5(1 - 0.08)1/2
(-0.08),+ . . .
+ ...) =
J
5(0.9592) = 4.796,
. which is correct to three decimal places.
EXERCISES X. D
Find the first four terms in the expansion of
1. (1
4. (1
7. (1
+ X)-I.
+ x) lI3.
+ X)-l!2.
+ X)lI6.
10. (1
13. (1 - 3X)O.3.
2.
6.
8.
11.
14.
(1 - X)-2.
(1 + X)2/3,
(1 - X)-1!3.
(1 + 2X)-1/2 •.
(1 + X)-O.7.
3.
6.
9.
12.
16.
.'
•
'
I
(1 + X)1!2.
(1 + X) 114.
(1 + X) 116.
(1 + X)-l.
(1 - 2x 2)-0,l.
Find approximations to the square roots of the following
numbers by using three terms of the binomial series:
16. 1.06.
19. 105.
22. 11.
26. 61.
17.
20.
23.
26.
1.03.
96.
5.
47.
18.
21.
24.
27.
0.98.
28.
37.
82.
Find approximations to the cube roots of the following numbers by using three terms of the binomial series:
28. 1.09.
31. 998.
34.25.
29. 0.98.
32. 7.
36. 65.
30. 1006.
33. 11.
36. 30.
Find approximations to the fifth roots of the following numbers by using three terms of the binomial series:
. 37. 1.04.
38.30.
39. 3000.
CHAPTER XI
Progressions
11.1. Arithmetic progression.
An arithmetic progression is a sequence of quantities,
called terms, each of which (after the first) can be obtained
from the preceding by adding to it a fixed quantity called
the common difference. Thus, the set of numbers 2, 5, 8,
( 1l, for example, is an arithmetic progression. Here the
common difference is 5 - 2 = 8 - 5 = 11 - 8 = 3.
11.2. The nth term of an arithmetic progression.
Let us denote the arithmetic progression by
If the common difference is represented by d, we have, by
definition,
as = al
+ 2d,
a4
= al
+ 3d,
. . . ,
and in general, for the nth term,
an = a1
+ en -
(1)
l)d.
Example.
Find the 10th term of the arithmetic progression 2, 5, 8,
11, . . . .
SOLUTION. alO =
2
+9.3
=
2
+ 27
=
29.
11.3. The sum of an arithmetic progression.
Let us denote the sum of n terms of an arithmetic progression by Sn.
To derive a formula for the sum; we write
173
[Ch. XI
PROGRESSIONS
174
it first in natural order and then in reverse order:
+ (al + d) + (al + 2d) + . . .
+ [al + (n - 2)d] + [al + (n an + (an - d) + (an - 2d) + ...
+ [an - (n - 2)d] + [an - (n -
Sn = al
Sn =
l)d],
l)d).
Adding, we get
2Sn = (al
+ an) + (al + an) + (al + an) +
+ (al + an) + (al + an) =
(since there are n terms).
Sn =
neal
+ an)
Therefore,
n
2' (al
+ an).
(1)
If in (1) we substitute an = al + (n - l)d, as given by
equation (1) of the preceding section, we obtain another
useful expression for the sum, namely,
Sn
n
=
'2 [2a l + (n
- l)d].
(2)
Example 1.
Find the sum of 10 terms of the arithmetic progression 2, 5,
8, . . . .
SOLUTION.
Substitute
alO
=
10
810 = "2 (2
29 (found above) in (1):
+ 29)
= 155.
Example 2.
Find the sum of 9 terms of the arithmetic progression 25, 21,
17, . . . .
SOLUTION. al
= 25, d = -4,
9
8 9 = 2 [2, 25
n
= 9. Substitute in (2):
+ 8( -4)]
= 81.
175
EXERCISES
11.4. Arithmetic means.
The terms between any two given terms of an arithmetic
progression are called arithmetic means between the given
terms.
Example.
Insert 3 arithmetic, means between 7 and 13.
SOLUTION. a1 = 7, as = 13 (n = 5, since the progression is
composed of the 3 means and the first and last terms). Substitute in formula (1) of § 11.2:
13 = 7
+ 4d,
d
= ll.
The progression is 7, 8t, 10, Ht, 13.
EXERCISES XI. A
Find a.. and 8 .. for the following progressions:
1.
3.
6.
7.
S.
9•
11.
12.
13.
14.
16.
16.
17.
18.
19.
20.
21.
22.
23.
24.
26.
26.
2, 7, 12, . . . (n = 10).
2. 37, 33, 29, . . . (n = 10).
3, 24, 45, . . . (n = 20).
4. 2t, 4, 5i,
(n = 12).
13, 9, 5, . . . (n = 15).
6. 3, 11, 19, . . . (n = 150).
0.6, 1.7, 2.8, . . . (n = 100).
5, 12, 19, . . . (n = 250).
1
1
1
• • • (n - 11). .
10 11",
1 4,
1 6,
1
• • • (n - 21)
T5",
T;f,
TO,
.
Given a1 = 5, d = 3t, an = 54; find nand 8 .. .
Given a1 = t, d = t, an = _1-/; find nand 8 .. .
Given a1 = 7, al8 = 35; find d and 8 n •
Insert six arithmetic means between 7 and 12.
Insert eleven arithmetic means between 3 and 30.
Given a12 = 37, d = 3; find a1 and 8 12 •
Given a65 = 100, d = i; find a1 and 8 6 5.
Given a] = 15, 8 10 = 120; find d and a10.
Given al = 120, 8 7 = 210; find d and a7.
Given au = 25, 8 13 = 39; find a1 and d.
Given r/ = 5, 8 g = 75; find al and ag.
Given d = It, 8 10 = 175; find al and alO.
Given a1 = 2, d = 8, 8 .. = 90; find n and a ...
Given a1 = 174, d = -6,8 .. = 1794; find n and a".
Given a1 = 20, d = -It, 8 .. = 108i; find n and an.
Given a1 = 8, a .. = 62, 8 .. = 210; find nand d.
PROGRESSIONS
176
27.
28.
29.
30.
31.
32.
33.
34.
36.
36.
[Ch. XI
Given al = 9, a" = 37, S" = 207; find nand d.
Given an = 200, d = 4i, Sn = 1952i; find nand al . . ' .,
Given as = 17, a12 = 122; find as.
Given al9 = 125, aao = 169; find a26.
Given a9 = 25, S9 = 153; find al2.
Determine x so that x + 2, 2x, 4x - 7 will be an arithmetic
progression.
Determine x so that 4x, 5x + 4, 3x 2 - 1 will be an arithmetic progression.
Derive a formula for the sum of the first n integers.
Derive a formula for the sum of the first n even integers.
Derive a formula for the sum of the first n odd integers.
Derive formulas for
37. al and Sn in terms of d, n, an.
38.
39.
40.
41.
42.
43.
44.
,
d and an in terms of al, n, Sn. '" . .
al and d in terms of n, an, S". .
i
al and an in terms of d, n, Sn.
d and n in terms of aI, an, Sn.
~
nand Sn in terms of al, d, an.
d in terms of al, n, an.
Prove formula (2) of § 11.3 by mathematiDal
~
'I
)~
~;
"
I
I
,I
induction.1:
,
". ;
11.5.
Geometric progression.
A geometric progression is a sequence of quantities,
called terms, each of which (after the first) can be obtained
from the preceding by multiplying it by a fixed quantity
called the common rati.o. The quantities 2, 6, 18, 54, for
example, form a geometric progression. Here the common ratio is ! = ¥ = ~ ~- = 3.
11.6.
The nth term of a geometric progression.
If the geometric progression is
and the common ratio is r, it follows from the definition
that
/
§ 11.7]
THE SUM OF A GEOMETRIC PROGRESSION
177
and in general,
(1)
Example.
Find the 5th term of the geometric progression 2, 6, 18,
SOLUTION. The ratio is
~
or
=
a5
-y.
= 3.
",
Therefore,
2 . 3 4 = 162.
11.7. The sum of a geometric progression.
Denoting the sum of n terms of a geometric progression
by Sn, we write
Sn = al
+ air + alr + ... + alr n - + alr n2
2
1
•
(1)
Multiply by r:
Subtract (1) from (2):
rS n - Sn = alr n - ai,
(r - I)S" = al(r n - 1).
Provided r
~
1,
Sn
rn - 1
=
(3)
a1 T- l'
If r is numerically less than 1, it is convenient to write
(3) in the form
'
'
1 - Tn
Sn = a1 1
.. '
(4)
/'
- r
! (
EXERCISE
Show that
Sn
= ran - a1
r-l
=
a1 -
ran.
1-r
(6)
Example.
Find the sum of 5 terms of the geometric progression 2, 6,
18, . . . .
PROGRESSIONS
178
[eh. XI
SOLUTION. By (3),
86
I
=
35 - 1
2 . 3 _ 1 = 242.
,
.!
I
11.8 Geometric means.
The terms between any two given terms of a geometric
progression are called geometric means between the given
terms.
J
Example.
,:1'
,:
'!'';!
Insert 3 geometric means between 16 and 81.
SOLUTION. al = 16, a5 = 81.
Substitute in the formula
81
16
r 4 =-,
r =
3
+ _.
-2
±ii, also, but only real values will be considered.)
are two progressions:
- (r =
16, 24, 36, 54, 81;
There
16, -24,36, -54, 81.
EXERCISES XI. B
Find an and 8 n for the following progressions:
1.
3.
6.
6.
7.
9.
10.
11.
12.
13.
14.
16.
2, 10, 50, . . . (n = 6).
2. 2,6, 18, . . . (n = 8).
4, -12,36, . . . (n = 7).
4. 16,24,36, . . . (n = 9).
768, 384, 192, . . . (n = 10).
-54,36, -24, . . . (n = 6).
i, t, -h, ... (n = 5).
8. 2, -j, t, ... (n = 6).
3, 0.3, 0.03, . . . (n = 11).
3, -0.3, 0.03, . . . (n = 11).
1, 1.02, (1.02)2, . . . (n = 5).
2, 2v'~ 10, . . . (n = 12).
30, -6, 60, ... (n = 12).
6 - 3y'3, 3, 6
30, ... (n = 6).
Given al = 27, r = Ii, an = 208t; find nand 8 n •
+
NOTE. Consider only real values in the following exercises.
16. Given
17. Given
al
al
= 128, a5 = 162; find rand 8 5•
= 32, a6 = 162; find rand 8 5•
\
§ 11.9]
INFINITE GEOMETRIC PROGRESSION
179
Given al = 7000, a6 = 2.24; find rand S6.
Insert three geometric means between 6 and 14,406.
Insert four geometric means between 6250 and - 486.
Given a5 = 16, r = i; find al and S5.
Given a7 = 729, r = i; find al and S7.
Given r = 2, S9 = 1533; find al and a9.
Given r = i, S6 = 1477; find al and a5.
Given al = 50, r = 0.2, Sn = 62.496; find n and an.
Given al = 6t, r = -It, Sn = 15ft; find n and an.
Given al = 11, an = 2816, r = 2; find nand Sn.
Given al = 6, r = 7, an = 705,894; find nand Sn.
Given al = 12Ii, an = 512, Sn = 1683i; find rand n.
Given al = -15, an = 405, Sn = 300; find rand n.
Given r = -4, an = 13,312, Sn = 10,660; find al and n.
Given r = -i, an = 1728, Sn = 12,501; find al and n.
Given al = 9, S3 = 279; find r and a~.
Given a1 = 1458, alO = 39,366; find al2.
Given a7 = 156, al9 = 7644; find an.
Given a4 = 3, al3 = 10; find a22.
Determine x so that x, 2x
7, lOx - 7 will be a geometric
progression.
38. Determine x so that 11 - x, 2x - 1, 9x + 3 will be a
geometric progression.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
+
Derive formulas for
39.
40.
41.
42.
43.
44.
45.
rand Sn in terms of aI, an, n.
al and an in terms of r, n, Sn.
al and Sn in terms of r, n, an.
an in terms of aI, r, Sn.
al in terms of r, an, Sn.
r in terms of aI, an, Sn.
Prove formula (3) of § 11.7 by mathematical induction.
11.9. Infinite geometric progression.
Consider a line segment 1 unit in length. (Fig. 11.1.)
Suppose that we take half of the line, then half of the
remaining part, then half of what still remains, and so
on. As we take more and more parts, their sum will
approach 1 in value, since what remains of the original
PROGRESSIONS
180
[Ch. XI
unit segment is shrinking to nothing. But the lengths of !.,
the segments which we are taking are the terms of the
geometric progression, t, i, ..g., •.• , whose ratio is t.
Thus, the sum of this progression,
as we take more and more t,erms,
o
% 7s 1 approaches the value 1; that is,
FIG. 11.1.
it can be made to differ from 1 by
an amount as small as we please. i
A geometric progression in which the number of terms
increases without limit is called an infinite geometric progression, or infinite geometric series. If the ratio is i
numerically less than 1, the sum of n terms of the progression, as n increases without limit, approaches a perfectly
definite value, called the limit, or sum, of the progression.
(Ob.';(,l'\"e that it is not a sum in the ordinary sense.)
For, from (4) of § 11.7, we see that
1 - rn
a1
a1
Sn = a1 ------- = - _ - - _ rn.
l-r
l-r
l-r
(1)
Now if the ratio is numerically less than 1 (Irl < 1), the
numerica 1 valuE' of rn decreascl'l as n increases, and by
taking n sufficiently large we ean make II nl, and consequently ialr"/(l - r)/, as small as we please. That is, as
n increases without limit, the last fraction in (1) shrinks
toward zero ill Y:llue, and the SUIll of n terms of the progression approaches the yaluc adC1 - r). In symbols, we write
S =
'"
a1
-~.
1 - T
(2)
Example.
Find the" sum " of the infinite geometric progression 2, j,
~,
....
-
SOLGTIOX.
-
'I
I-
I
REPEATING DECIMALS
§11.10]
181
EXERCISES XI. C
Find the" sums" of the following infinite geometric progressions:
____ 1
1.
3.
6.
7.
9.
11.
13.
14.
15.
17.
18.
2, t, -,fi,
25, 10, 4, . . . .
~-,
1, t, 7J\, . . . .
98, 42, 18, . . .
24, -16, 10-§-, ..
5, _2b'\ lh\ .. .
10. 5, - .y5, I, .. .
2.
4.
6.
8.
-Yo, 5'\, . . .
27, -18, 12, . . .
5, y'5, I, . . . .
I':j
Y6
60,4,
8 ""
12. y'6, -.y3,
6,.
2
3v177
5 + 20,1 + 30, 8 - 2.y3, . . . .
4 + -yi2, 8 - 5.y2, 44 - 31V/2, . . . .
O.S, 0.05, 0.005, . . . .
1EL '2, -0.0'2,0.000'2,,1, (1.04)-\ (1.04)-2, . . . .
(1.02)-1, (1.02)-2, (1.02)-3,
tM4~Xt~·.~..•
If
11.10. Repeating decimals.
A repeating decimal is an infinite geometric progression. Thus, 0.444 . . . (sometimes written 0.4) is the
progression
0.4
+ 0.04 + 0.004 + 0.0004 +
in which al = 0.4, r = 0.1. Likewise, the repeating
decimal 0.235,235,235 . . . (written 0.235) is the progression
0.235
+ 0.000,235 + 0.000,000,235 +
in which al = 0.235, r = 0.001.
Example 1.
Find the limiting value of 0.444
SOLUTION.
s =
00
_!!.!:_
1 - r
=
0.4
0,4
4
1 - 0.1 = 0.9 = ~r
This may be checked l>y dividing 4 hy 9.
PROGRESSIONS
182
[Ch. XI
Example 2.
, i
Find the limiting value of 0.743.
SOLUTION.
0.74:3 = 0.7
+ 0.043 + 0.00043 +
This is composed of 0.7 and a geometric progression in which
al = 0.043, r = 0.01. Thus,
Ii 0743 = !_ + 0.043 =!_ + 0.043 ~-'""-'-'-\ j
m .
10
1 - 0.01
10
0.99
7
43
736
368
.'/ J'i
i
= 10 + 990 = 990 = 495'
,,j" ~
Although the formula for S", can be used to find the
limiting value of a repeating decimal, perhaps the best
method is that used in developing the formula for the
sum of a geometric progression.
'I
• ; ..,.
~,
!
.i"
Example 3.
Find the limiting value of 0.444 . . . ,
SOLUTION.
i
Let
MUltiply by 10:
(2) - (1)
" ';<
x = 0.444 . , , .
lOx = 4.444 . " ' ,
9x = 4,
4
X =-.
9
Example 4.
(1)
(2)
/
1/
"y
,
'
Find the limiting value of 0.743.
//./
SOLUTION. Let
x= 0.74343 ' , ..
100x
= 74.34343 ' . . .
Multiply by 100:
(4) - (3)
99x = 73.6,
73.6
736
368
x = 99 = 990 = 495'
'I
(3)
(4)
/'
EXERCISES XI. 0
Find the limiting values of the following repeating C:ecimals:
1. 0.444 .
4, 0.16.
2, 0.888 .. , .
3. 0.999 .
6.0.25.
6.0.52.
/
HARMONIC PROGRESSION
§11.11J
8.
11.
14.
17.
20.
7.0.64.
10. 0.23i.
13. 0.285714.
16. 0.603.
19. 0.303.
0.40.
0.23i.
0.571428.
0.030.
0.4004.
183
9.
12.
16.
18.
21.
0.23i.
0.2573.
0.5363.
0.300.
O. i2345.
:- 11.11. Harmonic progression.
A harmonic progression is a sequence of terms whose
reciprocals are in arithmetic progression. For example,
t, t, t, t is a harmonic progression, since 2, 4, 6, 8 is an
arithmetic progression.
Example 1.
Find the 7th term of the harmonic progression t, t, -1, . . . .
SOLUTION. Form the corresponding arithmetic progression:
4, j, -1, . . . .
d = -2.5,
N ow, proceeding as in § 11.2, we write
a7 = al
+ 6d = 4 + 6(-2 ..5)
=
-11.
Take reciprocal: - 111'
The terms of a harmonic progression between any two
given terms are called harmonic means.
Example 2.
Insert 4 harmonic means between
SOLUTWN.
-! and +,.
In the corresponding arithmetic progression,
al = 2,
a6 = 17.
17 = 2 + 5d,
d = 3.
Arithmetic progression: 2,5,8, 11, 14, 17.
Harmonic progression: i, i, t, ft, -h,
n.
There is no simpler method of finding the sum of a
harmonic progression than to add ·the terms one by one.
PROGRESSIONS
184
I
[Ch. XI
I
EXERCISES XI. E
1. Find the 24th term of the harmonic progression t,
t, t,
2. Find the 32nd term of the harmonic progression, -I, t,
-y\, . . . .
3. Find the 10th term of the harmonic progression 6, 3, 2,
4. Find the 16th term of the harmonic progression 12, 6,
4, . . . .
5. Find the 15th term of the harmonic progression i, t, t, ....
6. Find the 8th term of the harmonic progression t, t, 5,
7. Insert seven harmonic means between 2 and 10.
8. Insert nine harmonic means between 5 and 30.
9. The 3rd term of a harmonic progression is 15, the 9th term
is 6.
Find (a) the 11th term, (b) the 14th term.
10. The 6th term of a harmonic progression is 12, the 10th
term is 6.
What is the 18th term?
5, 2x will be a harmonic
11. Determine x so that x, x 2
progression.
12. Determine x so that x-I, -l-(2x + 3), 3x + 1 will be a
harmonic progression.
13. Derive a formula for the harmonic mean of two numbers
a and b.
MISCELLANEOUS EXERCISES XI. F
1.
2.
3.
4.
6.
6.
7.
8.
9.
.
'.
Find as and 8 8 for the progression 5, 12, 19,
Find a5 and 8 5 for the progression 27, 45, 75, ..••
Find a6 and 8& for the progression 162, 108, 72, ..•
Find a15 and 8 15 for the progression 100, 82, 64, .•
Find a25 and 8 25 for the progression I, t, 1, . . . .
Find a5 and 8; for the progression t, t, H, ... .
Filld the next term of the progression 24, 12, 8, .•
Find a4 for the progression 12, 16, 24, . . . .
What is the next term of the progression 16, 8, .•. if it
is (a) a<'ithmet,ic? (b) geometric? (c) harmonic?
10. Insert (a) three arithmetic means, (b) three geometric
means, (c) three harmonic means between 2 and 162.
11. A particle sliding dmvn an inclined plane travels 2 feet the
first second. In any ",econd after the first it slides 4 feet
farther than it did in the preceding second. (a) How far
//
EXERCISES
12.
13.
14.
16.
16.
17.
18.
19.
20.
185-
will it slide in the 7th second? (b) How far will it slide in
7 seconds? (c) How long will it take to slide 128 feet?
A rotating wheel coming to rest makes 12.7 revolutions the·
first second. In any second after the first it makes ! as.
many revolutions as it did in the preceding second. How
many revolutions does it make (a) in the 5th second'? (0) in
5 seconds? (c) before stopping?
The bob of a pendulum is released, and on the first swing
it travels 100 centimeters. On each subsequcnt swing it,
travels 0.99 as far as it did on the preceding swing. How
far does it travel before coming to rest?
The tickets in a raffle are numbered 1, 2, 3, and so on.
The price of a ticket is the number of cents eqllal to the
number on the ticket. If the raffled article cost $100, what
is the least number of tickets that must be sold so that thm;e
conducting the raffle will not lose money?
Find the sum of all of the odd integers between 100 and
1000.
Find the sum of all of the even integers between 999 and
9999.
Find the sum of all of the positive integers less than 1000that are exactly divisible by 7.
A circle has a radius of 16 inches. In it is imicribed an
equilateral triangle. In the triangle is inscribed another
circle, in the second circle another triangle, and so on.
Find (a) the area of the 6th circle, (b) the sum of the areas,
of all of the circles, (c) the sum of the areas of all of the
triangles.
The side of a square is 16 inches. A second square is
form,~d hy joining, in the proper order, the midpoillts of
the d,:, 0 " [ the first square. A third square is formed by
joinin~ the midpoints of the second square, and so 011.
Find t,ll" side of the eleventh square.
Two lines meet at a certain angle. From a point Oil the
first line, I 0 inches from the vertex of the angle, a IWI'pendicular is dropped to the second line. From the foot
of the perpendicular another perpendicular is dropped
back to the first line, then another perpenuicular is dl"oPlwd
to the second line, and so on. Find the slim of tIl(' IplIgths
186
21.
22.
23.
tI.
~;
••
';,'
'-I
,
26.
27.
PROGRESSIONS
[Ch. XI
of all of Ule perpendiculars when the angle is (a) 45°, (b)
30°, (<') 60°.
A machine costing $1250 depreciates in value 20 per cent
each year. What will it be worth at the end of 4 years?
A colony of bacteria is increasing at the rate of 40 per cent
per hour. If there were originally 25,000 bacteria in the
colony, how many will there be at the end of 5 hours?
A 16-gallon container is filled with pure acid. Fou:- gallons
are drawn off and replaced with water. Then 4 gallons
of the mixture are drawn off and replaced with water, and
so on until 5 drawings and 5 replacements have been made.
Find the amount of acid in the final mixture.
For a cylindrical cistern 4 feet in diameter filled with water,
the amount of ,\\rork done in pumping out the water is 392
foot-pounds for the first foot of depth, 1176 foot-pounds for
the next foot, 1960 for the third foot, and so on. If the
cistern is 10 feet deep, how much work is done in pumping
out all of the water?
When a rectangular plate is submerged in water so that
the upper edge is in the surface of the water, the force on
(one side of) the plate is ilw for the first foot of depth,
where l is the horizontal dimension, in feet, of the plate
and w is the weight of 1 cubic foot of water (62.4 pounds,
approximately). The force on any foot after the first is
lw pounds more than that on the preceding foot. Find
the force on (one side of) such a plate which measures 6 feet
horizontally and 12 feet vertically.
The pitch of the musical note given out by a vibrating
string depends on the number of vibrations per second made
by the string. For a note one octave higher than the given
note the number of vibrations per second is twice as great
as that for the given note. If the number of vibrations
per second that gives middle C is 256, what is the number
of vibrations per second that will give (a) a note 5 octaves
above middle C? (b) a note 4 octaves below middle C?
Five weights are in geometric progression, the first being
1 pound and the ratio being 2. They are placed on a rod,
whose weight may be neglected, the first being at a distance
of 1 inch from a fulcrum, the second 2 inches from the
l'
'.
j
187
EXERCISES
fulcrum, the third 4 inches, and so on, these distances being
in geometric progression. How far from the fulcrum, on
the other arm of the rod, must one place a weight, equal to
the sum of the other weights, to effect a balance?
28. The sum of three numbers in geometric progression is 52.
If 8 is added to the middle number, the other two being
left unchanged, an arithmetic progression is obtained.
Find the geometric progression.
29. The sum of three numbers in arithmetic progression is 33.
If the numbers are increased by 2, 1, and 6, respectively,
the new numbers will be in geometric progression. Find
the arithmetic progression.
SUGGESTION.
Let the numbers be x - d, x, x
+ d.
30. The sum of four numbers in arithmetic progression is 42,
the sum of their squares is 566. Find the numbers.
31. The geometric mean and the harmonic mean of two numbers are 12 and 7i respectively. What are the numbers?
32. Three numbers are in harmonic progression. If the third
number were decreased by 4 they would be in arithmetic
progression. If the third number were decreased by 3 they
would be in geometric progression. Find the numbers.
33. Prove that if a 2, b2 , c2 is an arithmetic progression, then
b + c, c + a, a + b is a harmonic progression.
34. The first and second terms of a progression are a and b
respectively. What is the third term if the progression i~
(a) arithmetic? (b) geometric? (c) harmonic?
•
CHAPTER XII
Complex Numbers
. I
12.1. Imaginary and complex numbers.
In §§ 5.16 and 5.17, of which the present chapter is a
re\'iew and a continuation, we introduced the imaginary ,
unit i haying the property i2 = -1, assuming that this
new number obeys all of the laws of addition and multiplication that hold for real numbers. Since i 3 = i2 . i =
-i,i 4 =(z:2)2=(-1)2=I,i 5 =i 4 'i= i,'" ,itisseen
that the successive integral powers of i run repeatedly
through the cycle i, -1, -i, 1.
A nurnLer of the form a + bi in \vhich a and b are real
numbers is called a complex number. The number a is
called the real component, or real part, of the complex
number; the number b is called the imaginary component,
or imaginary part. Note that the imaginary component
is a real number.
Two complex numbers a + bi and c + di are equal if,
and only if, a = c and b = d, that is, if, and only if, their
real components are equal and their imaginary components are equal.
If b ~ 0, the complex number a + bi is called an
imaginary number. If b ~ and a = 0, the complex number reduces to the form bi, which is called a pure imaginary
number. If neither a nor b is zero, a + bi may be called
a mixed imaginary number.
If b = 0, the complex number reduces to the real number a (which may be zero).
The two complex numbers a + bi and a - b'i, which
differ only in the signs of their imaginary parts, are called
°
188
THE NUMBER SYSTEM OF ALGEBRA
912.21
189
conjugate imaginary numbers, either being the conjugate
of the other.
12.2. The number system of algebra.
The set of all complex numbers constitutes 01(' p.'lmber
system of algebra. The system is divided and !:luhdidded
as here shown:
.
{Integers
Rahonal numbers R at'lOna1rae
f t '10ns
Reahiumbers
{AlgebraiC irrs.{
(b - 0)
I fib
tional numbers
rra lOna num ers 1Transcendental
Complex numbers
t numbers
(a
bi; a, b real)
Pure imaginary numbers (a = 0)
Imaginary numbers Mixed imaginary numbers
(
+
(b ;;C- 0)
(a ;;C- 0)
The numbers a and b may be positive or negative.
An integer is a whole number or zero, that is, anyone
of the numbers 0, ± 1, ±2, ±3, . . . .
A rational fraction is a fraction in which the numerator
and denominator are whole numbers, of which the numerator is not exactly divisible by the denominator, e.g. iAn irrational number is any real number \yhich is not
rational.
An algebraic number is one ,vhich i:- the root of an _"
algebraic equation of the type
aoX n
+ ajx n - 1 + . . . + an_IX + an
= 0
(1)
in which the a's are rational numbers and n is a po~itivl~
integer. An example of an algebraic number is V2, since
it is a root of the equation X2 - 2 = O. It is not to be
inferred that all algebraic numbers are irrational, since every
rational number alb is a root of an equation bx - a = O.
An example of a complex number is 2 + V3 . i, which is
a root of the quadratic equation x 2 - 4x + 7 = O.
A transcendental number is a real number which cannot
be the root of an algebraic equation of the type (1), e.g. 7r, e.
190
,
COMPLEX NUMBERS
[Ch. XII
+ bi,
a and b are not
In an imaginary number a
restricted to be algebraic.
EXERCISES XII. A
Give five examples of
1.
3.
6.
6.
Pure imaginary numbers. 2. Mixed imaginary numbers.
Rational numbers.
4. Irrational numbers.
Transcendental numbers.
Complex numbers in which the real part and the imaginary
part are both integral.
Give an example of
7. A complex number in which the real part is irrational and
the imaginary part is integral.
. S. A complex number in which the real part is a rational fraction and the imaginary part is irrational.
9. A complex number in which the real part and the imaginary
part are both irrational.
10. A complex number in which the real part is transcendental
and the imaginary part is an algebraic irrational number.
12.3. Operations with complex numbers.
We add or subtract complex numbers by adding or
subtracting their real parts and imaginary parts separately. Thus,
(a
(a
+ bi) + (e + di)
+ bi) - (e + di)
+ e) + (b + d)i,
e) + (b - d)i.
= (a
= (a -
We multiply complex numbers according to the laws
for real numbers, simplifying results by making use of
the relation i2 = -1. Thus,
(a
+ bi)(e + di)
= ae
=
+ adi + bci + bdi
(ae - bd)
2
+ (ad + be)i.
Division of complex numbers can be accomplished by
writing the quotient in fractional form and multiplying
'
EXERCISES
191
numerator and denominator by the conjugate of the
denominator. Thus,
+ bi
c + di
a
a + bi c - di
ae - adi + bci - bdi 2
= e + di . e - di =
e2 - d 2i2
(ae + bd) + (be - ad)i
=
e2 + d 2
ae + bd
be - ad .
.
= e2 + d2 + e2 + d 2 ~ (e + d't ~ 0).
EXERCISES XII. B
Perform the indicated operations and reduce to the form
a
+ bi:
1.
3.
5.
7.
9.
10.
11.
12.
(5 + 7i) + (8 + 2i).
2. (3 + 4i) + (6 + 5i).
(11 + 2i) + (3 - 14i).
4. (19 - i) + (2 - 13i).
(15 + 2i) - (6 + 4i).
6. (20 - lli) - (25 + 2i).
(-10 + 9i) - (3 - 7i).
8. (6 - 14i) - (-10 + 17i).
(1 + iy'3) - (7 + 20 . t").
(0 + iy'3) - (6y2 - 70· i).
(2 + 6i) + (6 - 5i) - (4 - Hi).
(6 + 3i)(2 + 4i).
13. (9 + 8i)(7 ::t 6i).
+ 3i)(2 + 5i).
(2 + i0)(5 - 60· i).
14. (6
16.
17.
18.
19.
21.
23.
25.
26.
27.
28.
29.
30.
32
.
34.
15. (5 - 3i)(7
~ 2i).
(70 - 60· i) (2y2 - 7"\/'5 . i).
(3 + 2i)(7 - 8i)( -2 + 9i).
(5 + 3i)2.
20. (3 - 2i)s.
U1,j'
(6 + 7i) + (4 - 3i).
22. (8 - 3i) + (9 + 41).
<'K'
(2 - Hi) + (3 - 2i).
24. 5i + (6 - 7i).
(0 - i0) + (20 30· i).
(8y'S + 210· i) + (20 - 70· i).
(v'6 + 600 . i) + (2V6 - 300 . i).
(0 + i0) + (0 + h/7).
(2 + 3i)(3 - 4i) + (4
5i).
(5 + 6i)2 + (9 - 2i).
31. (-5 + 8i) + (3 + 2i)2.
33 (10i - 3)(4i + 3).
8 + 7i.
(5 + 6i) (5 - 2i)
.
9i - 8
(1 + 2i)(2 + 3i)
(3 - 5i)(7 + 4i)
(3 + 4i)(4 + 5i>'
35. (5 + 3i)(6 - i) .
+
+
192
COMPLEX NUMBERS
[Ch. XII
36. Show that the sum of two conjugate imaginary numbers
is real.
37. Sh0\Y that the difference of two conjugate imaginary numbers is a pure imaginary number.
38. Show that the product of two conjugate imaginary numbers is real and positive.
39. Show that the real part of the quotient of two conjugate
imaginary numbers is numerically less than or equal to 1.
40. Show that the imaginary part of the quotient of two conjugate imaginary numbers is numerically less than 1.
12.4. Geometric representation of complex numbers.
The complex number x + yi may be represented by the
point P whose coordinates are (x, y). (See Fig. 12.1.)
When complex numbers are so
represented, the x-axis is called
the axis of real numbers and
the y-axis the axis of imaginary
11
numbers. All real numbers lie
on the axis of real numbers, all
pure imaginary numbers lie on
the axis of imaginary numhers.
-2i
y'
The complex number x -+- yi
FIG. 12.1.
may also be represented by
the line OP going from the origin to the point P. Such lit
line, having length and direction, is called a vector. Vectors are important in physics and mechanics.
12.5.
Geometric addition and subtraction of complex numbers.
Let the complex numbers a + bi and c + di be represented by the points 11,1 and N respectively, and their sum,
(a + c) + (b + d)i, by the point P. (See Fig. 12.2.)
Draw OM, ON, 1~1P, NP; drop NQ, MR, PS perpendicular
to OX; and draw MT perpendicular to PS. Then MT =
RS = OS - OR = (a + c) - a = c = OQ, TP = SP -
I
I
!
EXERCISES
193
ST = (b + d) - b = d = QN. Therefore the right triangles MTP and OQN are congruent, and MP = ON.
Also LTPM = LQNO and NIP is parallel to ON. Quadrilateral 0 MP N is a parallelogram, since two sides are both
,equal and parallel.
Thus, to add two complex numbers graphically, or geometrically, cornplete the parallelograrn having as adjacent
s~'des the lines drawn frorn the origin to the points representing
the two nurnbers. The fourth vertex of y
this parallelogram will be the point representing the sum of the two numbers. *
If we think of the complex numbers
a + bi and c + di as represented by
the vectors OM and ON in Fig. 12.2,
the sum of the numbers will be the
FIG. 12.2.
vector OP.
To subtract c + di from a + bi graphically, we may
add a + bi and -c - di. Note that the negative of a
complex number represented by the point P can be
obtained by extending PO to a point pI such that OP and
OP' are equal in length but opposite in direction.
EXERCISES XII. C
Perform the indicated operations geometrically:
1. (6 + 2i) + (2 + 5i).
2. (3 - i) + (8 + 9i).
3. (-4 - 3i) + (4 + 7i).
4. (2 + 3i) + (4
6i).
6. (3 + 7i) - (5 + i).
6. 2i + (10 - 9i).
7. 5 - (2 + 6i).
8. (12 - l5i) - (4 - 8i).
9. (7 + 3i) - (3 + 7i).
10. (7
2i) - (7 - 6i).
11. (14 - 6i) - (7 - 3i).
12. (15 - Hi) + (10 + Hi).
13. 10 - (4 + 3i).
14. Wi - (4
3i).
15. (7 + 3i) + (8
i) + (5 - 4i).
16. (5 + 2i) + (4 - 6i) - (3 + 7i).
17. (10 + 3i) - (5 - 5i) - (8 - 2i).
+
+
+
+
* If the origin 0 and the points M and N lie in the same straight line, the
point P will lie in this straight line and OP will be equal to OM + ON.
COMPLEX NUMBERS
194
[Ch. XII
iI
12.6. Trigonometric form of complex numbers. *
II
Let the complex number x + yi be represented by the
point P in Fig. 12.3. Let OP = r, and let the angle XOP
be designated by A. Then
'
x
=
y = r sin A,
r cos A,
and the complex number may be written
x + yi = T(COS A + i sin A).
(1)
This last form, which may be conveniently abbreviated
T cis A, is called the trigonometric, or polar, form of the
number. The angle A is called the amplitude, or argup(X+y1..) ment, of the • number, the length
y
r = vix 2 + y2 IS called the modulus,
or absolute value, of the number.
Note that tan A = y/x. For A we
X
.
usually select the smallest angle satlsFIG. 12.3.
fying the relation 0° ;;:; A < 360 0 •
The definition of absolute value of a complex number is
consistent with that given for thE; absolute value of a real
number. (See § 1.4.) For if x + yi is real, y is zero and
viX2 + y2 becomes VX2, which is equal to Ix!. The absolute'
value of x + yi, namely vi X2 + y2, is also written Ix + yil.
EXERCISES XII. Dt
Reduce to trigonometric form:
1. -5
},
+ 5i.
"
.
\
SOLUTION.
r
, f~:,
= y(-5)2
tan A =
-5
+ 5i =
~5 =
+ 52 =
-1,
5y2 (cos 135°
50.
..'
A = 135°.
i
/
+ i sin 135°).
(The student is advised to plot the point as an aid i1\
determining the angle.)
* The remaining sections of this chapter presuppose a knowledge of trlgonometry.
t Table VIII at the back of the book may beuscd in solving these exercises.
I
MULTIPLICATION AND DIVISION
§12.7]
2. 3
+ 4i.
195
.' ;:.
SOLUTION.
T
=
tan A =
3
y'3 2
4
3=
42
1.3333,
A == 53.1° (from trigonometric tables).
5(cos 53.1° + i sin 53.1°).
,I
,
+ 4i =
3. 7 + 7i.
6. V3 - i.
9. Si.
12. 7 + 24i.
16. y'5 + iV2.
18.
+ = 5.
4.
7.
10.
13.
16.
19.
t + ti.
1 - iV3.
-6 - 6y'a· i.
12 - 5i.
-9 - 9i.
i.
v'7 -
i - -hi.
5.
8.
11.
14.
4 + 2i.
-4.
-S + 15i.
-21 - 20i.
17. 0 + 20 . i.
20. -lr; + ii.
'.;
+ bi):
S(cos 30° + i sin 30°).
5(cos 225° + i sin 225°).
Reduce to rectangular form (i.e., the form a
21.
23.
26.
27.
28.
29.
31.
33.
35.
37.
lO(cos 60° + i sin 60°).
22.
6(cos 45° + i sin 45°).
24.
4(cos 135° + i sin 135°).
26.
v'2 (cos 315° i sin 315°).
V3 (cos 300° + i sin 300°).
cos (-60°) + i sin (-60°). 30.
5(cos 270° + i sin 270°).
32.
lOO(cos 100° + i sin 100°). 34.
9(cos 128° + i sin 12S0).
36.
20(cos 180° + i sin 180°). 38.
+
2(cos 120 0
+i
sin 120°).
3(cos 90° + i sin 90°).
6(cos 240° + i sin 240°).
12(cos 256°
i sin 256°).
5(cos 500° + i sin 500°).
15 (cos 345° + i sin 345°).
+
12.7. Multiplication and division of complex numbers in
trigonometric form.
Complex numbers expressed in trigonometric form can
be multiplied by a very simple formula. For example,
rl(cos Ai + i sin Ai) . r2(cos A2 + i sin A2)
= rlr2[cos Ai cos A2 - sin Al sin A2 + i(sin Al cos A2
+ cos Al sin A 2 )]
= rIr2[cos (AI + A 2 ) + i sin (AI + A 2 )].
(1)
That is, the product of two complex numbers is a complex
number whose absolute value is the product of their absolute
values and whose amplitude is the sum of their amplitudes.
196
[Ch. XII
COMPLEX NUMBERS
Division is the inverse of multiplication, and it can
readily be shown that, inversely, the quotient of two complex
numbers is a complex number whose absolute value is the
absolute value of the dividend divided by the absolu.te value
of the divisor and whose amplitude is the amplitude of the
dividend minus the amplitude of the divisor. That is,
rl(cos Al
rz(cos A2
+i
+i
sin iiI)
sin A 2 )
!
.
rl [cos (AI - A 2 )
r2
=
--,"
,I
'. J
"
i
+i
"'j
sin (A 1 ; - A,)]; (I)
I
I
I
EXERCISES XII. E
Perform the indicated operations, first reducing to trigonometric form any numbers not already so expressed:
'
1.
2.
3.
4.
6.
7.
8.
9.
10.
12.
13.
+
+
+
+
2(cos 25°
i sin 25°) . 5(cos 75°
i sin 75°).
8(cos 18°
i sin 18°) . 6(cos 24°
i sin 24°).
15(cos 140°
i sin 140°) . 4(cos 280°
i sin 280°).
(1 + iy3) (4 + 4i).
5. (6 - 6i)( -7 + 70· i).
2(cos 15°
i sin 15°) . 3(cos 70°
i sin 70°)
. 4(cos 65°
i sin 65°).
lO(cos 125°
i sin 125°) + 5(cos 85° + i sin 85°).
3(cos 130°
i sin 130°) + 6(cos 217°
i sin 217°).
i(cos 250°
i sin 250°) + Hcos 140°
i sin 140°).
(4 + 40· i) + (6 - 6i). 11. (1 - i) + (1 + i).
(-va + 3i) + (y6 + 30· i).
(6va - 6i) + (y'2 + iy'6).
+
+
+
+
+
+
+
+
+
+
12.8. Powers of complex numbers.
P(l+i)
~
o
X
Raising to a power is a special case of
multiplication, and it follows by a repeated
application of (1) of the preceding section that
[r(cos A
+ i sin A)]n =
rn(cos nA
+ i sin nA),
FIG. 12.4.
where n is a positive integer.
The relation (1) is known as De Moivre's Theorem.
(1)
. j.
ROOTS OF COMPLEX NUMBERS
§12.9]
197
Example.
Find the value of (1 + i)5.
SOLUTION. Plot the number 1 + i. (See Fig. 12.4.)
absolute value is 0 and the amplitude is 45°.
(1
+ i)5 = [0 (cos 45° + i sin 45°)]5
= 4 0 (cos 5 . 45° + i sin 5 . 45°)
= 4 0 (cos 225°
+i
sin 225°) = 4( -1 - i) .
+ i)6 by
.. The student may check this result by expanding (1
the hinomial theorem.
12.9.
The
Roots of complex numbers.
To proye De Moivre's Theorem for the case in which the
exponent is the reciprocal of a positive integer, take the
expression
[r(cos A
+i
I,et A = nB.
rlln(cos nB
sin A)Fln
rlln(cos A
=
+i
sin A) lin.
(1)
Then the right side of (1) reduces to
+ i sin nB)1ln
+ i sin B)n)1/n
rlln(cos B +-Uiln.8),
= rlln[(cos B
=
or
[recos
A
+i
sin A)]lIn
=
,
r11n (cos
A
n
+i
sin
A). ( (2)
n
Since for any whole number k,
cos (A
+ k· 360°)
= cos A,
sin (A
+ k· 360°)
= sin A,
we have in the same way
[r(cos A
+i
sin A)]lIn
+ k· 360°) + i sin (A + k· 360 )}]lrn
. A + k . 360°) .
cos A + k . 360° + . SIn
0
= [rlcos {A
= r lin
(
n
~
n
(3)
COMPLEX NUMBERS
198
[Ch. XII
Thus, by giving to k integral values from 0 to n - 1
inclusive, we obtain n distinct roots of the number
r(cos A + i sin A). If we give k any
y:
other positive integral value "ve get
x~·-r-~-i-:::-'-.X one of the n numbers already obtained.
Therefore any complex number r(cos
A + i sin A), other than zero, has exactly
n distinct nth roots.
Example.
Find the fourth roots of -1 - iV3.
FIG. 12.5.
Plot the number -1 - iV3 and note that
SOLUTION.
-1 - iV3 = 2(cos 240°
+ i sin 240°),
(-1 - iV3)114
i
_ 2114 (
240°
cos
+4k . 360° + . sm. 240 + 4k . 360°)
= ~2 [cos (60
+
0
'l,
k . 90°)
+ i sin
(60°
+
k . 90°)].
Giving k successively the values 0, 1, 2, 3, we find for the four
distinct fourth roots of -1 - iV3:
°
v_'1-2 (cos 60 + tsm60)
0"
v_~j-2 (cos150° + isin,150
.
0
)
.0\ = '2v
1 _'/0 i _'I2 2+'1,2)
v_'1-(1
2 +2v 18,
i)
= v_"j-2 ( - 2y3 + '2
= _ ~ ~18
.i
jI
=
+ ~ ~'2, '.
~.,
~'2 (008.240°+, iein~O)
= ~'2 (- ~ - i v'3\
"
2
2)
,
, '1\
'J
~ (cos 330° + i sin 330°)
= -
=
_1
-';-2 __i2 v
-'/18 ,
2v
; c! '
"\.1
V'Z' ( ~3 -~) = ~ ~ -
",,"c
;
~.
In Fig. 12.6, the point P represents the complex number
2(cos 240° + i sin 240°); Pi, P2, P 31 P 4 represent the
four roots whose amplitudes are 60°, 150°, 240°, 330°,
respecti vely.
i
!-
i
I
EXERCISES
199
Note that the roots can be found graphically as follows:
Draw a circle with center at the origin and with a radius
equal to the numerical fourth root of the absolute value
of the number whose fourth roots are to be found; that is,
a radius equal to v2. Take one-fourth of the amplitude
of the original number ct· 240° = 60°). This locates
the point Pi on the circle. The
y
four roots all lie on the circle and
are spaced at equal intervals of
90°. Thus we can find P 2 , P a, P 4 •
In general, the nth roots of the X'
complex number recos A + i sin A)
can be found as follows: Draw a
circle whose center is the origin
and whose radius is the numerical '7:, p
nth root of r; divide the angle A ' ./
FIG. 12.6.
by n, the index of the root. Now
divide the circumference of the circle, from A/n to A/n +
360°, into n equal parts. The n points of division will be
the required roots.
EXERCISES XII. F
Use De Moivre's Theorem to raise to the indicated powers:
1.
3.
6.
7.
9.
11.
12.
13.
+
[4(cos 12°
i sin 12°)]3.
2.
[y'3 (cos 25° + i sin 25°) ]6. 4.
(1 - i)lO.
6.
(-2V3+2i)8.
8.
[5(cos 16° + i sin 16°)]-4. 10.
[cos (-10°)
i sin (-10°)]6.
(cos 10°
i sin 10°)-6.
(y'3
i)-10.
14.
+
+
+
+
[3(cos 18°
i sin 18°)]4.
[2( cos 20° + i sin 20°) ]8.
(1 + iy'3)9.
(y'3-i)12.
[0' (cos 7° i sin 7°)]-7.
+
(-1 - i)-12.
Find all of the
16. Square roots of 36(cos 50° + i sin 50°).
16. Square roots of 49(cos 140°
i sin 140°).
17. Square roots of 25(cos 100°
i sin 100°).
+
+
18.
19.
20.
21.
22.
23.
i f
Cube roots of 8(cos 72° + i sin 72°).
Cube roots of 216(cos 27° + i sin 27°).
Square roots of 1 - iV3.
Cube roots of 1 - iV3.
Square roots of -1 - iy'3.
Cube roots of 1.
SUGGESTION. 1 = cos 0°
24.
25.
26.
27.
28.
29.
30.
31.
32.
[Ch. XII
COMPLEX NUMBERS
200
+i
I
',:
sin 0°. '
Cube roots of 8.
Cube roots of -1.
Cube roots of i.
Cube roots of 3 + 4i. "
Fourth roots of - 5 + 12i.
Fifth roots of 1.
>---. ','
Sixth roots of -27i.
Fifth roots of 160 (-1 + i).
Seventh roots of - y'3 - i.
, j~ !
"
:u
,: !,
Obtain all of the roots of the following equations:
33.
35.
37.
39.
41.
x3
343 = O.
+ 1 = O.
x 1 - 1 = o.
x 9 - x = O.
3
2
X4 + x + x +
-
x5
X
+ 1 = O.
34. x 3
36. x 5
38. x8
40. X4
+ 216 =
-
O.
32 = O.
+ 16 = o.
+ x2 + 1 =
-7
0.:,
'.
SUGGESTION. Multiply by x-I, solve the
equation, then discard the extraneous root x = 1.
42.
43.
x3 + x2 + 2x 3 + 4x 2
X4 X4
+ 1 = O.
+ 8x + 16 =
X
O.
r
! if'
resu1t~
',"
CHAPTER XIII
Theory of Equations
J.:
,j
;
/.l
'
13.1. Polynomial.
A polynomial of degree n in the variable x is the function
f(x)
=
ooxn
+ 01Xn-l + . . . + On-1X + Om
(1)
in which n is a positive integer, and the a's are constants,
of which ao ~ O. For example,
2x 5
-
3x 4
+ 7x
3
-
X
+ 10
is a polynomial of degree 5. The polynomial (1) is also
called an integral rational fun.ction of degree n. (Cf. §§
1.13 and 3.1.) If we set it equal to zero we have an
integral rational equation of degree n. A root of the equationf(x) = 0 is called a zero of the functionf(x).*
Although the coefficients in (1) may be any complex
numbers, we shall restrict our discussion to polynomials
and integral rational equations having real coefficients.
However, in any theorem and its proof, the coefficients
may be complex unless otherwise stated.
13.2. Remainder theorem.
If a polYlwm7:al f(x) is divided by x - r, the remainder is
equal to the value of the polynomial when r is substituted for x.
Divide the polynomial by x - r until the remainder,
which may be zero, is independent of x. Denote the quoti~nt by Q(x) and the remainder by R. Then, according to
the meaning of division,
f(x) == (x - r)Q(x)
+ R.
* This definition is not restricted to integral rational functions.
201
THEORY OF EaUA TIONS
202
[Ch. XIII
Since this is an identity in x, it is satisfied by all values of
x, and if we set x = r we find that
fer)
(r - r)Q(r)
=
+R
=
o· Q(r) + R
=
R.
Here it is assumed that a polynomial is finite for every
finite value of the variable. Consequently, since Q(x) is a,
polynomial, Q(r) is a finite number, and 0 . Q(r) = O. -
I
Example.
f(x) = 2x 3
2x 3
-
-
5x 2
4x 2
-
+
X
x~
-
x2
+ 2x
x
- 3x
- 3:r
f(r) = f(2) = 2.2 3 - 5.2 2
= 16 - 20 - 2
10(x - 2 (= x - r)
2x 2 - X - 3 (= Q(x))
+ 10
+
'i
:.
6
4 (= R)
2 + 10 = 2· 8 + 10 = 4 = R.
-
5· 4 - 2
;
+ 10
13.3. Fador theorem.
If r is a root of f(x) = 0, then x ~ r is a factor of f(x).
By the pneceding section, the remainder obtained by
dividing f(x) by x - r is R = fer). But fer) = O~ since r
is a root of f(x) = O. Therefore,
f(x)
=
(x - r)Q(x)
+R
=
(x - r)Q(x),
or x - r is a factor of f(x).
13.4. Converse of the factor theorem.
Conversely, if x - r is a factor of f(x) , then r is a root of
f(x) = O.
Since x - r is a factor of f(x), we can write
f(x)
=
(x - r)Q(x).
Therefore,
fer) = (r - r)Q(r) = 0 . Q(r) = O.
!i
/
SYNTHETIC DIVISION
§ 13.51
t03
13.5. Synthetic division.
The division of a polynomial by a binomial of the type
x - r can be effected simply and quickly by a process
called synthetic division, which we shall illustrate by
means of the example 2x 3 - 3x 2 - l3x + 5 + x - 3.
By the ordinary process of long division we get
2x 3 -- 3x 2 - 13x + 5 (~_-___£
2x 3 - 6x 2
2x 2 + 3x - 4
3x 2
3x 2
-
13x
9x
4x
4x
+ 5
+ 12
-
(
7
. Much of this is quite superfluous. In the first place it is
unnecessary to write the first term in each line to be subtracted. Omitting these terms, and not bringing down
the terms of the dividend, we have
2x 3 - 3x 2 - 13x + 5 (x - 3
-:- 6x 2
2X2 + 3x - 4
9x
4x
+ 12
-
7
Next we write only the coefficients, * and push these up
into the more compact form shown below. The quotient
has been omitted since the coefficients of the terms of the
quotient, as well as the remainder, appear in the third lioo
(provided we bring the first coefficient, 2, down into this
line).
2 - 3 - 13 + 5 (1 - 3
- 6 - 9 + 12
2+3- 4- 7
• It is unnecessary to write the x's, since the power of x is indicated by
the position of the coefficient. The coefficient farthest to the right is the
constant term, the next is the coefficient of :l;, and so on.
I
I
THEORY OF EQUATIONS
204
[Ch. XIII
i
Finally, to replace subtraction by addition, we replace
(in the divisor) -3 by +3 and omit the 1:
2 - 3 - 13 + 5
+ 6 + 9 - 12
2 + 3 - 4 (-7)
Quotient =
2X2
+ 3x
@
- 4, remainder = -7.
RULE. To divide f(x) by x - r, arrange f(x) according to
descending powers of x, being sure to supply each missing power
with a zero as its coefficient. The work can be conveniently
arranged in three lines.
In the first line write the coefficients of f(x) in order, thus:
ao al a2 . . . an. (Zero must be written for any missing power.)
Write ao in the first place in the third line. Multiply ao by r,
place the product in the second line under al and add, placing the
sum in the third line. Multiply this by r, place the product in
the second line under a2 and add. Continue this process as far as
possible.
The last sum in the third line will be the remainder, and the
preceding numbers, reading from left to right, will be the coefficients
of the powers of x in the quotient, arranged according to descending
powers of x.
Example.
Divide 3x 4
5x 2
-
+ 2.
+
5
1 - 20
-6+12-14+26
3 - 6
Quotient = 3x 3
20 by x
-
+0 -
3
SOLUTION.
+X
-
+
6x 2
//(-2
_e
"Ie
7 - 13 (+ 6)
+ 7x -
13, remainder = 6.
EXERCISES XIII. A
Use synthetic
1. (x 3
6x 2 3
2. (x - 4x 2 3. (x 3 -- 7x 2
4. (2x 3 + 10x 2
+
+
6. (x 3 6. (6x 3
division in the following exercises: /
7x 5x 2x
+ 5x
+
3x
25x + 12)
+ 12x2 + X - 20)
2
-
ex -
8) -;2).
6) -;- (x + 3).
4) -;- (x - 1).
+ 11) -;- (x + 5).___~_______,/'
+ 4).
ex + 5).
-;- (x
-;-
!
I
§ 13.6J LOCATION OF THE REAL ZEROS OF A POLYNOMIAL 20!)
7.
8.
9.
10.
11.
_12.
13.
14.
16.
16.
17.
18.
19.
(5x 3 - 6x? - 17x - 9) -;- (x - 7).
(7x 3 - X + 24) -;- (x - 6).
(X4 + 4x 3 - 23x 2 + lOx - 12) -;- (x - 3).
(X4 - 2x 3 + 14x2 - 30x - 63) -;- (x - 3).
(x 4 - 7x 3 + 2x 2 - 15x - ·1) -;- (x + 4).
(2x 4 + 3x 3 + 4x 2 + 5x + 6) -;- (x + 2).
(3x 4 + 4x 3 - 7x + 22) -;- (x + 5).
(4X4 - 120x2 + 18x - 720) -;- (x - 6).
(x 5 + x 3 + X + 1) -;- (x + 1).
(2x 3 - 7x 2 - 5x + 4) -;- (x - i).
(4x 3 - 4x 2 - llx + 6) -;- (x + j).
(3x 3 + 2x 2 - 15x - 10) -;- (x - 0).
(x 3 - 15x 2 + 60x - 46) -;- (x - 7 + 0).
Use synthetic division and the remainder theorem in the following exercises:
20.
21.
22.
23.
24.
+ 5x
= x3
Given f(x)
Given f(x)
Given f(x)
Show that
Show that
2
3x - 2; find f(3), f( -3).
2
6x + 5; find f(2), f( -5).
= X4 - 2x 3 - 3:1:') + 7x - 8; find f(4), f( -4).
x + 4 is a factor of 2x 3 + 5x 2 - 32x - 80.
(x - 2)2 is a factor of 3x 4 - lOx 3 - x 2 + 28x -- 20.
--
= 2X4 - 3X3
+ 7x
13.6. Location of the real zeros of a polynomial.
Synthetic division is useful in graphing a polynomial,
and consequently in locating its real zeros. For examplc:',
consider the polynomial
f(x)
= 2x 3
-
7x 2
lOx
-
+ 20.
The remainder theorem tells us that the remainder
obtained when dividing by x - r is fer). By synthetic
division we can then construct the following table:
[~~r
-3
I -2
-1
0
I
1
II
2 '
3 I
~_I~I
Itwll-=67I~-;-·W5~--=W -=-=B
If the values shown in this table are plotted we have a
graph of the polynomial (Fig. 13.1), We note that it
crosses the x-axis between -2 and -1, between 1 and 2
THEORY OF EQUATIONS
206
[Ch. XIII
I
I
and between 4 and 5. Since f(x) = 0 wherever the graph
crosses the x-axis, we have located three roots of the
equation f(x) = O.
Here we have made use of the following general principle, which applies to any polynomial f(x):
PRINCIPLE. If f(a) and feb) have opposite signs, then
f(x) = 0 for at least one value of x between a and b. That is,
there is at least one real root of f(x) = 0 between a and b.
This principle depends upon
the fact that a polynomial is a
continuotls function and that
its graph is a continuous curve,
which means essentially that
the graph is not composed of
disconnected parts. If f( a) and
feb) have opposite signs, the
points A and B on the graph,
FIG. 13.1.
f(x) = 2x 3 - 7x' - lOx + 20
corresponding to x = a and x =
b respectively, will be on opposite sidei' of the x-axis. Consequently the curve must cross
the axis at least once, and in any case an odd number of
times, between A and B. But to each crossing there
corresponds a root of f(x) = O.
As a consequent'e of this fundamental principle we have
the following useful propositions:
If an integral rational equation of odd degree with real
coefficients has ~he coefficient vf its term of highest degree
positive, the equat1:on has at least one real root whose sign is
opposite to that of the constant term.
Let the equation be
f(x)
=
aoxn
+ ... + an
= 0,
ao> O.
(1)
Since n is odd, f(x) will be positiye for large positive values
of x and negative for large negative values. * It will have
* This is becausc thc sign of f(x) is determined, for sufficiPlltly large
nllffiPrical values of x, by the sign of the term of hiJ!;hest degrpc. For
t'xampie, x 3 - SOx' - 50x - lOUO will certainly be positive if x is lUO or
more.
§13.71
201
UPPER AND LOWER BOUNDS FOR ROOTS
the value an for x
=
O.
Symbolically,
f( - co) = - co ,
f(
+ co)
=
+ oc .
H an is positive there will be a root between - co and 0,
:that is, a negative root. Similarly, if an is negative, there
will be a positive root.
If an integral rational equation of even degree with real
coefficients has the coefficient of its highest power positive.
and has its constant term negative, the equation has at least
one positive and at least one negative root.
H, in (1), n is even and an < 0, thenf(x) will be positive
for large positive and large negative values of x, but will
be negative when x = O. It follows that there will be at
least one posi~ive and at least one negative root.
1 3.7. Upper and lower bounds for roots.
If r is positive (or zero), and upon dividing a polynomial
f(x) by x - r (synthetic division) we find that all numbers
in the third line are positive or zero, then r is an upper bound
to the roots of the equation f(x) = O. That is, no root can
be greater than r.
For any number larger than r will make the numbers in
the third line still larger, and consequently the remainde.::cannot be zero. Hence no such larger number can be a
root.
Example 1.
Find an l~per bound to the roots of
f(x)
=
2x 3
-
7x 2
-
lOx
+ 20 = O.
,SOLUTION. 'Ve have already found, in the preceding section,
that there is a root between 4 and 5. Let us divide the polynomial by x - 5:
7 - 10 + 20
+ 10 + 15 + 25
2 + 3 + 5 + 45
2 -
(~
All the numbers in the third line are positive, hence 5 is an
upper bound to the roots.
[Ch. XIII
THEORY OF EQUATIONS
!08
To find a lower bound to the roots, use the equation
f( -x) = 0, whose roots will be negatives of those of
f(x) = O. (Th:1t is, if f(x) = 0 has a root x = -5, then
f( -x) = 0 will have a root x = 5.) If r (a positive nurnber) is an upper bound to the roots of f( -x) = 0, then -r '
is a lower bound to the roots of f(x) = O.
The following rule, which is not difficult to demonstrate,
may also be used in determining a lower hound to the roots
of an integral rl1tional equation, f(x) = 0: If r is negative, .
and upon dividing f(x) by x - r (synthet1:c division) we find
that the nurnbers in the third line alternate in sign, * then r is a ;
lower bound to the roots of f(x) = o.
I
Example 2.
Find a lower bound to the roots of
f(x) = 2x 3
7x 2
-
-
lOx
+ 20 -
O. '
i
I-
SOLUTION.
fe-x)
~
2(-X)3 - 7(_X)2 - 1O(-x)
,,:: -2x 3 - 7x 2
lOx
20.
+
+
+ 20
I
,I
I·
Note that replacing x by - x changes the signs of all terms ol'odd
degree but does not affect terms oj even degree.
'.
Set f( - x) e<\u0.l to zero and ehan.!be. S.i.JbllS.:
2x 3
+ 7x
2
-
lOx - 20 = O.
We have found in § 13.6 that f(x) has a zero between -1
and - 2; consequently f( - x) will have a zero between 1 and 2.
Let us show that 2 is an upper bound to the roots off( -x) = o.
+ 7 - 10 - 20
+ 4 + 22 + 24
2 + 11 + 12 + 4
2
(2
All the numbers in the third line are positive; 2 is an upper
bound to the roots of f( - x) = 0, and - 2 is a lower b("''1d to the
roots of f(x) = o.
* A 0 may be counted as either + or
-.
- /.
NUMBER OF ROOTS
§13.81
209
EXERCISES XIII. B
Draw graphs of the following polynomials. Give exact values
of their integral zeros and locate their other real zeros between
consecutive integers.
1. x 3
3.
6.
7.
9.
11.
13.
16.
16.
3x 2 3x + 13x 2
x3
4x 2 S
X - 7x 2
6x 3 - x 2 2x 3 - 7x 2
X4
2x 3 X4
2x 3 16x4
3x 3
-
3
+
+
+
lOx + 24.
+ 2x - 8.
7x - 28.
2. x 3 +
4. 2x 3 6. x 3 8. 2x 3
10. 2x 3 12. 5x 3
14. X4 -
+ 3.
+
+
+ +
+
+
3x 2 - 13x - 15.
x 2 - 16x + 15.
7x + 3.
X 10.
11x 2 + 18x - 14.
29x 2 + 29x - 7.
2x 3 - 4x 2
6x
3.
32x
20.
22x
13.
5x 2
6x - 24.
x 2 - 2x
10.
37x 2
4x - 60.
+
+
+
+
+
Find, by the method of § 13.7, upper and lower bounds
(nearest integer above or below, respectively) for the roots of
the following equations:
17. x 3 + x 2 - 3x + 4 = O.
18. x J - 4x 2 - 2x - 7 = O.
3
2
19. x + 5x - X + 4 = O.
20. 2x 3 - 7x 2 - 4x + 3 = O.
2
21. X4 - 4x s + 2x - 7x - 3 = O.
22. X4 + 3x 3 + 2x 2 + 5x + 1 = O.
23.
2X4 -
24.
X4 -
+
5x 3
7x 2 4x 3 + 6x 2 +
X
X -
+8 =
150
O.
O.
=
13.8. Number of roots.
We assume the following theorem, which is called the
fundamental theorem of algebra: Every integral rational
equation has at least one root. (For proof see Garrett
Birkhoff and Saunders MacLane, A Survey of Modern
Algebra, The Macmillan Company, revised edition, 1953,
p. 107.)
Then we can prove:
THEOREM.
f(x) = aoxn
Every integral rational equation of degree n,
+ alXn-1 + . . . + an
has exactly n roots.
= 0,
ao ~ 0,
(1)
THEORY OF EQUATIONS
210
[Ch. XIII
By the fundamental theorem, f(x) = 0 has at least one
root. Let this root be rl. Then, by the factor theorem,
x - rl is a factor of f(x) and we can write
where Ql(X) is a polynomial of degree n - 1, having aoX n- 1
as its term of highest degree.
Applying the fundamental theorem again, we find that
Ql(X) has at least one zero, say r2, and x - r2 is a factor
of Ql(X). Thus,
Ql(X) = (x - r2)Q2(x),
f(x) = (x - rl)(x - r2)Q2(x),
and
where Q2(X) is a polynomial of degree n - 2, whoee tel1ll
of highest degree is aoX n - 2 •
• t
Continuing this process n times we get
I
.
f(x) = (x - rl)(x - r2) ••• (x - rn)Qn(X),
where Qn(x) is a polynomial of degree n - n = 0, whose
term of highest degree is aoxo = ao. That is, Qn is the
constant ao. Thus,
f(x)
=
ao(x - rl)(x - r2) ••• (x - rn),
(2)
and f(x) = 0 has the n roots rl, r2, ••• , rn.
The equation can have no other roots, for no other number when substituted for x will make any of the factors
on the right side of (2) equal to zero.
It should be noted that the numbers rl, r2, . . . ,rn do
not have to be real. Also they are not necessarily all distinct. For example, we might have x - r occurring m
times as a factor of f(x), in which case r is called a multiple
root of f(x) = 0 of order m, or a root of multiplicity m.
A root of multiplicity 2 is called a double root, a root of
multiplicity 3 is a triple root, and so on. A root corresponding to a linear factor which occurs only once is called
a simple root.
IMAGINARY ROOTS
§13.91
211
As a corollary of the foregoing theorem we have the
following:
If two polynomials in the same variable, each of degree not
greater than n, are equal in value for more than n distinct
values of the variable, then the polynomials are identical.
Let the polynomials be
aoX n
+ alx n - l + . . . + an,
boXn + blxn - l
+ . . . + bn.
1£ they are equal for more than n values of x, their difference is zero for more than n values of x. This difference
may be written in the form:
(ao - bo)x n
+ (al -
bl)x n- l
+ ... + (an -
bn ) = O.
(3)
But if one or more of the coefficients (ao - bo), (al - bl),
. . . , (an - bn) in (3) were different from zero, we should
have an equation of degree n, or less, with more than n
distinct roots. This contradicts the foregoing theorem
and is therefore impossible. Thus, each of the coefficients in (3) must be zero, and we have
ao = bo,
In other words, the two polynomials are identical; their
values are equal for all values of x.
13.9. Imaginary roots.
If an imaginary number a + bi is a root of an integral
rational equation f(x) = 0 with real coefficients, the conjugate
imaginary number a - bi is also a root.
Since a + bi is a root of f(x) = 0, then by the factor
theorem, x - (a + bi) is a factor of f(x). The method of
proof will be to show that x - (a - bi) is also a factor of
j(x) by showing that the. product of these two linear factors is a factor of f(x). Denoting this product by P(x),
we have
P{x) = [x - (a + bi)][x - (a - bi)]
= x 2 - 2ax + a 2 + b2 •
THEORY OF EQUATIONS
212
[Ch. XIII
Divide f(x) by P(x) until the remainder is of degree not
higher than the first (that is, until it is linear or constant).
Denoting the quotient by Q(x) and the remainder by
Rx + S, we have the identity
f(x)
== P(x) . Q(x)
+ Rx + S,
!
!
(1)
in which Rand S are real because P(x) has real coefficients.
In the above identity set x = a + M. By hypothesis,
f(a + bi) = O. Since x - (a + bi) is a factor of P(x), it
follows by the converse of the factor theorem that i
pea + bi) = O. Therefore from (1) we obtain the relation·
o=
O· Q(a
or
Ra
+ bi) + R·
(a
+ bi) + S,
(2)
+ S + Rbi
=
O.
(3)
an
Since
imaginary number cannot be zero unless both _
its real part and its imaginary part are zero (see § 12.1),
it follows from (3) that
Ra
+S
=
0,
Rb = O.
(4)
But, since a + bi is imaginary, b ~ O. From the second
equation of (4) it follows that R = 0, and then, from the
first equation of (4), that S = O. That is, the remainder
Rx + S in (1) is zero, and f(x) = P(x) . Q(x). In other
words, P(x) is a factor of f(x).
But x - (a - bi) is a factor of P(x), and consequently.
of f(x). Therefore, a - bi is a root of f(x) = O.
13.10. Quadratic surd roots.
A radical like v'3 or ~5, in which the number under the
radical sign is rational but the radical itself is irrational,
is called a surd. A surd is called quadratic, cubic, and so
on, according as its index is two, three, and so on.
A theorem for quadratic surds, similar to the theorem
in the foregoing section, is as follows:
I
§13.111
GRAPH OF A FACTORED POLYNOMIAL
213
If a quadratic surd a + vib (a rational) is a root of an
integral rational equation with rational coefficients, the conjugate surd a - vi) is also a root.
The proof follows a line of reasoning similar to that used
_. for the theorem on conjugate imaginary roots, and will not
be given here.
EXERCISES XIII. C
1. Solve the equation X4 - x 3 - 16x 2 + 59x
that one of its roots is 3 + 2i ..
SOLUTION. Since 3
fore the expression
(
[x - (3
+ 2i is a root, 3
+ 2i)][x -
(3 - 2i)]
=
+ 13 = 0,
- 2i is a root.
x 2 - 6x
given
There-
+ 13
is a factor of the left side of the given equation. Dividing by
this factor, we get x 2 + 5x + 1 = 0, a quadratic equation
whose roots are i( -5 ± y'2I). The complete solution is
x = 3
±
i( -5 ± y'2I).
2i,
2. Solve the equation X4 - 9x + 25x 2 - 19x - 6 = 0, given
that 2 + y'S is a root.
3. Solve the equation 6x 4 - 35x 3 + 58x 2 + 23x - 22 = 0,
given that 3 - iV2 is a root.
4. Solve the equation x 6 + 2x 5
X4
4x 3 - x 2 2x - 1 =
0, given that i is a double root.
5. Solve the equation 2x 6 - x 5 - 29x 4 + 14x 3 + 112x2 - 49x
- 49 = 0, given that y'7 is a double root.
6. Solve the equation x 6 + 16x 5 + 86x 4 + 144x 3 - 191x 2 832x - 676 = 0, given that - 4 + y'3 is a double root.
7. Solve the equation x 6 - 13x 5 + 66x 4 - 152x 3 + 108x 2
140x - 200 = 0, given that 3 + i is a double root.
3
+ +
+
+
13.11. Graph of a factored polynomial.
When a polynomial, or integral rational function, f(x),
is expressed in factored form, its graph can be quickly
sketched. At each real root of f(x) = 0 the graph meets
the x-axis. At each simple root the curve crosses the axis
THEORY OF EQUATIONS
214
,
[Ch. XIII
at an angle, at each multiple root of even order it is tangent
to the axis and does not cross it, at each multiple root of
odd order it is tangent to the axis and crosses it. A graph
of the polynomial
y
=
aoCx - rl)(x - r2)3(x - rs)2
(al
> 0),
in which the r's are all real, would appear as in Fig. 13.2.
If the polynomial contains '
any imaginary factors they
~
TaX occur in conjugate pairs
(§ 13.9). These imaginary,
FIG. 13.2.
factors do not affect the
y = ao(x - Tl)(X - T2)'(X - Ta)!
manner in which the curve
meets the x-axis, but do affect the shape of the curve,
Proofs of these statements are not difficult but will be
omitted here.
~.J."
\
EXERCISES XIII. 0
Sketch the graphs of the following polynomials:
+
+
1. (x
3)(x - 1)(x - 4).
2. x(x - 3)(x
5).
3. (x + 2)(x - 2)2(X - 6).
4. -2(x + 4)(x - 4)(2; 6. (2x - 5)(x
3)(x - 4)2,
6. (x
4)2(X - 5)2,
7. i(x
2)(x - 1)3(X - 3)2.
8. (x
2)2(X - 5)4,
9. -x 2(x
5)3(x - 2).
10. (3x
2)(2x - 3)2.
11. x 3 (3 - x)(5 - X)2,
12. (x 2 - 9)(x - 9)2.
13. (x - 1)(x - 2)2(X - 3)3(X - 4)4.
14. (x
5)(1 - x)(3 - x)2(5 - x)3.
+
+
+
+
+
+
7).
+
13.12. Descartes' rule of signs.
Two consecutive terms of a polynomial in x, with real
coefficients, and terms arranged according to descending
powers of x, are said to have a variation in sign when one
is plus and the other minus. Thus, the polynomial
2x 5
-
3x 4
-
4x 3
+ 3x
- 7
has three variations in sign. (Note that some powers of z
may be missing.) Descartes' rule of signs states:
DESCARTES' RULE OF SIGNS
§'13.12J
215
An integral rational equation f(x) = 0, with real coefficients, has as many posit'ive roots as it has variations in
s.ign, or fewer by an even number. (A root of multiplicity
m is counted as m roots.)
Suppose that a polynomial P(x) is represented by the
~- following scheme, in which the dots after a sign indicate
that there may be more signs of the same kind, but that
no change in sign can occur before the next explicitly
written sign:
P(x)
=
+ . . . - . . . + ....
Multiply P(x) by x - r (r > 0). Schematically the multiplication may be represented by
f
/~
xP(x) =
-rP(x) =
(x - r)P(x) =
+ ..
_ ... -
+
+±... - ±
+ ...
++±
The ambiguous sign ± indicates that the sign of the corresponding term is undetermined-we do not know whether
it is plus or minus. But no matter what signs these doubtful terms may have, there is at least one more variation in
sign in the product (x - r)P(x) than in P(x), since there
is an additional variation at the end.
The product may contain more than one additional
variation; for successive terms of like signs, for example,
+ + + or - - -, in the original polynomial, which are
replaced in the product by ambiguities may actually be
replaced by + - + or - + - respectively. But such
changes always increase the number of variations by an
even number.
Hence the number of variations in (x - r)P(x) exceeds
that in P(x) by 1, or by 1 plus an even number, that is,
by an odd number.
Now suppose that the product of all the factors corresponding to negative and complex roots of f(x) has been
formed into a polynomial P(x). Since P(x) = 0 has no
[Ch. XIII !
THEORY OF EQUATIONS
216
I
positive roots its first and last terms must have like signs.
(Otherwise it would be negative for x = 0 and positive for
large values of x, or vice versa, and would have a positive
root.) Therefore, P(x) must have an even number of variations in sign, say 2k.
If a positive root is introduced by multiplying P(x) by
x - 11 (r1 > 0), the number of variations is increased by an
odd number, say 2k1 + 1. Suppose that we have introduced m positive roots, r1, . . . ,rm • We have increased
the number of variations to
2k
+ 2k1 + i + . . . + 2km + 1
= 2(k + k1 + . . . + k
+ m.
m)
That is, the number of positive roots, m, is the number of ' -variations decreased by 2(k + k1 + ... + km ), which is
an even number (zero if all the k's are zero). But this is
Descartes' rule.
1-i
Information concerning the number of negative roots of i
an hluation may be obtained by applying Descartes' rule :
tof(-x) = o.
:,',
Example 1.
I
"
/'
Discuss the nature of the roots of
"
lex) = x 3
SOLUTION.
+ 3x
- 5
= O.
1 variation; therefore 1 positive rQot.,;
!( -x) = -x 3
3x - 5 = O.
-
No variation; therefore no negative roots of the original
equation.
The equation is of third degree and must have three roots. ,
Thus, two roots must be imaginary.
Example 2.
Discuss the nature of the roots
lex) = 2x· - 3x 4
-
of/~-
4x 3
+ 3x -
7 = O.
RATIONAL ROOTS
§13.131
217
3 variations; therefore 3 or 1 positive roots.
fe-x) = -2x 6 - 3x 4 + 4x 3 - 3x - 7 = o.
SOLUTION.
2 variations; therefore 2 or 0 negative roots.
The various possibilities for the kinds of roots are shown below.
Positive
Negative
Imaginary
Total
3
2
3
0
0
2
5
5
1
2
2
--
5
1
0
4
-5
EXERCISES XIII. E
(
Give all of the information obtainable from Descartes! ~
of signs about the roots of the following equations:
1. x 3 + 4x - 1 = O.
2. x 3 - 8x 2 - 9 = O.
3
2
3. x + 2x + 5 = O.
4. x 3 - 5x 2 + 2x - 3 = O.
3
2
5. 2x - 3x - X + 9 = O.
6. 3x 3 - 2x 2 - 7x - 5 = O.
3
2
7. x + 3x + 7x - 2 = O.
8. 2x 3 - 5x 2 + 2x + 6 = O.
3
2
9. 5x + x - 4x + 1 = O.
10. 4x 3 - 7x 2 - X - 15 = O.
3
2
11. x + 2x - 5x - 8 = O.
12. 2x 3 + 3x 2 + 4x + 5 = O.
2
3
13. 8x - 10x + 4x - 9 = O.
14. x 3 - 17x 2 - llx + 14 = O.
15. 22x 3 - 19x 2 - 15x - 12 = O.
16. X4 - 10x 3 + x 2 + 6x + 11 = O.
17. 2X4 + 3x 3 - 5x 2 + 2x - 22 = o.
18. 5x 4 - 2x 3 + 3x 2 - 4x + 10 = O.
19. X4 - 17x.1 - 14x2 + X - 5 = O.
20. 3x 4 - 4x 3 - 7x 2 - 18x - 22 = O.
21. 2X4 - ,1X 3 - 8x 2 - 3x + 1 = O.
22. X4 - 20x 3 - lOx 2 + 3x + 24 = O.
23. 5x 4 + 4x 3 - 7x + 11 = O.
24. 3x 4 - 5;r.2 + 19x - 15 = O.
25. x 6 + X4 - X + 12 = O.
26. 6x 4 + 3x 3 - 7x 2 - 20 = O.
6
2
27. x + 4x - 30 = O.
28. 4X4 + 2x 2 + 1 = O.
5
29. x - 2 = O.
30. x 6 - 12 = o.
13.13. Rational roots.
A rational number is one which may be exactly
expressed as the ratio of two whole numbers. (See § 12.2).
THEORY OF EQUATIONS
218
[Ch. XIII
.i
If a rational number blc, a fraction in its
lowest terms or an integer, * is a root of the integral rational
equation
THEOREM.
aoxn
+ alXn-1 + . . . + an-1X + an
= 0
(1)
with integral coefficients, then b is a factort of an and c is a
factor of ao·
. .,-~-To prove this, substitute blc in (1):
_,'
tlo
(~Y + al
GY-I + . . . +
an_l
~ + ~" - o.
;
(2)
Multiply both sides of the equation by cn :
aob n
+ a1bn-1c + . . . + an_1bc n - 1 + ancn =
O.
(3)
Transpose the last term to the "'ight side and factor b from
the expression remaining on the left:
b(aob n - 1
+ a 1bn- c + ... + an_ICn-I)
2
=
-anc n •
(4)
Since b is a factor of the left side it is a factor of the right
side. It cannot be a factor of c, since blc is in its lowest
terms. It must therefore be a factor of an.
Similarly, by transposing the first term in (3), we get
c(aIb n-
1
+ ... + a n_ 1bcn- + anc n- 1)
2
=
-aob n •
(5)
Arguing as before, we see that c must be a factor of the
right side, and since it is not a factor of b it must be a factor
of ao.
COROLLARY.
xn
Any rational root of the equation
+ alXn - 1 + . . . + an-1X + an =
0,
(6)
in which the a's are integers, is an integer and is a factor
of an·
Rational roots can be found by synthetic division.
• If b/c is an integer, then c is to be taken as 1.
t When we say that b is a factor of a.. we mean here that the integer b is
contained in the integer a.. a whole number of times.
\ .
RATIONAL ROOTS
§13.131
219
Example.
Find the rational roots of
f(x) = 2X4
+ 9x 3 +
15x 2
+
13x
+ 6 = O.
SOLUTION.
Possible numerators (factors of 6): 1, 2, 3, 6.
Possible denominators (factors of 2); 1, 2.
Possible rational roots; ± (1, 2, 3, 6, t, j).
By Descartes' rule there are no positive roots of any kind.
consequently we only need to try negative values.
We find
(
f( -x) = 2X4 - 9x 3
+ 15x
2 -
13x
+ 6:
4 variations i therefore 4, 2, or 0 negative roots.
We test the possible roots in order, beginning with
smallest. *
2 - 9 + 15 - 13
1- 4
+ 6 (i:
2-8+11
2 - 9 + 15 - 13 + 6
2- 7+ 8-5
2-7+ 8- 5(+1)
the
(l
Note that it is unnecessary to complete the synthetic division
for 1/2. For 1/2 X 11 is a fraction, and it would be impossible
to come out with a zero remainder.
2 - 9 + 15 - 13 + 6
3- 9+ 9-6
2 - 6 + 6 - 4 (+0)
(1
i
is a root of f( - x) = 0,
- j is a root of f(x) = o.
We now use the depressed equation, 2x 3 - 6x 2 + 6x - 4 =
0, represented by the numbers 2 - 6 + 6 - 4, which is the
quotient obtained by dividing f( -x) = 0 by x - j. This can
be simplified, by division by 2, to x 3 - 3x 2 + 3x - 2 = O. It
has the same roots as f( -x) = 0, with the exception of 3/2,
which has been divided out.
/
* By beginning with the numbers which are smallest numerically, we can
sometimes find bounds for the roots, doing away with the necessity of testing any larger values.
THEORY OF EQUAnONS
220
[Ch. XIII
i '
For the depressed equation we have
I
i
Possible numerators: 1,2.
Possible denominators: 1.
Possible rational roots: 1,2.
~
1
\
i
f
,i
Try 2:
2 is a root of the depressed eq...~,,~
-2 is a root of f(x) = o.
.- 1
The new depressed equation is x 2
the quadratic formula, we get
x=
± y'1="4
2
=
X
-
1
+1=
± iV3
2
O.
Solving by
.
The roots of f( x) = 0 are negatives of these. The original equation was of degree 4, and we have found all 4 roots; they are:
3
- 2'
I
I
.I
We have already tried 1 in f( -x).
1-3+3-2
2-2+2
1 - 1 + 1 (+0)
t
-2,
-1
± i'Va
2
\
.
Note that even though a certain number has already
been found to be a root it may also be a root, of the
depressed equation. This will be the case if it is a multiple root of the original equation.
EXERCISES XIII. F
Find the rational roots of the following equations:
1.
3.
6.
'1.
9.
10.
11.
12.
13.
x3
+x
2
5x - 2 = O.
x 3 + X + 10 = O.
2x 3 + 7x 2 - 19x + 6 = O.
5x 3 + 3x 2 + 13x - 6 = O.
6x 3 - 19x 2 + X + 6 = O.
6x 3 - 29x 2 + 39x - 10 = O.
2x 3 - 3x 2 + X - 6 = O.
2x 3 + 13x 2 + 12x + 3 = O.
12x 3 - 32x 2 - 33x - 7 = O.
-
2.
4.
6.
8.
x 3 - 3x 2 2x 3 - 3x 2 2x 3 + x 2 +
3x 3 + 4x2 -
8x - 10 = O.
7x - 6 = O.
x - I = O.
16:1: + 8 = O.
.,
§ 13.14]
221
+ 50x 2 -
31x - 15 = O.
4x + 48 = O.
X4 3x~ - 21x2 - 43x + 50 = O.
12x 4 - 20x 3 - 57x 2 + 50x + 75 = O.
28x 4 - 3x 3 - 154x 2 - 54x + 36 = O.
X4 x 3 - 9x 2 + llx + 6 = O.
4
8x - 24x 3 + 5x 2 + 52x - 15 = O.
21. 6x 4 - 31 x 3 - 44x 2 + 371x - 392 = O.
22. X4 + 10.1':3 + 20x 2 + 20x - 44 = O.
23. In Fig. 13.3 the tangent TP is 3 inches in length.
diameter T R is extended 1 inch to
the point Q. The tangent QS,
extended, passes through P. Find
the radius of the circle.
14.
15.
16.
17.
18.
19.
20.
I
IRRATIONAL ROOTS
24x 3
X4
+x
3 -
16x 2
The
13.14. Irrational roots.
Irrational roots of any type of
Q
equationf(x) = 0 can be found by a
FIG. 13.3.
process of successive approximations,
using the general principle stated in § 13.6, that if f(a) and
feb) have opposite signs there is a root of f(x) = 0 between
a and b. If the equation is an integral rational equation
we may employ synthetic division to obtain values of f(x),
or we may obtain them by actual substitution. For
other types of equation, such as *
log x - x
+2
=
0,
we must use actual substitution.
Example.
Find an irrational root of J(x) = x 3 +
X -
5 = O.
SOLUTION. By Descartes' rule the equation has one positive
root and no negative root, hence two imaginary roots.
By synthetic division, or by actual substitution, we find that
J(I) = -3, f(2) = 5. Thus, the curve y = f(x) crosses the
• The symbol log x means the logarithm of x.
(See Chapter XIV.)
THEORY OF EQUATIONS
222
[Ch. XIII
x-axis between x = 1 and x = 2. (See Fig. 13.4.) A first
approximation to the root is x = 1.
To get a better approximation we
add hI to 1. To find hI we use similar
triangles: *
1
2
AD
DE
AC = CB'
2
1
x
or
hI = 0.4 approximately.
A second approximation to the root
is x = 1.4.
D
c
We now find f(x) for x by tenths,
FIG. 13.4.
using synthetic division or straight
substitution. We shall use the latter, as it is quite general
in its application, synthetic division being applicable only to
polynomials. Using tables of cubes or a calculating machine,
we find
)'<
(1.4)3 =
2.744
1.4
4.144
-5
-f(1.4) = -0.856
(1.5)3 =
3.375
1.5
-4.875
-5
f(1.5) = -0.125
(1.6)3 =
4.096
1.6
5.696
-5
f(1.6) = 0.696
,'"
__
We see that the curve crosses the x-axis between 1.5 and 1.6.
To get a better approximation to the root we add h2 to 1.5.
FIG. 13.5.
FIG. 13.6.
* This assumes that the graph of the equation y = f(x)
straight line.
from A to B is a
Actually it is a curve, of which AB is a chord.
,_
j
EXERCISES
223
Again we find, by similar triangles,
h2
0.125
0.125
=
=--,
0.1
0.125 + 0.696
0.821
-
h2
= 0.015+.
To carry the result still further we proceed as before, ta~~'
by hundredths.
. ," !~.!
(1.51) 3 =
3.442951
1.51
4.952951
~5
-5
---
f(1.51) = -0.047049
3.511808 .
1.52
5.031808
(1.52)3 =
---
f(1.52) =
0.031808
hs
0.047049
0.01 = 0.047049
0.031808
0.047049
0.078857' ,.
hs = 0.0060-.
+
Our next approximation to the root is
x = 1.51 + 0.0060
. <
, r·,
= 1.5160.
The process can be continued as far as desired,
successive steps become more laborious.
altho'i'ith the!
fJ
-
F.XERCISES XIII. G
Find, correct to three decimal places, the irrational roots of
the following equations:
1.
3.
6.
7.
9.
11.
13.
16.
17.
18.
+
x3
x - I = O.
x 3 + x 2 - '3 = O.
x 3 - 3x
1 = O.
2x 3 + 9x 2 9 = O.
x 3 + 3x 2 + 4x - 2 = O.
x 3 - 2X2 - X
1 = O.
4x 3 - 9x
6 = O.
x3
3x 2 - 3x - 3 = O.
X4
x 3 - 2X2
3x - 1 =
X4 3x 2
6x - 2 = O.
+
+
+
+
+
+
+
+
2. x 3 4. x 3 6. 2x 3
8. X s 10. x 3 +
12. x 3 14. x 3 16. x 3
12:z: - 20 = O.
X - 5 = O.
x~ - 7 = O.
9x
12 = O.
6x~ + 6x - 10 = O.
9x~ + 21x - 15 = O.
9x~
31x - 15 = O.
3x~ - 2x - 3 = O.
+
+
O.
+
+
ii4
THEORY OF EQUATIONS
13.15.
[Ch. XIII
Transformation to diminish the roots of an equation
by a ~xed amount.
In order to develop a more systematic but more technical method of finding irrational roots of integral rational
equations, it is necessary to be able to transform an equation into a new equation whose roots are less, by a given
amount, than those of the original equation.
Suppose that the original equation is
If we replace x by x' + h, we get a new equation in x', and
if x = r is a root of the original equation, then x' = r - h
will be a root of the transformed equation. That is, each
root of the new equation will be less by an amount h than
. the corresponding root of the old equation.
Setting x = x' + h in (1) we get
ao(x'
or
+ h)n + alex' + h)n-l + ...
+ an_lex' + h) + an = 0,
aoX'n + Alx'n-l + ... + An_IX' + An = 0.
:;;
(2)
(3)
We can determine the A's by the artifice of changing back
to (1) again by setting x' = x - h in (3). This gives
ao(x - h)n
+ AI(x -
h)n-l
+ ...
+ An_l(x -
h)
+
An
= 0,
(4)
which is merely j(x) arranged in powers of x-h.
It will be noted that An is the remainder ifjex) is divided
by x - h, A n - l is the remainder if the resulting quotient
is divided by x - h, and so on.
Example.
Find 3,n equation whose roots are
X4 - 5x S + 20x - l{$ = C?
less
.~-~",,_../
by
3 than
those of
HORNER'S METHOD
§13.16]
1 - 5+ 0+
3 - 6 6+
1 3 + 3 1+1 3+~
SOLUTION.
2-
31-
,~'~
"f
1
+:1+
20 - 16
18 + 6
10
9.
7
(~
21-
!
~
.
_',
';
The trai)J£ormed equation is
+ 7X'3 + 9X'2 -
~ ~
1",L. '~ . :;
1+7
X'4
225
.
.~<:)
) {:."
.
d j:l'{ ~
,',
.'
i
7x' - 10 ...
O~
EXERCISE
Check the result by verifying that the roots of the origjnal
equation are -2, 1, 2, 4 and those of the transformed equation
--5, -2, -1, 1.
EXERCISES XIII. H
Obtain equations whose roots are equal to the roots of the
following equations diminished by the number indicated.
3x 2 + 2x - 7 = 0, 3.
5x 2 + 7x - 2 = 0, 1.
3
2x
5x 2 - 6x - 4 = 0, 2.
3
2x - 7x 2 - 6x - 1 = 0, 4.
x 3 + x 2 - 20 = 0, 3.
3x3 + 7x2 + 5x + 4 = 0, -2.
4x3 - x 2 - 6x + 5 = 0, -1.
x 3 + 3x2 + 4x - 2 = 0, 0.3.
8x 3 + 8x - 7 = 0, 0.6.
X4 - 2x 3 - 13x 2 + 24x + 12 = 0, 3.
X4 - 12x 2 + 24x - 5 = 0, - 4.
2X4 + 4x 3 + 7x 2 - 6x - 15 = 0, 1.
1. x 3
2. x 3
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
-
+
13.16. Horner's method.
This method of finding an irrational root of an integral
rational equation is best explained by means of an example.
226
THEORY OF EQUATIONS
[Ch. XIII
Example.
Find an irrational root of
f(x) =
X3
+X
-
5 = O.
(1)
i
By synthetic division (or by substitution) we rindf(1) = -3,
f(2) = 5. Thus, f(x) has a root between 1 and 2. Transform ,
(1) into an equation whose roots are less by 1 than those of
(1), so that the new equation will have a root between 0 and 1: '
1+0+1-5
1+1+2
1+1+
3
1+2
(1
21-
1
+ ~I+ 4
.) j
1+3
i
The transformed equation is
fl(XI) = x~
+
3xi
+ 4XI
-
3 = O.
(2) -
When Xl is small, x~ is small and we can get a fairly good
approximation to the root of (2) by solving the quadratic equation 3xi + 4XI - 3 = 0, getting Xl = 0.5+. (The positive
root is the only one considered.) By synthetic division, we
find/1(0.6) = 0.696,/1(0.5) = -0.125. The fundamental principle which must be kept in mind is that if f(a) and feb) have
opposite signs, there is a root of f(x) = 0 between a and b.
Transform (2) into an equation whose roots are less by 0.5:
1 + 3.0 + 4.00 - 3.000 (0.5
0.5 + 1.75
2.875 1 + 3.5 + 5.75 - 0.126
0.5 + 2.00
~+7.76
+
1 + 4.6
The transformed equation is
h(X2) = x~
+ 4.5x; + 7.75x2 -
0.125 = O.
(3)
To get an approximation to the root ofA3), neglect x~ + 4.5x;
and set 7.75x2 - 0.125 = 0, getting X2 = 0.016. We find
) 10.02) = 0.0:31808, f(0.01) = - 0.047049.
HORNER'S METHOD
§13.161
227
Transform (3) into an equation whose roots are less by 0.01:
(0.01
,
. !
----
The next transformed equation is
fa(xa) = x~
+ 4.53xi +
Setting 7.8403xa - 0.047049
The root of f(x) = 0 is
X =
7.8403xa - 0.047049 = O.
(4)
= 0, we get Xi = 0.0060.
1 + 0.5 + 0.01 + 0.0060 = 1.5160.
This is certainly correct to two decimal places and very probably to three. The work can be continued further if desired.
The solution may be compactly exhibited as follows:
1+0+1-5
1+1+2
(!
1 + 1 + 2/- 3
1+2
+ ~I+ 4
1 + 3.0 + 4.00 - 3.000
0.5 + 1.75 + 2.875
1 + 3.5 + 5.75 - 0.126
1
(0.5
ax}
+ 4xl -
3 = 0,
0.5.
Xl=
0.5 + 2.00
~+7.76
1 + 4.60 + 7.7600 - 0.126000 (0.01 7.75x2 - 0.125 = 0,
0.01 + 0.0451 + 0.077951 X2 = 0.016.
1 + 4.51 + 7.7951 - 0.047049
0.01 + 0.0452
7.8403xs - 0.047049 = 0,
1 + 4.521+ 7.8403
Xs = 0.0060.
0.01
x = 1.5160.
1 + 4.63
THEORY OF EOUA liONS
228
[Ch. XIII
In the present example, the remainders in the synthetic
division process have always been negative. This is
because, in passing through the value zero (with increasing
x), f(x) changes from negative to positive. In the contrary case, we should want the remainders to be positive.
To find a negative root of f(x) = 0 by Horner's method,
find the corresponding positive root of f( -x) = 0 and
prefix a minus sign.
13.17. Suggestions k· Anding the real roots of a numerical
equation.
Obtain as much information as possible about the na£ure
of the roots from Descartes' rule of signs.
Remove all rational roots, depressing the equation each time
that such a root is removed.
Always take advantage of any information revealed about
"
upper and lower bounds for the roots while looking for either rational or irrational roots.
IAfter all rational roots have been removed find the irra~J
tional roots of the depressed equation by the method of § 13.14
il
or by Horner's method.
;11
If it is possible to reduce the equation to a quadratic equa.
lion, the solution can be completed by solving this quadratic.
Negative roots may be found by solving f( -x) = O.
I
EXERCISES XIII. I
Find, correct to three decimal places, all of the real roots of
the following equations;
1. x 3
3. x 3
+ 2x
- 5 = O.
+ 6x +
3x s + 4X2 -
5.
7. x 3
9. x 3
11. x 3
2
+
+
-
13. x 3 14. 3x 3 -
X
-
9
= O.
5 = O.
18x + 30 = O.
2x 2 - 6x - 2 = O.
+
2. x 3
4. x 3
6. x 3
8. x 3
10. x 3
3x 2
12x - 16 = O. 12. x 3
2
3x - 2x
5 = O.
18x2
27 x
4 = O.
+
+
+
-
+
-
+
X - 3 = O.
3x 2 + 7x - 11 = O.
7x 2
17 x
13 = O.
Sx + 1 = O.
6x 2 + 18x - 22 = O.
3x2 - 6x
4 = O.
+
+
+
229
EXERCISES
15.
16.
x 3 - x 2 + 2x - 2 = O.
4x 2 - 8x - 4 = O.
+ 2x 3 - 13x2 - 24x + 12 = O.
+ x 2 - 1 = O.
19. X S
• ''''',>II.;\iIt
X4 X4
17.
X4
18.
XS
+
x3
-
1
~~1'
=
O.
Extract the following roots, correct to three decimal places,
by using Horner's method or the method of § 13.14:
20.00.
SUGGESTION. Find the real root of the equation x 3
21.05.
22. {flO.
-
20.
23.~.
Solve the following simultaneous equations:
24.
26.
27.
28.
29.
30.
+
+
y4 = 97 J
25. x 2 + y = 4,
x
y = 1.
x + y2 = 2.
The length of a rectangular box is 2 inches more than its
width and 5 inches more than its height. Its volume is 100
cubic inches. What is its height?
A tray is to be made from a rectangular sheet of metal 15
by 20 inches by cutting equal squares from the corners and
bending up the sides so that they are
3'
perpendicular to the base. What size
of square must be cut out if the tray is to
contain 300 cubic inches?
The edges of a rectangular box are 5,
6, and 12 inches, respectively. What ~
6'
equal increase of each dimension will
increase the volume 200 cubic inches?
A rectangular box is 6 feet long and 3
feet wide. A board 6 inches wide is
fitted into the bottom of the box in the
3'
position shown in Fig. 13.7. Find the
FIG. 13.7.
length of the board.
A stone post at the entrance to a driveway is in the form
of a right prism whose base is a square and whose altitude
is 4 feet, surmounted by a sphere whose diameter is equal
to the side of the base of the prism. If the total volume is
20 cubic feet, what is the diameter of the sphere '!
X4
[Ch. XIII
THEORY OF EQUATIONS
230
13.18. The general cubic.
The general cubic equation is
aox 3 + alx 2
or
+ a2X + a3
x 3 + bx 2
0,
=
+ ex + d
40
= 0,
vi< 0,
(1)
(2)
,:,
where
d
a,
=-.
ao
(3)
Set
x
y -
=
b
3'
(4)
. I
and (2) becomes:
y3
+ Cy + D
v
= 0,
(5)
in which
be
D = d - -
3
3
2b
+-..
;
27
(6)
Equation (5) is called the reduced cubic.
To solve the reduced cubic we introduce two unknowns,
u and v, whose sum is to be a root of the reduced cubic;
that is, we set
(7)
y = u + v,
and substitute in (5).
placed in the form
u3
The resulting equation may be
+ v + (3uv + C)(u + v) + D
3
= O.
(8)
Since we have substituted two unknowns, u and v, for
the single unknown y, we can impose a condition on them.
If we impose the condition
3uv
+C
= 0,
(9)
equation (8) reduces to the simpler form
u3
+v +D
3
= 0.
(10)
i;
!
THE GENERAL CUBIC
§13.181
231
Solving (9) for v and substituting in (10), we obtain
u3
CS
-
_,')7 u 3
+D
(11)
0,
=
or
(12)
This is a. quadratic in us, and we find
_ D
2 +
- ~2 VD2
+ 4cs/27.
(13)
Taking.
(14)
( we find from (10) that
v3 = NOTE.
D 1
2 - 2 v'152+
If we take u 3 = -
that v3 = -
4C3j27. ,
(15)
D 1
-:2 - 2 V D2 + 4C 3/27,
~ + ~ VD2 + 4C3/27;
i.e.,
U
we find
and v have simply
been interchanged. This is to be expected, since they enter
symmetrically into the various equations such as (7), (8), and
(9).
There are three cube roots of (14) and three cube roots
of (15). They can be found by the Illethods of § 12.9.
Let U and V be cube roots of (14) and (15) respectively
which satisfy condition (9), namely, 3UV = -C. The
other cube roots of (14) are wU and w~U, and the other
cube roots of (15) are wV and w2V, where wand w2 are the
imaginary cube roots of 1 (see exercise XII.F, 23), namely,
1
.
1
.
w = - 2 + ~ v'3,
w2 = - 2 - ~ v'3.
These roots may be paired to satisfy (9) as follows:
3'wU'w 2 V = -C,
3'w 2 U'wV
=
-C.
THEORY OF EQUATIONS
232
[Ch. XIII
If these pairs of values are substituted in (7), we obtain
for the roots of the reduced cubic,
Yl = U
+ V,
Y2
+ w2V,
= wU
Ya = w2 U
+ wV.
(16)
,,
II
Making use of (4) and (3), we find for the roots of (1),
Xl
= U
+V-
a1
-3
ao '
(17)
Xs
=
(,)2U
+ (,) V
i
al
- 3-'
ao
These expressions for the roots of a cubic are called
Cardan's formulas.
Example 1.
Solve the equation x 3
+
3X2
+
15x
+1=
(18)
O.
SOLUTION. Set x = y - 1.
yS
Set y = u
+ 12y -
(19)
12 = O.
+ v in (20):
u + v + (3uv + 12)(u + v)
3
3
(20)
-
12 = O.
(21)
Impose the condition
3uv
reducing (21) to
+ 12 =
US
0,
+ VS -
or
uv
= -4,
12 = O.
(22)
(23)
Eliminate v between (22) and (23):'
u 6 - 12u s - 64 = O.
u 8 = 16.
va = -4.
From (22),
Take
U=~=2¢,
V=--0.
(24)
(25)
(26)
(27)
These values satisfy (22).
Yl
Xl
=
-0, Y2 = w20 - w 0, Ys = w 20 2\Y2 - -0 - 1, X2 = w2\Y2 - w .y'4 - 1,
=
2V!z -
2
2
w0;
2
Xa
= w22\Y2 - w.y'4 -
1.
-
233
THE GENERAL QUARTIC
§13.191
Example 2.
Solve the equation x 3
-
SOLUTION. Set x =
+ v in (28):
U
3
U
(28)
12x - 8 = O.
+ v + (3uv
3
+ v)
- 12)(u
- 8 = O.
(29)
Impose the condition
3uv - 12 = 0,
or
Uti
1=
(30)
4,
1-
reducing (29) to
u3
+v
3
-
8
= O.
':p ~P)
':'.
(31)
/: Eliminate v between (30) and (31):
From (30),
US - 8u 3 + 64 = O.
u 3 = 4 + 4 0 . i = 8 cis 60°.
v3 = 4 - 40· i = 8 cis (-60°).
(32)
(33)
(34)
As cube roots of (33) and (34) respectively take
U
= 2 cis 20°,
(35)
These values satisfy (30).
Xl
X2
X3
= U + V = 4 cos 20° = 3.759,
= wU + w2 V = 2 cis 140° + 2 cis 220°
= -4 cos 40° = -3.064,
= w2 U + wV = 2 cis 260°
= -4 cos 80° = -0.695.
+ 2 cis 100°
13.19. The general quartic.
The general quartic, or fourth-degree equation is
aox4
or
+ alx 3 + a2x2 + a3X + a4 =
X4 + bx 3 + cx 2 + dx + e =
0,
O.
(1)
(2)
The first step in the solution of (2) is to write it in the
form
(3)
[Ch. XIII
THEORY OF EQUATIONS
234
Complete the square of the left side by adding b2x 2/ 4 to
both sides:
(4)
We now introduce a number y, whose value is to be
determined later, and add
(X2 +
b;) y + ~2 to both sides
of (4), getting
bX)2 (
bX)
y2
( X2 + "2 + X2 +"2 y + "4
2
b
= ( "4
-
c
~,
+ Y) X2 + (bY
"2 -
(
d)
y2 - e.
x + "4
(5)
The right member of (5) is quadratic in x and will be a
perfect square if its discriminant is zero, that is, if
("2by - d)2 - 4 (b2"4 -
c
+ y )(y2)
"4 - e
=
o.
(6)
Equation (6) simplifies into*
y3 _ cy2
+ (bd
- 4e)y - b2e
+ 4ce
- rJ,2 = O.
(7)
We now determine y so that (7) will be satisfied; that is,
we find a root of the cubic (7). Let this root be y = r.
Substitute it in (5), whose right side will then be the square
of some linear function, px + q. Equation (5) will then
have the form
(X2
+ b; + ~y
= (px
+ q)2,
(8)
from which we get
bx
r
X2 + 2" + 2 = ± (px + q).
(9)
The two quadratic equations in (9) can be solved by the
quadratic formula or otherwise, yielding altogether four
roots.
* Equation (7) is called the resolvent cubic for (1), or (2).
I
EXERCISES
235
It can be shown that no matter which one of the three
roots of (7) is used, the final results will be the same.
The foregoing method is due to Ferrari.
Example.
------ Solve the equation
+ 6:1
4x 3
X4 -
-
+3
12x
O.
=
(10)
SOLUTION.
Add
4X2
Add (x 2
3.
to both sides:
X4 -
I
+ 12x -
4x 3 = - 4x 2
X4 -
+ (x 2 -
(X2 - 2X)2
(x
2
2x
-
+ 4x 2 =
12x - 3.
,(
+ ~2 to both sides:
2x)y
-
4X 3
2x)y
+~
+2y)2 =
= (x 2 - 2x)y
yx 2 -, 2(y - 6)x
+ ~ + 12z +"4~ - 3.
3,
(11)
Equate to zero the discriminant of the right side:
4(y - 6)
2- 4y (~2 - 3)
y3 - 4y2
+ 36y -
The roots of this last equation are 4,
Set y = 4 in (11):
x
2
x2
-
(x 2 - 2x
2 = 2x
1,
+
4x + 1 =
2x
x
+
0,
= 2
+ 2)2
= 4x 2
x2
"=
0,
144 = O.
± 6i.
+ 4x + 1.
2x + 2 =
-
X2
± 0.
x
-
(2x
= -3,
=
± i0.
EXERCISES XIII. J
Solve by the methods of §§ 13.18-13.19:
1. x 3
3. x 3
5. x 3
= O.
= 0.
+ 9x + 6 = O.
-
6x - 6
3x + 1
2. x 3
4. x 8
6. x.8
-
-
-
+ 1),
12x - 20 = O.
6x - 2 :;:::: O.
12x - 8 = O.
THEORY OF EQUATIONS
236
7.
9.
11.
13.
16.
16.
17.
18.
19.
20.
21.
[Ch. XIII
2x 3 - 6x - 5 = O.
8. 9x 3 - 30x - 82 = O.
3
2
x + 6x + 6x - 10 = O. 10. x 3 - 6x 2 + 21x - 20 = O.
x 3 - 6x 2 - 3x - 8 = O.
12. x 3 - 9x 2 - 9x - 15 = O.
3
x - 3x + 0 = O.
14. x 3 + 3x 2 + 9x + 9 = O.
2
3
X4 + 6x + 8x + 8x 16 = O.
X4 8x 3 + 20x 2 - 16x - 21 = O.
x· + 2x 3 - 8x 2 - 6x - 1 = O.
X4 8x 3 + 9x 2 + 8x - 10 = O.
3
X4 + 5x 25x - 25 = O.
x· + 10x 3 + 35x 2 + 46x + 10 = O.
X4 6x 8 + 16x 2 - 54x + 63 = O.
: ,(
i
!
I,i
i i
13.20. Algebraic solution of equations.
The solutions of an integral rational equation,
aoX n
+ aIx n - 1 + . . . + an-IX + an
=
0,
(1)
are functions of the coefficients. If these functions
involve no operations other than a finite number of addi- tions, subtractions, multiplications, divisions, and extractions of roots, the solution is algebraic.
We have already obtained algebraic solutions of the
general quadratic, cubic, and quartic (n = 2, 3, 4, respectively, in (1», and it might be surmised that the general
/
equation of any degree could be similarly solved. However, it can be proved that the general equation of degree
higher than four has no algebraic solution.
I
13.21. Coefficients in terms of roots.
If rl, rz, . . . ,rn are the roots of an equation, then the
equation may be written
(X - rl)(x - T2)·· •• (x - rn) = O.
Multiplying the factors together, we get
xn - (Tl + r2 + ... + rn)X n- l
+ (rlrZ + rlTa + ... + Tirn + Tzra + ...
- (rlr2ra + ... + rn_2rn_lrn)Xn-a +
+ (_1)nrlr2 ••• Tn = O.
/'
(1),
+ rn--lrn)Xn-t
(2)
.•
EXERCISES
237
Comparing (2) with the form
~+b~_1+b~-2+b~n~+
. + b_lx + bn
= 0,
(3)
'.,\-'
we see that
'1"Of;,',
bi = -sum of roots, ",,~
b2 = sum of products of roots taken two at a time,
ba = - sum of products of roots taken three at a time,
. .... . . . . . . . . . . . . . .. .. ......... . .
bn
=
(-l)n X product of roots.
H the equation is given in the form
(4)
(
it is merely necessary to divide through by ao to reduce it
to the form (3).
EXERCISES XIII. K
Find the sum of the roots and the product of the roots of the
following equations:
1. x 3
3.
4.
5.
6.
7.
8.
+ 4x + + 9 =
+
2
X
O.
2. x 3 - 5x 2
6x
12x 2 - 17x - 21 = O.
2x 3 - 23x - 18 = O.
6x·
2X2 - 7x
15 = O.
7x 3 - 19x 2 - 14x - 1 = O.
,...,'
2X4 + 3x· - 7x 2 + 9x - 8 = O.
"f
, ,
5x 4 + 10x 2 + 11x + 30 = O.
,I "
. I
10x4 - 15x 3
4.1;2 - X = O.
+8
= O.
3x 3 -
+
9.
10. tx 5 -
+
+
ix' + 1-X 3 -
/..
tx2 + t = O.
11. Solve the equation x 3 - 3x 2 + kx - 12 = 0, given that
the produut of two of its roots is - 6. What must be the
value of k?
12. Solve the equation 7x 3 - 37x 2 + kx - 12 = 0, given that
the sum of two of its roots is 5. Find the value of k.
13. Solve the equation 6x 3 - llx 2 + kx - 24 = 0, given that
two of its roots are reciprocals of each other. Find k.
14. Solve the equation 9x 3 - 21x2 + kx + 25 = 0, given that
it has a double real root.
[Ch. XIII
THEORY OF EQUATIONS
238
16. Solve the equation 8x 3 + kx 2 - 18x + 9 = 0, given that
one root is the negative of the other.
16. Solve the equation 12x4 - 44x 3 + cx 2 + dx + 3 = 0, given
that it has two double rational roots. Find e and d.
17. Solve the equation X4 - 8x 3 + ex 2 + dx + 169 = 0, given
that it has an imaginary double root. Find e and d, which
are assumed to be real.
18. Solve the equation 4X4 + 2x 3 - 3x 2 + dx + e = 0, given
that it has a triple root. Find d and e.
19. Show that if the equation x 3 + ex + d = has a double
root, then
(~y + (~y
°
=
0, and conversely.
20. Show that if the equation x 3
triple root, then d =
21. Show that if d =
x~
(~)
~r
+ bx 2 + ex + d =
+ bx + ex + d = ° has
2
a
3.
\
'.
and 3e = b2, then the equation
°has a triple root.
I
l
, , !', (
, F
J'
CHAPTER XV
Compound Int,"""and Annuities
\
'
'i'
.
15.1. Compound amount.
;;\
..
f
j
.i
Interest is money paid for the use of borrowed money.
A sum of money placed at interest is called the principal.
If the interest is added to the principal at the ends of equal
intervals of time the interest is said to be compounded,
or to be converted into principal. The interval of time is
called the conversion, or interest, period. This is usually
a year, half-year, or quarter. If no period is stated it will
be understood to be a year. The sum to which the principal accumulates is called the compound amount, or
simply the amount. The rate is the interest paid on one
unit of capital for one period. (In other words, it is the
ratio of the interest, for one period, to the principal.)
For example, a rate of 3 per cent (i.e., per hundred) means
that $3 per period would be paid on $100, or $0.03 per
period on $1. In problems it is best to write the rate as a
decimal. In this example the rate would be written 0.03.
If interest is compounded oftener than once a year, it is
customary to quote the rate on an annual basis. For
example, a rate of 3 per cent compounded semiannually
means actually a rate of It per cent for each half-year'
period. Such a rate is called a nominal rate of 3 per cent
compounded semiannually.
If P is the principal and r the rate, the interest due at
the end of the first period is Pro The amount at the end of
the first period is P + Pr, or P(1 + r). It is seen that the
amount at the end of one period can be obtained by multiplying the principal P by (1 + r). For example, the
263
264
COMPOUND INTEREST AND ANNUITIES
[Ch. XV
amount, at the end of one year, of $1000 at 3 per cent is
$1000 X 1.03 = $1030.
The principal for the second period is P(l + r). The
amount at the end of the second period is found by multiplying this principal, P(l + r), by (1 + r), getting
P(l + r)2.
Similarly, the amount at the end of three periods is
P(l + r)3.
In general, the amount at the end of n periods is
A
=
P(l
+ r)n.
(1)
For example, the amount of $1250 at 2 per cent for 3 years
is $1250(1.02)3.
15.2. Present value.
The present value of a sum of money due at a future
date is the principal which, invested at interest, will -,
accumulate to the specified sum at the end of the time.
It can be found by solving for P in equation (1) of the
preceding section. Thus, the present value of an amount
A due in n periods, when money can be invested at rate r is
!
P = (1
A
+ r)n
= A(l
+ r)-n.
(1)
Thus, if the interest rate is 4 per cent, the present value of
$2500 to be paid 3 years hence is $2500/(1.04)3.
Problems in compound interest can best be solved by
means of tables of compound amounts and present values.
When no such tables are available logarithms may be
resorted to, although results will not ordinarily be very
accurate unless logarithms to a large number of places
are employed. Results to any desired degree of accuracy
can be obtained by using the binomial formula.
Example 1.
Find the amount of $1000 at 4% interest compounded
annually for 5 year!;.
i
PRESENT VALUE
§15.21
SOLUTION BY COMPOUND AMOUNT TAllLES.
A
265
"
-
= P(l + r)n = $1000(1.04)6.
In Table III at the back of the book we look in the column
headed 4%, and opposite n = 5 we find 1.2167. That is, (1.04)6
= 1.2167. Therefore,
A
= $1000 X 1.2167 = $1216.70.
SOLUTION BY LOGARITHMS.
A = $1000(1.04)6.
log 1.04 = 0.0170
X5
0.0850
log 1(JOO = l!..___
log A = 3.0850
A = $1216 (approximately).
SOI,UTION BY BINOMIAL FORMULA.
(1.04)6
5·4
5·4·3
+ 1="2
(0.04)2 + 1 . 2 . 3 (0.04)1 +
1 + 0.20 + 0.016 + 0.00034 -t- . . . = 1.21664,
= 1 + 5(0.04)
=
.
A = $1000 X 1.21664 = $1216.64.
This is correct to within one cent, thEl accurate value being
$1216.65 to the nearest cent. One more term of the binomial
formula would give this accurate value.
Example .2.
Find the present value of $1250 due in 3 years if the rate of
interest is 4 %.
SOLUTION BY PRESENT VALUE TABLEEl.
P = A(l
+ r)-n =
$1250(1.04)-3.
In the 4% column of Table IV we find, opposite n = 3, the
value 0.88900. Thus,
P
= $1250 X 0.88900 = $1111.20.
COMPOUND INTEREST AND ANNUITIES
266
[Ch.XV
I
I
SOLUTION BY LOGARITHMS.
log 1.04 = 0.0170
X3
"-.
j
log (1.04)3 = 0.0510
log 1250
log (1.04)3
log P
P
3.0969
0.0510
= 3.0459
= $1112.
=
=
SOLUTION BY BINOMIAL FORMULA.
P
=
$1250(1.04)-3.
(1.04)-3 = 1 - 3(0.04)+ -~(.~4)(0.04)2+ -3i-:-~).(;5)(0.04)3
+ -3( -4)( -5)( -6) (0.04)4 _ . . .
1·2·3·4
= 1 - 0.12 + 0.0096 - 0.00064 + 0.00003840 = 0.8889984.
P =
~1250
X 0.8889984 = $1111.25.
i
~
i
EXERCISES XV. A
1. Find the amount of $1000 for 6 years at 4% compounded
(a) annually, (b) semiannually, (c) quarterly.
2. Find the present value of $1000 due in 5 years, the rate of
interest being 4% compounded (a) annually, (b) semiannually, (c) quarterly.
3. Find the present value of $1250 due in 4 years, the rate of
interest being 3% compounded semiannually.
4. Find the amount of $800 invested for 7 years at 5% compounded semiannually.
o. What is the amount of $500 at the end of a year if the
interest rate is 1 % per month?
6. What is the present value of $500 due one year hence if the
interest rate is 1 % per month?
7. How much must be deposited in a savings bank paying t!%
to amount to $2000 in 5 years?
8. How much must be deposited in a savings bank paying 2 %
compounded semiannually to amount to $1500 in 4 years?
9. In how many years will $100 amount to $160.10 at 4%?
i
.1
§15.31
f
ANNUITY
267
10. In how many years will $100 amount to $1.41.30 at 5 %
compounded semiannually?
11. In how many years will $100 ar ount to $116.10 at 4%
compounded quarterly?
12. If $100 amounts to $126.90 in 16 years what is the rate?
13. If $100 amounts to $121.84 in 8 years what is the rate?
14. If the present value of $100 due in 7 years is $84.13 what is
the rate?
15. If the present value of $100 due in 10 years is $74.41 what
is the rate?
16. What rate, compounded semiannually, will cause $100 to
amount to $126.90 in 8 years?
17. What rate will cause $1250 to amount to $1500 in 5 years?
18. If the present value of $2000 due in 6 years is $1600 what
is the rate?
19. A United States Savings Bond, Series E, costs $75 and is
redeemed 9 years 8 months later for $100. What rate of
interest does it yield? (See exercise 22.)
20. How long would it take a sum of money to double itself
at (a) 1%? (b) 2%? (c) 3%? (See exercise 22.)
21. What rate will cause a sum of money to double itself in
(a) 15 years? (b) 20 years? (c) 30 years?
22. The compound amount for a fractional part of a year is
defined by the usual formula S == P(l + r)", in which
fractional values of n are permissible. Find the compound
amount of $100 at 4% (compounded annually) for (a) 6
months, (b) 3 months, (c) 1 month.
23. What single sum of money paid at the end of 3 years will
fairly discharge two debts, one of $3000 due in 2 years and
another of $2500 due in 5 years, if the interest rate is 3 %?
SUGGESTION. The fair amount would be the compound
amount of the $3000 for 1 year plus the value of the $2500
2 years before it is due.
15.3. Annuity.
A series of equal periodic payments is called an annuity.
Thus, a series of equal annual payments on a piece of property is an annuity. Likewise, the monthly payments of
I
268
COMPOUND INTEREST AND ANNUITIES
[eb.XV
rent for a house or an apartment constitute an annuity.
However, unless otherwise stated, it will be assumed that
the payments are annual and that the first payment is
made at the end of the first year. For example, a threeyear annuity of $1000 would consist of three payments of
$1000 each, one payment made at the end of one year,
another at the end of two years, and the last at the end
of three years.
!,
._,
.
,;:
15.4. Amount of an annuity.
We shall derive the formula for the amount of an annuity
(that is, the sum to which the periodic payments accumulate at compound interest) on the basis of payments of
one dollar each. If each payment is R dollars, instead of
one dollar, the accumulation will be R times as much.
If the annuity continues for n years there will of course
be n payments. Since the first payment is made at the
end of the first year, it will be at interest for n - 1 years,
and will, by formula (1) of § 15.1, amount to (1 + r)n-t.
The second payment will be at interest for n - 2 ~ ears,
and will amount to (1 + r)n-2. The next to the last parment will be at interest for one year, and will amount to
(1 + r). The final payment of one dollar is made at the-'-end of the n-year term, and its value at that time will consequently be 1. The total amount of the annuity will be
the sum of the amounts of the separate payments. Writing these in reverse order, we have for their sum,
s
= 1
+ (1 + r) + ... + (1 + r)n-z + (1 + r)n-t.
(1)
This is a geometric progression of n terms whose common .
ratio is (1 + r). Its sum, by formula (3) of § 11.7, we
find to be
(1 + r)n - 1
s = (1 + r) - 1 '
or
s=
(1
+ r)n
r
- 1
.
(2)
§15.51
PRESENT VALUE OF AN ANNUITY
!69
If each payment is R instead of 1 the amount of the
annuity will be Rs, or
S
= R (1
+ r)n -
1.
(3)
r
The amount of an annuity of $500 a year for 3 years at
2t per cent, for example, is
$500 X
(1.0~~J;5-
1. ","
15.5. Present value of an annuity.
The present value of an annuity is the sum of the present
values of all of the payments. The present value of the
first payment of an annuity of one dollar is (1 + 1')-1; the
present value of the second payment is (1 + 1')-2; and so
on. The last payment is made at the end of n years, and
has a present value of (1 + r)-n. The total present value
of the annuity is
p = (1
+ 1')-1 + (1 + 1')-2 + ... + (1 + r)-n.
(1)
This is a geometric progression of n terms with common
ratio (1 + 1')-1. By formula (4) of § 11.7, we find
p
I
= (1
l'
)-1. 1 - (1
1 - (1
= _1_ . 1 - (1
1
or
+
+ l'
1 - (1
+ 1')-1'
+ 1') 1
+ r)-n,
+ r)
1
p=l-(\+r)-n.
(2)
If the annual payment is R instead of 1 the present value
of the annuity will be Rp, or
P = R 1 -
(1
+ r) -n.
r
(3)
For example the present value of an annuity of $500 a
year for 3 years at 2t per cent is
$500 X 1 - (1.025)-3.
0.025
COMPOUND INTEREST AND ANNUITIES
270
[Ch. Xv
Problems in annuities are best solved by means of
annuity tables. When these are not accessible we can use
logarithms, or, if greater accuracy is desired, the binomial
formula.
Example 7.
Find the amount of an annuity of $1000 a year for 5 years at
t!.%.
SOLUTION BY ANNUITY TABLES.
S = R (1
+ r)n r
1 = $1000 X (1.04)6 - 1.
0.04
,
In Table V,. in the 4% column, we find opposite n = 5 the value
5.4163 for [(1.04)6 - 1]/0.04. Thus,
S = $1000 X 5.4163
= $5416.30.
SOLUTION BY LOGARITHMS.
S = $1000 X (1.04)6 - 1
0.04
log 1.04 = 0.0170
X5
log (1.04)6 = 0.0850
(1.04)5 = 1.216.
I
S
= $1000
X
,';",
, 1\
1.2~~0; 1 = $5400 (a~~te1y).
SOLUTION BY BINOMIAL FORMULA.
(1.04)5 = 1 + 5(0.04)
.'
1:
+ 10(0.04)2 + 10(0.04)3 + 5(0.04)4
+ (0.04)6
= 1 + 0.20 + 0.016 + 0.00064 + 0.00001280
+' )000001024
= 1.2166529.
S = $1000 X
1.216~~g:
-
1 = $5416.32.
Example 2.
Find the present value of an annuity of $1250 a year for 3
years at 4%.
I
I
, !
EXERCISES
271
SOLUTION BY ANNUITY TABLES.
P = R 1 - (1
r
+ r)-n
= $1250 X 1 - (1.04)-3.
0.04
In Table VI, in the 4% column, we find opposite n = 3 the
---- value 2.7751 for [1 - (1.04)-3]/0.04. Thus,
P = $1250 X 2.7751 = $3468.90.
SOLUTION BY LOGARITHMS.
> ~Jl"
P = $1250 X 1 - (1.04)-8.
0.04
log 1.0·1 = 0.0170
X -3
log (1.04)-3 = -0.0510 = 0.9490 - 1
(1.04)-3 = 0.8892.
P = $1250 X 1
-0~O~892 =
$3462 (approximately).
SOLUTION BY BINOMIAL FORMULA. In example 2 of § 15.2
found by the binomial formula (1.04)-3 = 0.8889984.
P
=
$1250 X 1 -
~~8489984
We
= $3468.80.
The correct value (found by using annuity tables to 6 decimal
places) is $3468.86. In order to obtain this value, the binomial
expansion used in § 15.2 for calculating (1.04)-3 should be
carried to one more term.
EXERCISES XV. B
1. Find tile amount of an annuity of $1000 a year for 6 years
at 3%.
2. Find the amount of an annuity of $1200 a year for 5 years
at 2~%.
3. Find the present value of an annuity of $1000 a year for 5
years at 4%.
4. Find the present value of an annuity of $800 a year for 3
years at 5%.
5. Find (a) the amount and (b) the present value of an annuity
of $500 per quarter for 6 years, the interest rate being 4%
compounded quarterly. .
272
COMPOUND INTEREST AND ANNUITIES
[Ch. XV
6. The purchaser of a piece of property pays $5000 down and
agrees to pay $1000 at the end of each year for 15 years.
What is the equivalent cash price of the property if the
interest rate is 5 %?
'1. A man buys a house, paying $8000 down and $1000 at the
end of each 6 months for a period of 10 years. Find the
equivalent cash price of the house, assuming an interest
rate of 5% compounded semiannually.
8. A man and his wife buy some furniture, paying $50 down
and $50 at the end of each month for a year. Find the I
equivalent cash price of the furniture on the basis of an
interest rate of 1 % per month.
9. A man has a debt of $5000 due in 4 years. In order to
pay it off he deposits a certain amount, at the end of each
year for the 4 years, in a fund bearing 2t% interest. How
much must his annual payment be? (A fund of this sort
is called a sinking fund.)
10. A man buys a house for $16,0'l0. He pays $3000 down and
agrees to pay the balance, including principal, and interest on the unpaid balance at 5 %, in 10 equal annual installments. What annual payment must he make?
11. A piece of property is purchased for $20,000, The purchaser makes a down payment of $5000 and alrees to pay
the balance, including principal, and interest on the unpaid
balance at 4%, in 12 equal annual installments. Find the
__.. ,. amount of each installment.
12· A quarry yields a net annual income of $10,000. It is
estimated that the rock will be exhausted in 10 years.
What price could an investor afford to pay for the quarry
:if he is to make 6% on his investment?
13. A man deposits $300 at the end of each 6 months in a bank
paying 2% compounded semiannually. How much will
he have to his credit at the end of 10 years if he makes no,
withdrawals?
14. Find the present value of the annuity in the preceding
exercise.
15. A bond has attached to it 20 coupons. At the end of each
half year a coupon is detached by the owner of the bond and
cashed for 32.50. At the end of 10 years he turns in the
EXERCISES
16.
-----
17.
18.
19.
20.
21.
22.
23.
273
bond and receives $100. What is the present value of the
bond (including the coupons) if the current interest rate is
4% compounded semiannually?
A manufacturing company has a machine which will be
completely worn out in 8 years. How much must be set
aside at the end of each year, in a fund invetsted at 3%,
to purchase a new machine costing $5000 when the old one
is worn out? (A fund of this sort is called a depreciation
fund.)
A son receives, by the terms of his father's will, an annuity
of $5000 a year for 15 years, beginning at the end of 1 year.
The state imposes an inheritance tax of 1 % on the present
value of the annuity. What is the amount of the tax if
money is worth 4% (Le., if the current interest rate is 4%)?
A man puts $1000 a year, at the beginning of each year
for 15 years, into a fund which is invested at 3 %. How
much will be in the fund at the end of the 15 years?
The beneficiary of a life insurance policy is to receive $2000
a year for 5 years, the first payment to be made at the time
of death of the insured. Find the value of the annuity
at the time of death of the insured, assuming the current
interest rate to be 4%.
A man puts $1000 at the end of each year for 5 years into a
fund which is invested at 3 %. If the fund is allowed to
accumulate for 5 years more, without any additional payments being made, what will be its amount at the end of
the lO-year period?
A student borrows $1000 at the beginning of each of his 4
years in oollege, signing a note to payoff the debt, with
interest at 2t%, in 5 equal annual installments, the first
installment to be paid 5 years after the first $1000 was
borrowed. Find the amount of each installment.
Solve the next two exercises by the method of § 13.14 or
that of § 13.16.
An annuity of $600 a year for 5 years amounts to $3200.
What is the rate?
The present value of an annuity of $500 a year for 4 years
is $1875. Find the rate.
/
CHAPTER XVI
I
Permutations and Combinations
16.1. Fundamental principle.
If one thing can be done in 'IE different ways, and {f after
it has been done in any of these ways a second thing can be
done in n different ways, the two things can be done in the
order stated in mn different ways.
This principle can obviously be extended to the case in
which more than two things are involved,
Example 1.
How many committees consisting of one boy and one girl
can be selected from a group of 3 boys and 2 girls?
SOLUTION. Let the boys be A, B, C, and the girls X, Y.
the possible committees are:
A,X
A,Y
B,X
B, Y
Then
C,X
C, y /'
There are 3 ways to choose the boy, and for each of these way8
there are 2 ways to choose the girl, so that there are 3 X 2 = 6
ways altogether to form the committee.
Example 2.
How many three-place numbers can be formed with the digits
1, 2, 3, 5 if each digit can be used only once?
SOLUTION. The hundreds' place can be filled in anyone of
4 different ways, the tens' place can then be filled in anyone of
3 different ways, finally the units' place can be filled in 2 different
ways. Thus, we can form 4 X 3 X z' = 24 numbers.
274
.
\
275
PERMUTATIONS
516.21
Example' 3.
How many three-place numbers can be formed with the d~
1, 2, 3, 5 if any digit can be repeated?
SO:r:UTION. The hundreds' place can be filled in 4 ways, the
---- tens" place can then be filled in 4 ways, and the units' place can
also be filled in 4 ways. We can form 4 X 4 X 4 = 64 numbers.
Example 4.
How many three-place numbers less tha.n 300 can be formed
with the digits 1, 2, 3, 5 if repetition of digits is not allowed?
(
SOLUTION. The hundreds' place can be filled in 2 different
ways (with 1 or 2 only, since 3 or 5 in this place would make the
number greater than 300). The tens' place can then be filled in
3 ways and the units' in 2. Thus, 2 X 3 X 2 = 12 numbers can
be formed.
Example S.
"
,'f
How many even numbers of three places can be formed with
the digits 1, 2, 3, 5, repetition of digits not allowed?
SOLUTION. The units' place can be filled in only 1 way (with
the 2). The tens' place can then be filled in 3 ways, and the
hundreds' in 2. There are 1 X 3 X 2 = 6 numbers.
16.2. Permutations.
Each arrangement in order of a set of things is called a
permutation of the set. Thus, the set of letters a, b, c, if
we use all of them, can be arranged in the following orders:
abc
acb
bac
bca
cab
cba
We see that there are six permutations. This can be
determined by noting that the first place can be filled by
anyone of the three letters, that is, in 3 different ways;
after the· first place is filled the second place can be filled
by either of the two remammg letters. The first two
276
PERMUT A nONS AND COMBINATIONS
i
I
(Ch. XVI
places can therefore, according to our fundamental principle, be filled in 3 X 2 = 6 ways. After the first two
places are filled the letter to go into the third place is
determined. Thus, the three letters can be permuted in
3 X 2 X 1 = 3 J = 6 ways.
If we have four letters, a, b, c, d, and take them two at a
time, we can form the permutations:
We note that the first place can be filled in 4 different
ways, the second in 3, and consequently the two places in
4 X 3 = 12 different ways.
In general, if we have n different things and take r of
them at a time, we can fill the first place in n different ways
After the first place is filled there are n - 1 things left,
and anyone of these can be put into the second place.
That is, the second place can be filled in n - 1 different
ways. The first two places can therefore be filled in
n(n - 1) different ways. The third place can be filled,
in n - 2 different ways, and so on to the rth place, which
can be filled in n - r + 1 different ways. Thus, the
number of permutations of n things taken r at a time is
given by the formula
nPr = n(n - 1) . . . to r factors
= n(n - 1) . . . (n - r
+ 1).
(1)
(2)
If we set r = n, we find that the number of permutations of n things taken all at a time is
nPn = n(n - 1) . . . 3' 2 . 1
'=
n 1.
(3)
16.3. Permutations of things some of which are alike.
Suppose that we have five letters a, a, a, b, c. Consider
any permutation of them such as abaac. If the a'8 were
..1
1
EXERCISES
277
different, say aI, a2, aa, we should have 3! permutations
similar to this, namely,
a 2balaSc
a 2ba aalC
a l ba 2a S C
a l ba aa2c
aabala2c
aab a 2a l c
These permutations would be indistinguishable if the subscripts were removed. Thus, to any given permutation
such as abaac, in which the a's are alike, there corresponds
a set of 3! permutations in which the a's are different.
That is, there are 3! times as many permutations when
the a's arc different as there are when the a's are alike.
But for five different letters aI, a2, aa, b, c there are 5!
permutations. Consequently, if the three a's are alike
the number of permutations is 5! + 3! = 20.
In general, if there are n things of which nl are alike of
one kind, n2 alike of another ,kind, and so on, the number
of permutations of these things taken all at a time is i ;p =
n!
nl ! nz !
This can be established by the same line of reasoning
employed for the special case.
EXERCISES XVI. A
1. In how many different orders can 5 persons be seated (a)
in a row? (~) about a round table?
2. In how many different orders can 6 books be arranged on a
shelf?
3. In how many different orders can 7 books be arranged on a
shelf (a) if a certain 3-volume work is not to be separated?
(b) if this 3-volume work is not merely not to be separated
but must be kept in the proper order?
4. In how many different orders can 4 men and 4 women be
seated in a row so that no 2 persons of the same sex will be
together?
5. In how many different ways can 4 persons be seated in
consecutiye seats in a row of 7 seats?
,J,
278
PERMUTATIONS AND COMBINATIONS
[Ch.XVI
6. In how many different ways can 5 students make a selection
of 1 book each from a shelf containing 15 books?
7. A shelf contains 5 books in red binding, 4 books in blue,
and 3 in green. In how many different orders can they
be arranged if all books of the same color must be kept
together?
8. A shelf contains 7 books bound in red and 5 bound in blue.
In how many different orders can they be arranged (a) if
books of the same color must be kept together? (b) if the
red books must be kept together, but the blue books may
be separated?
9. How many different 3-place numbers can be formed from
the digits 1 to 9 inclusive (a) if repetitions are allowed?
(b) if repetitions are not allowed?
10. How many different automobile license numbers can be
formed by using 1 to 6 digits preceded by a letter if the
digit immediately following the letter cannot be zero and
the letters 0 and I are excluded?
11. How many different 5-place even numbers can be formed
with the digits 1, 2, 3, 4, 5, 6 (a) if repetitions are not
allowed? (b) if repetitions are allowed?
12. How many different numbers less .than 400 can be formed
with the digits 1, 2, 3, 4, 5 (a) if repetitions are not allowed?
(b) if repetitions are allowed?
13. How many different numbers greater than 400 can be
formed with the digits 1, 2, 3, 4, 5 if repetitions are not
allowed?
14. There are 5 positions on a flagstaff and 4 different colors
of flags (at least 5 of each color). How many different
signals can be made by displaying 5 flags simultaneously?
16. There are 6 positions on a flagstaff and 6 flags, of which
R are red, 2 blue, and 1 white. How many different signals
can be made by displaying all of the flags simultaneously?
16. In how many different ways can 7 linemen be placed in
the 7 line positions of a football team if 2 of the men can
play end only?
17. In how many different ways can 4 baseball players be
assigned to the positions of the infield if only 2 of the men
can play shortstop?
-
COMBINATIONS
§16.41
279
18. How many different permutations can be made of the letters
of the following words? (a) "dependent," (b) "incision,"
(c) "calculable."
19. In how many different ways can 5 persons be seated in an
automobile having places for 2 in the front seat and 3 in
the back if only 2 can drive and one of the others insists
on riding in the back?
20. Six persons wish to select 1 book each from a shelf containing 10 books, 5 of which are in English, 3 in German, and
2 in French. How many different selections are possible
(8.) if one of the persons cannot read German and must
select a book in English or French? (b) if two of the persons
must select books in English?
16.4. Combinations.
A set of things without reference to the order in which
they are arranged is called a combination. Thus, abc, acb,
bac, cab, are the same combination, although they are different permutations. A formula for the number of combinations of n different things taken r at a time can readily
be derived from the formula for the number of permutations of n things taken r at a time. For, corresponding to
each combination of the r things there are rl permutations.
Therefore, if nCr denotes the number of combinations of
n things taken r at a time,
(1)
nPr = r 1nCr,
from which we get
nPIr'
nCr = r .
..
(2)
I
"-
or
nCr =
n(n - 1) ••• to r factors
1, 2 ••. T
= n(n -
1)
. (n T1
/
T
+ 1).
(3)
(4)
no ...... .
PERMUTATIONS AND COMBINATIONS
[Ch. XVI
If we multiply numerator and denominator by (n - r) !
we reduce (4) to the form
C =
n r
T
n !
.
l(n - T) !
'"
r i , ',n
(6)
•
We note that the formula (5) is also ·the fonnula for
That is,
,.<
r"
llCn -
(6) ,
or in words: The number of combinations of n things taken r
at a time is the same as the number of combinations of n
things taken n - r at a time.
This can also be established by the following reasoning:
To each combination of r things which we take from the
n things, there corresponds a combination of n - r things
which we leave.
16.5. Binomial coefficients.
By referring to § lOA, we see that the coefficient of
term number r + 1 of the binomial formula, is (4) of the
preceding section, namely, nCr. The binomial formula
may therefore be written
(a + b)n = an + nC1an-1b + nC2a n- 2b2 + ...
+ nCran-rbr +
+ nC"b".
This can also be shown as follows:
(a
+ b)"
= (a
+ b)(a + b)
. . . (a
+ b), n factors.
Each term of the expanded product is the product of n
letters, one from each of the n factors. Thus, every term
involving arb n- r is obtained by taking a from any r factors
and b from the remaining n - r factors. Therefore, the
number of terms involving arb n- r is the number of ways in
which r things can be selected from n, namely nCr.
i 6.6. Total number of combinations.
If we set a = b = 1 in the last equation,
(1
+ 1)"
= 2" = 1
w~ get
+ nC + nC + . . . + nC",
l
2
(1)
EXERCISES
181
from which we find
nC l
+ nC + ... + nCn = 2n 2
1,
(2)
which gives the total number of combinations of n. things
taken 1, 2, . . . , n - 1, or n at a time. For example,
the total number of combinations of the three letters a, b,
c, taken 1, 2, or 3 at a time, is 2 3 - 1 = 7, as can readily
be verified.
EXERCISES XVI. B
1. How many combinations can be made of the letters a, b ,
c, d, e, f, g, h taken 4 at a time?
,
(
-
2. How many different subcommittees of 3 can be chosen from
a committee of 11 persons?
8. How many different kinds of bouquets of 3 kinds of flowers
each can be made from 8 varieties of flowers?
4. How many different basketball teams can be formed from
a squad of 15 men?
6. A shelf of reserved reference books contains 25 different
6.
7.
8.
9.
10.
11.
12.
books. If a student is allowed to take out 2 books overnight how many selections can he make?
A student wishes to use 4 different colors to make a map.
In how many different ways can h8 do this if he has 10
colors to select from?
How many different combinations of 5 cards each can be
dealt from an ordinary deck of 52 cards?
How many different sums of money C9,n be formed with a
penny, a nickel, a dime, a quarter, and a half dollar?
How nl&ny different committees of 5 can be selected from
15 Rer-u~cltns and 10 Democrats (a) if it must contain 3
Republicans and 2 Democrats? (b) if it must contain at
least 3 Republicans?
How many different connections must a telephone exchange
be able to set up if it has 5000 subscribers?
There are 8 baseball teams in a league. How many games
will be played if each team plays each of the other teams
40 times?
How many different combinations can be formed from the
letters a, b, c, d, e, f, g, h taken one or more at a time?
i82
PERMUl A liONS AND COMBINA liONS
[Ch. XVI
13. The tennis squad of one college consists of 12 players, that
of another consists of 10 players. In how many ways can
a doubles match between the two institutions be arranged?
14. Ten golfers are at the first tee waiting to start. (a) In how
many ways can a foursome be formed, not considering the
way in which the players are paired as partners? (b)
Mter the first foursome has been formed, in how many
ways can the second be formed?
lI. From a list of 15 exercises on combinations and 1] exercises
on permutations how many different examinations can be
made which will include 4 questions on each of these topics?
16. How many different combinations consisting of 5 black
balls and 6 white balls can be selected from a box containing
8 black balls and 10 white balls?
17. In how many ways can one make a selection of 4 black balls,
5 red. balls, and 2 white balls from a box containing 8
black balls, 7 red balls, and 5 white balls?
18. In how many ways can one make a selection of 3 novels.
2 biographies, and 5 detectiv~ stories from a shelf containing
10 novels, 8 biographies, and 10 detective stories?
19. In how many ways can one make a selection of 8 books
from the shelf in the preceding exercise if he chooses at
least2 novels and at least 3 biographies?
ZOo Prove that nC.. nC +1 = n+lCr+l' (Note that Pascal's tri. angle, § 10.3, exemplifies this formula.)
+
T
MISCELLANEOUS EXERCISES XVI. C
1. In how many different orders can 7 persons be seated (a)
in a row? (b) about a round table?
2. In how many different orders can 5 keys be arranged on a
ring?
3. How many different committees of 6 can be selected from
12 seniors and 20 juniors (a) if the committee is to consist
of 4 seniors and 2 juniors? (b) if the committee is to consist
of at least 4 seniors?
4. How many different basketball teams can be formed from
a squad of 12 men if only 2 of the men can play center and
these two can play in no other position?
/ .
!
, I
I
!-
EXERCISES
283
6. Given 5 flags of different colors, how many different signals
can be made by displaying them in a vertical line, using
any number at a time?
6. From a collection of 5 different colors of flags (at least 5
flags of each color) how many different signals can be made
~
by displaying them in a vertical line, using any number of
flags from 1 to 5?
. '1. How many distinct lines can be drawn through 15 points
no 3 of which are in the same straight line?
8. In how many points do 10 lines, no two of which are parallel
and no 3 of which are concurrent, int~rsect?
9. In one set of parallel lines there are m lines; in another set of
lines which are parallel to each other but not to the first set,
there are n lines. How many parallelograms are formed?
(
10. In how many different ways can the letters of the word
"equation" be arranged without changing the order of the
vowels?
11. In how many different orders can the letters of the word
"quartic" be arranged if the first place is to be filled with
a consonant and no consonants are to be adjacent to each
other?
12. How many different baseball teams can be formed from a
squad of 18 men if only 4 can pitch, only 2 can catch, and
these 6 men can play in no other position?
13. In how many different ways can 6 obj{lcts be separated into
2 equal groups?
14. In how many different ways can 7 obj~cts be separated into
groups of 4 and 3 respectively?
16. In how many ways can 7 persons be fmated in a row (a) if
a certain couple must be together? (b) if a certain couple
must not be together?
16. In how many ways can 7 persons be !Seated about a round
table (a) if a certain couple must be together? (b) if 8; certain couple must not be together?
1'1. From 8 different pairs of gloves, in how many ways can a
right-hand and a left-hand glove be selected without taking
a pair?
18. In how many ways can a selection of 3 letters be made from
the letters a, b, c, d, e, f, g, if repetitions are allowed?
PERMUTATIONS AND COMBINATIONS
284
[Ch. XVI
19. How many different arrangements of 5 letters each, con-
sisting of 3 consonants and 2 vowels, can be made from the
letters a, b, c, d, e, if repetitions are allowed?
20. How many different 3-place numbers can be formed by
using 2 odd digits and 1 even digit if zero is excluded but
repetition of digits is allowed?
21. How many different 3-place numbers can be formed by
using 2 odd digits and 1 even digit, it being permissible to
use the digit zero, but obviously not in the hundreds' position? Repetition of digits is allowed.
";
•
J
".~ ;
CHAPTER XVII
Probability ~:'
>"!,
',)"?i .'~,; \:
ft"')
"'L
17.1. Probability.
(
If the ways in which an event can either happen or fail
are all equally likely, the probability that it will happen
in a given trial is the ratio of the number of ways in which
it can happen to the total number of ways in which it can
either happen or fail. For example, in drawing a card
from an ordinary deck, the probability of obtaining an ace
is 4/52 = 1/13, since there are four aces out of a total of
fifty-two cards.
In general, if the event being tried can happen in h ways
and fail in 1 ways (all equally likely), the probability that
it will happen is given by
h
p=--,
h +1
and the probability that it will fail is given by
q = _f_.
h +1
It is readily seen that p + q = 1. Thus, in the carddra\ving example cited above, the probability of failing
to draw an ace is ~ ~ = H- = 1 - -h.
It is frequently impossible to analyze an event into the
number of equally likely ways in which it can happen or
fail. In such cases, however, it may be possible to observe
a large number of trials of the event and to record the
number of happenings. The relative frequency of occurrence of the event is the ratio of the number of happenings
to the number of trials. If the number of trials has been
285
286
PROBABILITY
[Ch. XVI!
large, we assume that this relative frequency is approximately the same as the probability and may be substituted for it.
For example, suppose we had a deck containing an
unknown number of cards and an unknown number of
aces. If 1000 draws (with replacement of the card drawn
before the next draw was made) have resulted in 96 aces
we should say that the probability of drawing an ace (as
. determined by statistical experiment) is 96/1000 or 0.096.
Here it would be possible to check our experimental result
with the mathematical probability by counting the number of cards and the number of aces.
i
17.2. Mortality tables.
One important application of probability in which it is
impossible to determine the probability in any other way
than by using relative frequency is in actuarial science, that is, the science of life insurance. The life insurance
companies have kept, over many years, records of the
. relative frequency of death of persons in different age
groups. These data have been compiled into tables
. called mortality tables. Essentially, a mortality table
exhibits the number of persons dying each year out of an
initial group of specified size.
Such a table is the 1941 Commissioners' Standard
Mortality Table (Table VII at end of book). This is
the legal reserve standard table for life insurance. It
traces a hypothetical group of a million males throughout
their lives. Age in years is denoted by x. The number
surviving at age x out of the original group is shown in the ,
column headed 1". For example, at age 30 there are
924,609 still living. The number of men 30 years old who
. die before they become 31 is 3292, as shown in the column
headed d". The probability that a 30-year-old man will
die during the year is dao/1 ao = 3,292/924,609 = 0.00356.
This probability is denoted by 1". The probability that a
i
1-
I
.!
EXPECTATION
§ 17.3]
287
man x years old will ~urvive for at least a year is den·,)ted
by pX' Obviously Px = 1 - qx' Values of px and qx are
not shown in Table VII.
Example 1.
What is the probability that a boy 10 years old will live to
be at least 40?
SOLUTION. The probability is equal to the number living at
age 40 divided by the number living at age 10, viz"
l40
l10
= 883342 = 0 908971
971804
'
.
Example 2.
!
Find the probability that a man 35 years old will die between
60 and 65 inclusive.
SOLUTION.
The number dying between 60 and 65 incl«sive is
l60 -
l66
= 677,771 - 554,975 = 122,796.
The required probability is found by dividing tllis number by
l35 = 906,554, and is 0.135454.
17.3. Expectation.
If p is the probability of success in a single trial of an
event, the expected number of successes in n trials is np.
The expected number of hearts in making 100 draws from
an ordinary d~ck of cards (replacing the card drawn before
making the next draw) is 100 X -t = 25, since the probability of obtaining a heart on a single draw is H = t.
If p is the probability that a person will win an amount
of money a, his expectation is defined to be pa. It is the
average amount that he would expect to win in the long
run. For example, if a person is to win $1.00 provided he
is successful in dra,ving a heart from a deck of cards, his
expectation is -t X $1.00 = $0.25, which means that if he
were to draw a card repeatedly under the same corrditions,
he would expect to win, on the average, $0.25.
PROBABILITY i
288
[Ch. XVII
EXERCISES XVII. A
1. What is the probability of drawing a face card (king, queen,
or jack), from an ordinary 52-card deck?
2. The letters of the words "college algebra" are written on
identical plastic disks, which are then placed in a bowl and
thoroughly mixed. If one of the disks is drawn at random
what is the probability that it has the letter "1" on it?
.
3. A number between 1 and 50 inclusive is selected at random. :
What is the probability that it is a prime?
4. A number between 1 and 10,000 inclusive is selected at
random. What is the probability that it is a perfect
square?
6. The rim of a wheel is divided into 30 equal parts, which are
marked with the numbers 1 to 30. A fixed pointer at the
side indicates one of the numbers when the wheel is at rest.
If the wheel is spun ,,,hat is the probability that, when it
stops, the pointer will indicate (a) the number I3? (b) a
number divisible by 4? (c) a two-digit number? (d) a number greater than 25? (e) a prime number? (f) a number which
is a perfect square? (g) a number which is a perfect cube?
6. What is the expectation of a person who is to get 25 cents
if he draws a black ace from an ordinary deck of 52 playing
cards?
7. A box contains 5 white, 4 black, and 3 red balls. What is
the expectation of a person who is to receive a dollar if he
draws (a) a red ball? (b) a white or a red ball'?
8. The prize in a lottf'ry in which 250 tickets are out is $50.
What is the expectation of a person holding 3 tickets?
9. A box contains 6 white balls and 4 black balls. Three balls
are taken at random from the box. (a) What is the probability that all are white? (b) What is the probability that
2 are white and 1 black?
(a) The number of ways that 3 white balls can
6·5·4
be drawn from the 6 is 6e a = 1 .2.3 = 20. This is the
SOLUTION.
number of ways in which the event of drawing 3 white balls
can happen. The total number of ways in which the event
can happen or fail is the number of ways in which 3 balls
EXERCISES
289
can be selected from the 10 in the box, viz.,
lOC 3
=
10·9· 8
1.2. 3
. d
b b'l' . aC 3
20
1
= 120. Th e reqUIre
pro a 1 Ity IS lOC = 120 = 6'
3
(b) The number of ways of getting 2 white balls is aC2 =
",
,i
~:~
= 15.
The number of ways of getting 1 black ball is
4C 1 = 4. The total number of ways of getting 2 white
balls and 1 black ball is the product of these two, and the
required probability is
aC 2 , 4C 1
lOC 3
15 . 4
1
= 120 ='2'
10. A box contains 10 white balls and 5 black balls.
H.
12.
'13.
14.
15.
16.
17.
If 4 balls
are drawn at random what is the probability that (a) all
are black? (b) 2 are white and 2 black? (c) 3 are white and
1 black?
What is the probability that of 5 cards dealt from a well
shuffled deck 3 will be hearts and 2 spades?
Each of the numbers from 1 to 15 inclusive is written on a
separate ticket. These tickets, which are identical in
appearance, are thoroughly mixed and 2 of them are drawn
at random. Find the probability that their sum will be
13.
If 8 persons are seated at random in a row what is the
probability that a certain couple (a) will be together?
(b) will not be together?
If k persons are seated at random in a row of n seats what
is the probability that they will occupy consecutive seats?
Five red books and 3 blue books are placed at random 011 a
shelf. Find the probability that the blue books will all
be together.
Five red books, 4 blue books, and 3 green books are placed
at random on a shelf. What is the probability that the
blue books will all be together and the green books all
together?
Certain manufactured articles are sold in boxes of 25 each.
They are inspected by taking a random sample of 5 from
each box and passing the box as satisfactory if no defective
PROBABILITY
[Ch. XVII
articles are found in the sample. Find the probability that
a box will be passed if it contains (a) 1 defective, (b) 2
defectives, (c) 3 defectives.
In the remaining exercises of this set use the 1941 Commissioners' Standard Ordinary Mortality Table (Table VII at end
of book).
18. (a) What is the probability that a man 25 years old will
die within one year? (b) What is the probability that he
will live at least one year?
19. (a) What is the probability that a man 80 years old will
die within one year? (b) What is the probability that he
will live at least one year?
20. Find the probability that a man 30 years old will live at
least 5 years longer.
SOLUTION. Table VII shows that 924,609 men (out of a
million male children at age 1 year) are living at age 30.
The number living 5 years later (i.e., at age 35) is 906,554.
The probability that a man 30 years old will live at least
5 years longer is 906,554/924,609 = 0.9805.
21. Find the probability that a man 30 years old will die within
5 years.
SOLUTION. The number dying between ages 30 and 35 is
924,609 - 906,554 = 18,055. The probability of dying in
this period is 18,055/924,609 = 0.0195. Note that the sum
of the probabilities in exercises 20 and 21 is one Why'?
22. What is the probability that a man 35 years old (a) will
live at least 10 years? (b) will die within 10 years?
23. Find the probability that a man 35 years old (a) will live
at least 15 years, (b) will die within 15 years.
24. What is the probability that a man 25 years old will die
between the ages 40 to 49 inclusive?
17.4. Mutually exclusive events.
Two or more events are mutually exclusive if not more
than one of them can happen in a given trial. Thus, the
drawing of an ace and the drawing of a king (in the same
INDEPENDENT EVENTS
~17.5\
291
draw of a single card) from a deck of cards are mutually
exclusive events. The drawing of an ace and the drawing
of a heart are not.
The probability that some one or other of a set of mutually
. exclusive events will happen in a single trial is the sum of
---- their separate probabilities of happening.
Let us for simplicity consider the case of two mutually
exclusive events. Suppose that the first can happen in ml
ways, the second in m2 ways, and that the total number of
ways in which either of the two events can happen or fail
is n. Then the probabilities of happening are Pl = mdn
and pz = mdn respectively.
The ml ways in which the first event can happen and the
mz ways in which the second can happen are mutually
exclusive. Consequently the probability that either the
first or the second event will happen is
m]
+
m2
n
=
ml
n
+ m2
n
=
Pl
+ P2.
As an illustration let us calculate the probability of
drawing an ace or a king from a deck of cards. According
to the theorem this probability is T:r + T:r = 123. As a
verification we note that since there are 4 aces and 4 kings,
the probability of drawing either an ace or a king is
4
+4
~ =
8
52
2
=
13·
It is not difficult to extend the reasoning so as to cover
the case of more than two events.
17.5.
Independent events.
The events of a set are independent if the happening of
anyone of them does not affect the happening of the
others.
The probability that all of a set of independent events will
happen in a trial is the product of their separate probabilities
of happening.
292
PROBABILITY
[Ch. XVII
------------------------------------------
For the case of two independent events, suppose that
the first can happen in ml out of nl possible ways of happening or failing, and that the second can happen in m2
out of n2 possible ways. The probabilities of happening
are PI and P2 respectively. By the fundamental principle
of § 16.1, the two events can happen together in ml1nZ ways,
and the total number of possible ways in which they can
happen or fail is nlnZ' Thus, the probability that both
will happen is
The extension to more than two events is obvious.
As an illustration, let us consider the probability of
throwing two aces with a pair of dice. According to the
theorem the probability is t . t = 31a. This is verified if
we note that there is only one way to obtain two aces
(both dice must show aces). There are however 6· 6 =
36 possibilities, since anyone of the six faces of either die
may occur with anyone of the six faces of the other.
Thus the probability is la.
;. (~~'lf '
"
17.6. Dependent events.
; .'"
If the happening of one event affects the probability of
occurrence of another, the second event is dependent on
the first. For example, in drawing two cards from a deck
the probability of obtaining anJ1Ce on the second draw
depends on whether an ace has been drawn the first time.
If the probability that an event will happen is PI, and if
after it has happened the probability that a second event will
happen is pz, the probability that the first event will happen
and will be followed by the second event is PI'P2.
Suppose that the first event can happen in ml out of ni
possible ways of happening or failing; then Pl = mdni'
Suppose further that after the first event has occurred the
second can happen in 1n2 ways and happen or fail in n2
ways; then P2 = mdn2. By the fundamental principle of
,
i
1-
EXERCISES
293
§ 16.1, the first event and then the second can happen
in mlm2 ways; an occurrence or failure of the first, followed by an occurrence or a failure of the second, can
take place in nlnZ ways. Thus, the probability that the
first event will happen and be followed by the second is
This proof admits of an easy extension to more than
two dependent events.
Example.
Find the probability of obtaining an ace on both the first and
second draws from a deck of cards when the first card is not
replaced before the second is drawn.
SOLUTION. The probability of obtaining an ace on the first
draw is PI = ~ =
If the first card drawn is an ace there
are only 3 aces remaining in the deck, which now consists of 51
cards. Thus, the probability of getting an ace on the second
draw is P2 = -IT =
The required probability is PIP2 =
n.
-is- . -1'7 =
rtT.
n.
The problem can also be solved by noting that the number of
ways favorable to the event of drawing 2 aces in 2 draws is
4C 2, while the total number of ways in which 2 cards can be
drawn is 52C 2. The required probability is
4C 2
;;C
2
4
2!2!
1
= ---s2 = 22(
2!50!
EXERCISES XVII. B
. 1. A box contains 15 white balls, 4 black balls, and 6 red
balls. What is the probability of drawing a black ball or a
red ball in a single draw?
2. What is the probability of drawing either a black ace or
the queen of hearts in a single draw from an ordinary deck
of cards?
294
PROBABILITY
ICh. XVII
3. Each of the first 9 letters of the alphabet and each of the
first 9 positive integers is written on a separate slip of
paper. These slips, which are identical, are placed in a
box and thoroughly mixed. What is the probability, in a
single draw, of obtaining either a vowel or an odd integer?
4. A box contains 6 white balls and 4 black balls. If 2 balls
are drawn in succession what is the probability that both
are black (a) if the first ball is replaced before the second is
drawn? (b) if the first is not replaced before the second is
drawn?
5. Find the probability that a man 25 years old will die either
between the ages 40-44 inclusive or 50-54 inclusive.
6. A is 30 years old, B is 40. Assuming that the probabilities
of their deaths are independent, find the probability (a)
that both will die within a year, (b) that just one of them
will die within a year.
7. A man draws 1 card from each of 2 separate decks. What
is the probability (a) that the card from the first deck is
an ace and the one from the second deck a face card (king,
queen, or jack)? (b) that the card from the first deck is red
and the one from the second deck is a spade? (c) that the
card from the first deck is a queen and the one from the
second deck a jack?
8. A man draws a card from an ordinary deck. Without
replacing it he draws another card. What is the probability
(a) that the first card is red and the second a black queen?
(b) that the first card is a black king and the second a red
king? (c) that the first card is a face card and the second a
seven?
9. What is the probability of throwing 4 aces in a single throw
with 4 dice?
10. If 10 coins are tossed what is the probability that all will
show heads?
11. A box contains 3 white balls and 3 black balls. If they
are drawn out at random, one at a time, what is the probability that they will be alternately white and black until
the box is empty?
12. One box contains 3 white balls and 5 black balls; another
contains 9 white balls and 3 black balls. A die js thrown
i
i-
§17.7]
PROBABILITY IN REPEATED TRIALS
295
and if it shows an ace a ball is drawn from the first box,
otherwise a ball is drawn from the second box. Find the
probability that the ball is white.
U. One box contains 7 white balls and 2 black balls, another
contains 6 white balls and 4 black balls. A car~ is drawn
from a well shuffled deck and if it is a spade 3 balls are
drawn from the first box, otherwise 3 balls are drawn from
the second box. Find the probability of getting 2 white
balls and 1 black ball.
14. Ten identical tickets are numbered from 1 to 10 inclusive,
placed in a bowl and thoroughly mixed. A man draws a
ticket at random. If it is numbered 5 or less he gets another
draw, otherwise not. Find the probability that he draws
a total of 8 or more (a) if the first ticket drawn is not
replaced, (b) if it is replaced.
<)1
17.7. Probability in repeated trials.
If p is the probability that an event will happen and q is
the probability that it will fail in any trial, the probability
that it will happen exactly r times in n trials is
nCrprqn-r
=
n!
prqn-r.
r I(n - r) !
(1)
As an illustration consider the probability of throwing ...~
exactly 2 aces in 5 throws of a die. The probability that
the event (throwing of an ace) will happen in any trial is
p =t and the probability that it will fail is q = t. The
probability that it will happen twice and fail three times
in a specified order, such as happening the first two times
and failing the last three, is by the theorem on independent.
probabilities,
~.~.~.~.~
=
(~) (~)
2
3.
But the 2 aces may be obtained on any 2 of the 5 throws,
that is, in 5C 2
=
5·4
1.2
=
10 ways.
If the happening of
PROBABILITY
296
[Ch. XVII
the event is indicated by h and the failure by f, these 10
ways may be indicated by
hhfff
fhfhf
hfhff
fhffh
hffhf
ffhhf
hfffh
ffhfh
fhhff
fffhh
The probability for anyone of these orders is (t)2(t)'~
and the total probability is
_J_.
6
10 GY (~y 3::8'
=
I
More generally, the probability that an event will happen in each of r specified trials and fail in the remaining
n - r is p'l'qn-'I'. But these r trials can be chosen from the
n trials in nCr ways. Consequently, since these ways
form a mutually exclusive set, the total probability is
nCrprqn-r.
It should be noted that this is term number r + 1 in the
binomial formula
(q
+ p)n
= qn
+ nC1qn-lp + nC qn-2 p 2 + .. .
+ nCrqn-rpr + ... + pn.
(2)
Example 1.
Find the probability of throwing at least 3 aces in 5 throws
of a die.
This is the probability of throwing 3, 4, or 5 aces
and is thus
&C 3p 3q2
i
2
In fact the terms of this expansion give the probabilities
that the event will happen exactly 0, 1, 2, . . . ,r, ... ,
n times in n trials.
The probability that an event will happen at least or at
most a certain number of times in a given number of trials
is found by taking the sum of the appropriate terms in (2).
SOLl7TION.
i:
+ ,C 4p 4q + pO =
10
(~y (~y + 5 (~y
250
=
"""66 +
25
6'
+
1
6'
276
G) + GY
23
= 7776 = 648'
/
297
EXERCISES
Example 2.
Find the probability of throwing at least 2 aces in 10 throws
of a die.
SOLUTION. This probability is the sum of the probabilities of
throwing any number of aces from 2 to 10 inclusive. Instead of
summing the nine corresponding terms of the binomial expansion
we find the probability of throwing no aces or 1 ace and subtract
from 1.
The probability of 0 or 1 ace is
5)10
( "6
(
+ 10
(5)9 1
"6
3 . 5 10
9765625
'"6 = 2 10 . 3 10 = 20155392'
The probability of throwing at least 2 aces is
9765625
10389767
1 - 20155392 = 20155392'
EXERCISES XVII. C
1. Six draws are made from an ordinary deck of playing cards,
the card drawn being replaced and the deck thoroughly
shuffled before the next draw is made. What is the probability of obtaining (a) exactly 4 spades? (b) 4 or more
spades?
2. Find the probability of getting exactly 2 double sixes in
5 throws with a pair of dice.
3. A baseball player has a batting average of 0.200. Assuming that this may be used as the probability that he will
get a hit, find the probability that he will (a) get exactly
2 hits in 4 times at bat, (b) get at most 1 hit in 5 times at bat.
4. A rifieman is able, on the average, to hit a target at a certain range 75 times out of 100. What is the probability
that in 10 shots he will (a) hit the target every time? (b)
hit the target 8 or more times?
6. According to records in the office of a college registrar, 4 per
cent of the students in a certain course fail to pass. What is
the probability that out of a random group of 4 students
of the course exactly 2 will fail?
298
PROBABILITY
(CII. XVII
6. Hospital records show that 20 per cent of the cases of a
certain disease are fatal. Find the probability that out of
5 patients admitted with this disease (a) all will recover,
(b) exactly 2 will recover, (c) at least 2 will recover.
7. In a certain manufacturing process it is found that, on the
average, 2 articles out of 100 are defective. What is the
probability that a random sample of 5 articles will contain
(a) no defective articles? (b) exactly 1 defective article?
(c) exactly 2 defective articles?
8. Find the probability, in throwing a die 6 times, of getting
(a) 3 aces and 3 non-aces, (b) a run of 3 aces followed by a
run of 3 non-aces, (c) the sequence a a a a a a, in which a
denotes "ace" and a "non-ace."
9. Find the probability that of a group of 3 fifty-seven-yearold men (a) exactly 2 will die during the year, (b) a specified
2 will die, but the other will live, (c) at most 1 will die during
the year.
10. A and Bare 88 years old. What is the probability that
(a) both will die within the year? (b) both will survive for
at least one year? (c) one will die and the other survive?
(d) A will die but B survive?
MISCELLANEOUS EXERCISES XVII. 0
1. Two dice are thrown simultaneously. Find the probability
2.
3.
4.
6.
that the number of spots uppermost is (a) 4, (b) 9 or more,
(c) 7 or 11.
.
On a key ring are 6 keys, only one of which will open a
certain door. If the first and second keys have been tried
and have failed to unlock the door, what is the probability
that it can be unlocked by the third key?
A man has a ticket in a lottery in which 2000 tickets are
out. The main prize is $500 and thBrp, are also three
prizes of $100 each and ten $10 prizes. What is his
expectation?
From a group of 6 seniors and 10 juniors a committee of
5 is chosen by lot. What is the probability that it ",ill consist of more seniors than juniors?
Six identical disks are marked with the integers 1 to 6
respectively and placed in a bowl. After they have been
EXERCISES
tr
6.
'1.
,)!
(,
S.
9.
10.
.n
11.
12.
13.
14.
299
thoroughly mixed three of them are drawn at random.
What is the probability that their average value is 4 (a)
if they are drawn simultaneously? (b) if each disk is replaced
before the next drawing is made?
Ten men are placed at random in a line. What is the
probability that A and B have 3 men between them?
A student taking a true-false test consisting of 15 questions
guesses at the answers. Assuming that he is equally likely
to make a correct or an incorrect guess on each question,
find the probability that (a) all of his answers will be right,
(b) 8 of his answers will be right and the rest wrong, (c)
all but 1 of his answers will be right.
A mUltiple-choice test consists of 10 questions. After
each question are listed 3 answers, one and only one of
which is correct. If a student attempts to guess the correct
ans,ver to each question, what is the probability that more
than half of his answers will be right?
The probability that a typist will make 1 or more mistakes
in typing a letter is 0.05. Find the probability that in a
set of 5 letters which she types exactly 4 will be free from
mistakes.
A machine is composed of 6 parts essential to its proper
functioning, The probability that anyone of these parts
will function properly is 0.99. If the functioning of each
part is independent of the functioning of any of the others,
what is the probability that the machine will function
properly?
A and B take turns throwing a die. The first one to throw
an ace is to receive $10. If A has the first turn what is the
expectation of each?
A and B are playing a tennis match. The first to win 3
sets wins the match. If the probability of winning an
individual set is t for A and f for B, what is the probability
of each for winning the set?
One thousand tickets, marked 1 cent, 2 cents, and so on up
to $10.00, are placed in a box and thoroughly mixed. What
is the expectation of a person who is to receive the amount
marked on the ticket which he draws?
In a clairvoyance experiment 6 cards are placed, face down,
300
15.
16.
17.
18.
19.
20.
PROBABILITY
[Ch. XVII
on a table. The subject on whom the experiment is being
tried knows that 3 of the cards are red and 3 black, but has
no other information regarding them. If he tries to designate which cards are red and which black, but is not clairvoyant, what is the probability that the number of cards
whose color he names correctly is (a) 6? (b) 5? (c) 4? (d) 3?
(e) 2? (f) I'? (g) OJ
In a mental telepathy experiment 4 different cards (ace,
king, queen, and jack) are used. A looks at the cards, one
at a time; B, who knows what cards compose the set of 4
but cannot see them, attempts to name each card as A
looks at it. If there is no telepathic communication from
A to B what is the probability that the number of cards
which B names correctly is (a) 4? (b) 3? (c) 2? (d) I? (e) O?
A man stands at a street corner and tosses a coin. If it
falls heads he walks 1 block north, if it falls tails he walks
1 block south. Starting at his new position he repeats
this procedure. What is the probability that after having
gone through the procedure 6 times he is (a) at his starting
point'? (b) 2 blocks from his starting point? (c) 4 blocks
north of his starting point?
Solve the preceding exercise for the case in which the man
throws a die instead of a coin and goes north if he throws
an ace or a deuce, south if he throws anything else.
A committee of 5 wishes to elect a chairman. Each member may vote for anyone except himself. If the balloting
is purely random what is the probability that a specified
individual will receive no votes?
Three checkers are placed at random on the squares of a
checkerboard. Find the probability that (a) they all lie
on a diagonal, (b) they lie at the vertices of a right triangle
whose legs are parallel to the edges of the board.
Four checkers are placed at random on the squares of a
checkerboard. Find the probability that (a) they all lie
on a diagonal, (b) they lie at the corners of a rectangle
whose sides are parallel to the edges of the board, (c) they
lie at the corners of a square whose sides are parallel to the
sides of the board.
CHAPTER XVIII
11 '
Determinants
')
m'
·Iu
. " jJ
'j
18.1. Determinants of order two.
Let us solve the pair of equations
(1)
MultiJ)!J ~~e., ~ by b2, the second by-b 1• ! mrls gives
,
k ,.
i
..... -
~.'
,:'
"i
,
. a l b2x
; , a 2 b lx
{~
+
+
bl b2 y
bl b2 y
=
=
k lb2 ,
k 2bl .
<.'
. if
'f
Subtracting, we get
(alb'}. - a'}.bl)x = k1b'}.'- k'}.b 1 ,
(2)
Similarly,
(3)
That these values satisfy equations (1) can be proved by
actual substitution.
The denominators in the expressions for' x and yare identical. We denote this common denominator by the symbol
(4)
301
DETERMINANTS
302
[Ch. XVIII
which is called a determinant. This determinant, which
has two rows and two columns, is said to be of order two.
The a's and b's are elements of the determinant. The
elements al and b2 which lie in the line from the upper lefthand corner to the lower right-hand corner of the determinant constitute the principal diagonal. It is seen that the
value of the determinant is found by taking the product of
the elements in the principal diagonal and subtracting
from it the product of the elements in the other diagonal.
Thus,
, ,IN! ~
I_~
'c
~
5 ' 3 - 2 ,4
~ 15 - 8 ~ 7,
I
= 6·3 - (-2) ·4 = 18
+8
,r
=
26:
i
1:
We can now write the solution (2) and (3) of the pair of;""
equations (1), by means of determinants, as fo11O\vs:
i.l
,!
I kl
X =
k2
bb1 \
1al
b11
a2
2
b2
Y
\ al
a2
kl
al
1
-I
a2
k2
bb
2
I
-l
I
'l
It may be noted that for the determinant in the denominator, the elements in the first column are the coefficients
of x in the original equations (1), and that the elements in
the second column are the coefficients of y. When we
solve for x, the determinant in the numerator is the same
as that in the denominator, except that the coefficients of
x are replace a by the right-hand members of equations
(1); when we solve for y, the determinant in the numerator
is the :,;ame as that in the denominator, except that the
coefficients of yare replaced by the right-hand members of
(1).
I"
DETERMINANTS OF ORDER THREE
§18.21
303
Example.
Solve the following pair of equations by determinants:
2x - 3y = 16,
5x + 2y = 2.
,"':",'
SOLUTION.
X=
11~
1
Y=
2
5
I;
-31 '
2 _ 16 . 2 - 2 . (-3}'iI:: 38
-31 ,2·2 - 5· (-3) ; 19
2
'
1~ I
2· 2 - 5·16 .
2· 2 - 5· (-3)
\! -;1
::::i '2'
t,
¥
-76
=19= -4.
:
;<
}.l
EXERCISES XVIII. A
Solve exercises 1-20 of II. C by means of determinants.
18.2. Determinants of order three.
Systems of linear equations in more than two unknown
quantities can also be solved by means of determinants.
To illustrate, let us solve the bystem of equations
+ bly + CIZ
+ b 2y + C2Z
aaX + baY + CaZ
alX
= kl'
a2X
= k2'
(1)
(2)
(3)
= k a•
The method of obtaining a given equation is shown at
the left; thus, our first step is to multiply equation (1) by
C2, which is indicated by C2 • (1).
(1)
CI • (2)
(4) - (5)
C2 •
Ca'
(2)
(alc2 -
+ b1C2Y + CIC2Z
+ b2CIY + CIC2Z
a2cl)x + (b1C2 - b2Cl)Y
= k1C2 -- k2Cl.
+ b2CaY + C2C3Z
(6)
(7)
alC2X
=
a2CIX
= k2Cl.
a2CaX
=
k 1c2.
k2C3'
(4)
(5)
DETERMINANTS
304
C2 . (3)
a3C2x
(a'l.clI - a~c'l.)x
+ b3C2Y + C2C3Z
+ (b'l.c?, -
(7) - (8)
[Ch. XVIII
=
b~c'l.)Y =
kaC2.
(8)
k'l.c?, - k?,c'),.
(9)
Next eliminate Y from (6)' and (9), obtaining
[(alc2 - a2cl)(b2Ca - baC2) - (a2Ca - aac2)(bIC2 - b2CI)]X
= (k 1C2 - k2CI) (b 2Ca - b aC2)
(k2Ca - kaC2) (b1C2 - b2CI).
Solve for x, and simplify:
- k l b3C2 + k2b3Cl - k 2 bl C a + k3b1C2 - k a b2cl
.
- a1b2Ca - a l b3c2 + a 2 baCI - a2b1c3 + a3b1c2 - a 3b2 c l
(10)
C
x- k 1b2 a
(It is assumed that the denominator is not zero.)
The denominator of the fraction in (10) may be put in
the form
(11)
We now introduce the new symbol,
a1
a2
aa
b1
b2
ba
C1
C2
'i
'
,.,
)'
,
(12)
Ca
whose value is defined to be (11). This symbol is a determinant of order three (it has three rows and three
columns).
The numerator of (10) is similarly transformed into
kl \
~:
C2 \ _
Ca
k2 \ bl
ba
Cl \
Ca
+ ka \ bbl
2
kl
= k2
i
ka
Cl \
C2
bl
b2
ba
Cl
C2 '
Ca
(13)
and we can write the value of x as the quotient of (13)
divided by (12). The unknowns y and z can be found as
/
DETERMINANTS OF ORDER THREE
§ 18.2]
305
similar quotients, and the complete solution of the linear
system (1), (2), (3) can be written as
X=
leI
bI
CI
al
kl
Cl
k2
b2
ba
bI
b2
bs
C2
a2
k2
C2
aa
al
ks
b1
Ca
Cl
ka
al
a2
, ."
.
as
--
Ca
Cl
Y=
C2
i,
a2
b2
C2
bs
Ca
b1
aa
kl
Ca
at
as
al
b'},
bs
bl
a2
b2
aa
ba
a2
-,
-
-;!.
;
"
~.)
in"
r
&
Iii:
k2
ks
Cl
C2
Ca
(14)
That (14) is a solution of the given system of equations can
be proved by direct substitution.
The common denominator is the determinant of the
coefficients of the system of equations (1), (2), (3); that is,
its elements are the coefficients of x, y, z in these equations
arranged in their respective positions. The numerator in
the expression for any unknown is the same as the denominator, except that the coefficients of this unknown are
replaced by the corresponding constant right-hand members of the system of equations.
The determinant method of solving a system of linear
equations eliminates all but one of the unknown quantities
at once. Thus, the foregoing solution of the set of equations (1), (2), (3), for the unknown x, is equivalent to
multiplying them by the determinants
respectively, and adding the results.
and z simultaneously.
This eliminates y
[eh. XVIII
DETERMINANTS
306
Example.
,:I ;j:'l:'~
~fj
,"':
_if..!
Solve by determinants:
4x - y
2x
3y
7x - 2y
+
: .\
, I
!,.,
,
, I
/
+ 4z = 2,
+ 5z = 2,
+ 6z = 5.
!
,I' !
SOLUTION.
-1
3
2
2
5
4
2
7
:1:=
-2
-1
3
-2
4
5
6
4
2
7
4
2
7
y=
4
5
6
2
4
2
5
5
6
4
5
6
-1
3
-2
4
2
7
4
Z=
2
7
\.
-.~-
-1
3
2
2
-2 5
-1
3
-2
4
5
6
The denominator has the value
41-; ~ 1-21 =~ : I+ 71- ~ :I
It,;;
; = 4(18
+ 10)
+ +
- 2(-6
8)
7(-5 - 12)
7 (-17) = -11.
= 4, • 28 - 2 . 2
+
The numerator of x has the value
21_;
~1-21=; :1+51-~
= 2 . 28 - 2 . 2
Thus,
x
=
+ 5( -17)
:1
= - 33.
-33
-11 = 3.
The student should complete the solution for y and z.
results are y = 2, z = -2.
J
The
";'
EXERCISES XVIII. B
Evaluate the following determinants:
1. 1 2
2
3
6
2
4
18 .
1
2. 1 2 3
4
7
5
o·
8
9
/'
----,,~.
3
-5
-1
2
6
1
!\.
31
i
EXERCISES
4.
7.
2 7 6
9 5 1
438
17
13
-19
86
(
abc
cab
b c a
x a
a 1:
a a
3
1
12
8.
18
22
11
• 12.
:
6.
-4
3
8
10.
-78
66
34
t
Factor:
13.
2
-5
10
20
-41
62
-83
11
-23
11.
5.
-5 14
6 21
-7
. 9.
*
301
8
3-7
5
0
4
-9
-2
2
-12
15
-7
5
14
26
-84
37
-18
35
82
12
a
-b
-c
b
c
a
d
-d
a
99
-34
64
",
14.
a
a
x
x3
y3
Z3
X
Y
1
z
1
.urrm
1
Solve by determinants:
15. 3x
19.
z
y - 3z
3y
4z
=
4,
= -3,
+
+ + = 9.
10,
5x + 6y + 4z =
6x + 7y - 3z = -12,
8x - 2y + 9z = -11.
2x + 5y - 4z = 3,
3x - 3y + 7z = 2,
2x
4x
17.
+ 2y +
5x
+ 8y
21. 5x
+ 5y
3x - 6y
x
+ 4y
- 3z
= 5.
- 7z =
+ 2z
1,
= - 2,
- 3z =
2.
16. 3x - 4y
2x
5y
3x - 2y
+
+ 8z
+ 14z
+ 7z
= 5,
= 10,
= 10.
18. 5x - lOy + 3z = -91,
3x + 4y - 7z =
33,
6x + 5y + 2z =
13.
20. 3x - 4y - z = 2,
6x + 6y + 3z = 7,
gx - 8y - 5z = O.
22.
4y + 8z = 1,
6y - gz = 13,
15x - lOy - z = -1.
5x
lOx
+
+
* The elements of this determinant form a magic square; that is, the,
sum of the elements in any row, any column, or either diagonal is the same.
t The elements of this determinant form a Latin square; each letter
appears once and only once in each row and each column. The Latin
square is used in designing scientific experiments whose results are tl) be
analyzed statistically.
DETERMINANTS
308
+ 3y - 7z
3x + 2y + 8z
[Ch. XVIII
+ 3y + 2z
+ 4y + 5z
= 5,
= 10,
8x + 7 y + 3z = 14.
....,
25. Solve exercises II. C, 25-30 by determinants.
= -2.0,
= 7.9,
2x - 5y - 3z = 0.6.
24. 2x
4x
23. 5x
I
18.3. Determinants of any order.
The symbol
D =
all
a2l
au
a22
aln
a2n
!(l)~
I
anl
an 2
~ J,
ann
in which there are n 2 elements arranged in n rows and n
~olumns, is a determinant of order n. The first subscript
of an element is the number of the row in which the element lies, the second subscript is the number of the column. (The element a12,-read "a sub one-two" or "a
one-two," not "a twelve"-for example, lies in the first
row and the second column.) The value of the determinant may be defined as follows:
(i) Write down all possible products of nfactors each that
can be obtained by selecting one and only one element from
each row and each column. (According to § 16.2 there will
be n! such products.)
(ii) In each product count the number of pairs of elements
in which one element appears to the right of and above the
other (when the two elements are in their respective positions in the determinant). Pairs of this type may be
called negatiV'e pairs, pairs of the other type positive
pairs. If there is an even number of negative pairs, prefix a
plus sign to the product; if there is an odd number, prefix a
minus sign. Since there are no negative pairs in the principal diagonal (the diagonal running from upper left-hand
to lower right-hand corner), and zero is an even integer,
the plus sign is always to be prefixed to this special
product.
f
§ 18.4]
PROPERTIES OF DETERMINANTS
309
The value of the determinant is the algebraic sum of these
products. This sum is usually spoken of as the expansion
of the determinant.
To illustrate the rule of signs, (ii) above, let us consider
_. _ the product a2blcs from the third-order determinant (12)
of § 18.2. There are three pairs of elements in this product, namely, a2 b1 , a 2b3 , blcs. Of these, only the first
IS a negative pair; consequently the product must be prefixed with a minus sign.
EXERCISE
Show that the value of the determinant (12) of § 18.2, as given
by (11) of that section, is precisely the same as the value obtained
from the definition of the present section.
18.4. Properties of determinants.
I. The value of a determinant is unchanged if corresponding rows and columns are interchanged.
Let
D
D'
The products obtained from D' are exactly the same as
those obtained from D, so that the terms of D' are the same
as those of D. However, it is readily seen that any pair of
elements which is a positive pair in D is also a positive
pair in D', and similarly for negative pairs. Consequently,
the signs of the corresponding terms in the expansions of
the two determinants are alike, &.lld the values of the two
determinants are thus identical.
It follows that for every theorem about the columns of a
determinant there exists a corresponding theorem about the
rows, and vice versa.
II. If two columns (or rows) are interchanged the sign of
the determinant is changed.
310
DETERMINANTS
[Ch. XVIII
, !
Let us consider first the effect of interchanging two
adjacent rows. Obviously the products composing the
terms of the determinant will be unchanged, since each is
composed of one and only one element from each row and
each column. The only pair of elements of any product
whose status is changed as far as the rule of signs is concerned is the pair in the rows interchanged. If this pair
was a positive pair before the interchange, it will be a
negative pair afterwards; if it was a negtLtive pair before,
it will be a positive pair afterwards. ,lIence the inter- .
change transforms the number of negtLtive pairs from !
even to odd or from odd to even. Thus, the sign of every
term in the expansion of the determinant is changed; that
is, the sign of the determinant is changed.
N ow let us consider the effect of interchanging any two
rows. Suppose these two rows have k other rows between
"
them. Their interchange may be effecbted by 2k + 1 - "interchanges of adjacent rows: k + 1 to ring the lower
I
row into the position previously occupied by the upper, k
.~
more to bring the upper row into the position previously
:.
occupied by the lower. But 2k + 1 is an odd number,
}
and an odd number of interchanges of adjacent rows
changes the sign of the determinant.
III. If two columns (or rows) are identical, the value of
the determinant is zero.
Let D be the value of the determinant. An interchange
of columns will give a determinant whose value is - D.
But if the identical columns are interchanged, the value
of the determinant will obviously be unaffected. Hence
D = - D or 2D = 0, and D = 0.
IV. If all the elements of a column (or row) are multiplied
by the same number m, the determinant is multiplied by m.
That is,
§ 18.4]
311
PROPERTIES OF DETERMINANTS
From the definition of the value of [L determinant, as
given in § 18.3, it is seen that each term in the expansion
of the new determinant is m times the corresponding
term in the expansion of the original determinant. That
is, the new determinant is m times the original.
V. If every element of a column (or row) is zero, the value
of the determinant is zero.
For each term in the expansion of the determinant contains the factor zero.
VI. If each element of any column (or row) is expressed
as the sum of two or more terms, the determinant may be
expressed as the sum of two or more deterrf1inants. Thus,
(
al
a2
+ a~
+ a~
b1
b2
For any term in the expansion of the determinant on the
left is equal to the sum of the corresponding terms of the
determinants on the right.
VII. The value of a determinant is unchanged if to the
elements of any column (or row) are added the corresponding
elements of any other column (or row) each multiplied by the
same number m. (N ote that m may be negative, also that
it may have either of the particular vahles ± 1.)
For example, consider
D=
al
a2
b1
b2
D'
al
a2
+ -mb
+ -mb2
1
. .
b1
b2
\'
By VI,
D' =
al
a2
b1
b2
The first determinant on the right-hand side of this last
equation is D; if by IV we factor m from the second, it
[Ch. XVIII
DETERMINANTS
312
has two columns identical, and hence by III is zero.
Therefore, D' = D.
18.5. Expansion of a determinant
by
,
r
minors.
!
If in any determinant the row and the column containing a given element are removed, the determinant formed
by the remaining elements is called the minor of the given
element. For example, in the determinant
,
,
r,\
"
the minor of b1 is
al
a2
as
Ia2as
b1
Cl
b2
C2 ,
ba
Cs
, I··
'I : ~l
,tj
~,
'
:)
C2!.
Cs
We shall now show how a determinant of any order can
be expanded by means of minors.
THEOREM I. If the element in the upper left-hand corner of a determinant is au, the sum of the terms involving all
in the expansion of the determinant is allA ll , where All is
the minor of all.
For each term of the determinant involving all is,
obtained by multiplying all by one and only one element
from each of the remaining rows and columns, that is, by
a term of its minor, A 11. Furthermore, the sign of each
term of the original determinant involving all is the same
as that of the corresponding term obtained by multiplying
all by the appropriate term of All, since all forms a positive pair with each and every element of All,
THEOREM II. If the element of the ith row and jth column
of a determinant is aij, the sum of the terms involving aij in.
the expanJion is (-l)i+iaijAij, where Aij is the minor of aij'
The element aij can be carried into the upper left-hand
corner of the determinant by interchanging the ith row
with each preceding row in turn, and then interchanging
the jth column with each preceding column in turn. This
will not disturb the elements in the minor A ij • The sign of
§18.51
EXPANSION OF A DETERMINANT BY MINORS
313
the determinant will have been changed (i - 1) + (j - 1)
times, since the process consists of this number of interchanges of adjacent rows or columns. Thus, if D is the value
of the original determinant and D' is the value of the deter. ______ minant which has aii in the upper left-hand corner, then
D' = (_l)i-l+i- ID = (-1)i+i- 2D = (-l)i+iD.
But, by Theorem I, the sum of all terms in the expansion
of D' involving aii is equal to aii times its minor in D',
which is A ii . Hence the sum of terms in D involving aij
is (-l)i+iaikAik'
Combining Theorems I and II, we have the following
method of expanding a determinant according to the elements of a column (or row):
Multiply each element in the given column (or row) by its
minor, and prefix a plus or minus sign according as the sum
of the number of the row and the number of the column in
which the element lies is even or odd. The value of the determinant is the algebraic sum of these products.
The value of a determinant can be expressed, thus, in a,
variety of ways. For example, if capital letters represent
the minors of the elements denoted by the corresponding
small letters, we have ,.
al
a2
as
bl CI
b2 C2
ba
Ca
: I.
alAI - a 2A 2 + aaAa or -biBI + b2B2 - b3B3
=
,
+ ..
or alAI - blBI + CIC I - • • • ,
or -a 2A 2 + b2B2 - C2C2 + ...
(1)
(2)
(3)
(4)
The expression (1) is called the expansion, or development, according to the first column: (2) is the development.
[Ch. XVIII
DETERMINANTS
314
according to the second column, (3) and (4) are developments according to the first and second rows respectively.
-!<"
,#. ': ~,i
Example 1.
bl CI d 1
al
b2 C2 d2
b2
a2
b3 C3 d3 = al b3
a3
b4
a4
b4
C4 d4
"
+ aa
~, ~ ~~
bl
CI
b2
C2
b4
C4
dl
d2
d4
-
C2
d2
C3
da
C4
d4
a4
-
a2
bl
CI
ba
C3
b4
C4
bl
Cl
d1
b2
C2
d2
b3
C3
da
dl
d3
d4
,...,.·i----....:_-
'!
•
The development can be completed by expandivg the thirdQrder determinants.
In a numerical case, we can often simplify the process of
evaluation by making use of some of the properties devel·
oped earlier in the chapter.
Example 2.
Evaluate the determinant
SOLUTION.
4
2
-3
3
6
-2
-5
1
1
o
6
4
2
7
5
4
Factor 2 from 1st column (Property IV):
4
2
6
-2
-3 1
3 6
-5 4
1 2
2
1
=2
5
3
-1
4
0
7
-3
3
1 0
6
7
-5 4 5
1 2 4
cSelect element 1 in 1st row as a convenient one to use in reducing to zero the other elements in that row. Subtract twice 3rd
column from 1st column (Property VII):
2
1
2
3
-1
-
2·1
2·6
2·4
2·2
-3 1 0
0
-11
3 6 7
=2
-5'4 5
-5
1 2 4
-5
-3 1
3 6
-5 4
1 2
0
7
5
4
.. / /
i
EXERCISES
315
Add 3 times 3rd column to 2nd column:
-3
3
-5
1
0
-11
2
-5
-5
+ 3·1
+ 3·6
+ 3·4
+ 3· 2
1
6
4
2
0
0
-11
7
=2
-5
5
-5
4
1
6
4
2
0
21
7
7
0
7
5
4
Expand according to 1st row. Only one term in this expansion
remains-as was planned-all other terms have the value zero.
-11
21 7
11
7 5 = -2 5
7 4
5
-5
-5
2
I'
Subtract 2nd row from 3rd:
(
-2
11 21
7
5
0 0
7
5·
21
7
7
4
~~
i(
7
5 = (-2)( -1)
-1
III
5
211
7 = 2(11 . 7 - 21 . 5)
= 2( -28) = -56.
EXERCISES XVIII. C
{":
Evaluate:
1.
3.
1
3
-1
5
1
4
7
8
6.
7.
0
1
-2
2
2
0
3
2
3
5 6
8 9
-1 0
2
",
1
4
4
2
' 1
4
-2
2
3
2
5
4.
2
2
3
5
2
-3
-4
1
3
a a a
1 a a
a a 1 a
a a a 1
6.
7
1
-4
-4
5
1
4
2
5
6
!"j
1
a
1
-3
1
5
2.
10
-6
1
5
-4
-8
8.
1
4
7
2
5
6
1
8
1
4
-5
3
4
1
a
-a
1
-a
-a
-a
-a
1
2
1
1
2
-3
0
5
3
4
1
0
-1
0
6
2
a ai
a a
1 a
-a 1
-4
0
-2
1
9.
4
2
15
4
8
-1
-5
4
2
0
20
1
11.
-2
3
11
6
13.
1 0
2
501
2 1 -3
2 4
2
-2 0
1
14. 1
,t,
[Ch. XVIII
DETERMINANTS
316
12.
11 36
3
609
30 20 38
19 17 22
1-1
592
4
7
2
10.
-2
4
30
2
-3
5
15
4
4-1
-3
2
4
2
4
0
-2
2·
5
7
6
-1
3
2
5
9
7
3
abc
d
x
0
u
z
v
0
0 w
o
o
o
21
18
5
19
y
7
6
-4
3
2
3
-5
11
0
-8
i!
I
\' i
32
14
20
13
, ! >,
fl . ,
:,1
"I
.1 -
15. Show that
abO
I,·
,
1.:
0
r
~ ~ ~ ~ I~ ~ 1·1 ~ I·
o
=
0
h
i
1
I ,f:
1, ;
18.6. Application to the solution of linear equations.
We have already seen that systems of linear equations in
two and three unknowns can be solved by means of determinants. (See §§ 18.1 and 18.2.) In order to show that
the method is perfectly general, we shall find it necessary
to use the following theorem:
THEOREM III. If in the expansion of a determinant
according to a given column (or row), the elements of the
given column (or row) are replaced by the elements of another
§ 18.6J
APPLICATION TO SOLUTION OF LINEAR EQUATIONS
317
column (or row), the resulting expression is identically equal
to zero.
For example,
in which the expansion is according to the first column.
If we replace the a's of the right-hand member by the corresponding b's, we have
blAI - b2A2
+ baAs -
I
which is the expansion of
This last determinant, however, has two columns alike,
and is consequently identically equal to zero.
Let us now consider the following system of n linear
equations in n unknowns:
alx
a2X
+ b1y + CIZ +
+ b 2y + C2Z +
(1)
To obtain the solution, if there is one, we multiply the
first equation by AI, the second by -A2, and so on, where
the A's are the minors of the a's. Adding, we get
(alAI - a 2A 2 +
+ (blAl + (clAl = klAl -
aaAa - .. ·)x
b2A2 + baAa - .. . )y
c2 A 2 + caAa --' .. ·)z +
k2A2 + . . . .
(2)
[Ch. XVIII
DETERMINANTS
318
The coefficient of x is the expansion, according to its first
column, of the determinant of the coefficients of the system
of equations, namely,
The coefficients of the other unknowns are zero by Theo- .
rem III. The expression on the right of (2) is the expan- I
sion of the determinant
which is obtained from D by replacing the coefficients of x
(that is, the first column) by the right-hand members ot
equations (1).
Under the assumption that D ~ we find
°
x
=
Dl
D'
z
= Da
-
. i'Y
,
Similarly,
D'
/
.• ,.
where D2 is the determinant obtained from D by repladng
the coefficients of y (second column) by the k's of (1), Da
is the determinant obtained from D by replacing the coefficients of z by the k's, and so on.
Thus, if there is a solution of the set of equations (1),
there is only one, and it is x = D1/D, y = D2 /D, z =
Dal D, .. ' . It can be proved by actual substitution
that this really is a solution.
18.7.
Inconsistent and dependent equations.
The solution given in the preceding section breaks down
if D = 0, since division by zero is meaningless. If D = 0
and any Di ~ 0, the equations have no solution.
§18.7]
INCONSISTENT AND DEPENDENT EQUATIONS
319
Suppose, for example, that D = 0, Dl ~ 0, and that we
assume the existence of a solution x = Xo, Y = Yo, Z = Zo,
We should then have xD = D 1 • But if D = 0
and Dl ~ 0 we have a contradiction, so that the assump. ____ tion of the existence of a solution is false.
A system of equations having no solution is said to be
inconsistent. It is obvious from the preceding paragraph
that equations (1) of § 18.6 are inconsistent if D = 0
and any Di ~ O. If D = 0 and every Di = 0, the equations mayor may not have a solution; that is, they may be
consistent or they may be inconsistent.
When a system of equations has infinitely many solutions the system is said to be dependent.
I
Example 1.
Show that the following equations are inconsistent:
2x - 3y
4x - y
3x - 2y
+ 5z
= 3,
+ z=
+ 3z =
1,
4.
SOLUTION.
D =
2
4
3
-3
--1
-2
5
1 = 0,
3
3
Dl = 1
4
-3
-1
-2
5
1 = 4.
3
It is not necessary to find D2 and D a, since if D is zero and even
one of the D; is different from zero the equations are inconsistent.
Example 2.
Solve the following equations and
pendent:
3x - y
2z =
x
y
z =
5x + y + 4z =
+
+ +
show that they are de-
7.
(1)
(2)
(3)
4 - 3z
x = -4-'
(4)
1,
3,
Solution. Add (1) and (2):
4x
+ 3z =
4,
DETERMINANTS
320
[eh. XVIII
Substitute in (2):
4-3z
-4-
+y +z =
8-z
y = -4-'
3,
(5)
These values of x and y in terms of z will be found to satisfy
the systems of equations (1), (2) (3). Since any number of
values may be given to z, we can obtain an infinite number of
solutions, and the equations are dependent. (The third may
in fact be obtained by multiplying the second by 2 and adding
it to the first.) Observe that the solution may be written
4 - 3c
x = --4-' Y
8 - c
= -4-'
= c,
z
Note that D = Dl = D2 = D3 =
c arbitrary.
o.
Example 3.
The following equations are inconsistent, even though D = Dl
= D2 = Da = O.
3x -
y
6x - 2y
9x - 3y
+
2z
+ 4z
+ 6z
= 1,
= 3,
= O.
This can easily be shown. For if we multiply the first equation by 2 we obtain 6x - 2y + 4z = 2, which is inconsistent
with the second equation. Similarly, if we multiply the first
equation by 3 we obtain 9x - 3y
6z = 3, which is inconsistent with the third equation.
+
18.8. Linear systems with more equations than unknowns.
If a linear system has more equations than unknowns, a
solution does not in general exist; that is, the system is
usually inconsistent.
However, if a solution of certain of the equations can be
found, and if this solution satisfies all of the remaining
equations, the system is consistent.
If the number of equations is one more than the number
of unknowns, a necessary condition for the consistency of
§ 18.8J
LINEAR SYSTEMS WITH MORE EQUATIONS
321
the system can easily be derived. For simplicity let us
consider the case of three equations in two unknowns:
alX
a2X
aax
+ bly + Cl
+ b2y + C2
+ bJy + Ca
=
0,
0,
=
U.
=
(1)
(2)
(3)
(Note that the constant terms have been written on the
left-hand sides of the equations.)
Suppose that equations (2) and (3) are consistent.
Their solution will be
x=
1/-
I-C2
-Ca
Iaaa2
Iaaa2
Iazas
2
1
a
=
2
b 1
bs
b
b
-C2!
-ca
bzl
bs
Ibb2a
Iasa2
.. I!
czl
Cs
Al
2 = Gl '
b 1
ba
a2
as
az
aa
ci>-j
C2!
ca
"",
bzl
ba
-Bl
;00:--
:i
G1
(4)
(5)
provided C l ~ O. Here AI, B l , C l are minors of aI, bl, Cl
respectively, in the determinant composed of the coefficients and the constant terms of equations (1), (2), (3),
namely,
bl
al
Cl
(6)
D = az
b2
Cz
b
as *J a
Ca
.
\
If the system (1), (2), (3) is consistent, the values of
x and y, as given by (4) and (5), which are solutions of (2)
and (3), must also satisfy (1). Substituting them in (1),
we get
Al
Bl
al G1 - b1 G1 + Cl = 0,
.. \
(7)
or
alAI - blBl
+ ClG l
=
O.
(8)
But the left side of (8) is the development of the determinant D according to its first row.
DETERMI NA NTS
322
[Ch. XVIII
If C I = 0, then it follows from the preceding section
that Al and BI must be zero in order for equations (2)
and (3) to have a solution.
,
Thus, in any case, if the system (1), (2), (3) of three linear 1equations in two unknowns is consistent, we must have
D = 0.
The corresponding theorem for n equations in n - 1 i
unknowns can be proved by the same method.
18.9. Homogeneous equations.
An equation in which all terms are of the same
degree is homogeneous, otherwise it is non-homogeneous.
Homogeneous linear equations are equations all of whose
terms are of the first degree, for example,
x
+ 2y
= 0,
3x - 2y
+ 7z
= 0.
I
A system of homogeneous li'1ear equations
~ ~!-
J
1
alX
a2X
anx
+ bly + CIZ +
+ b 2y + C2Z +
+ bny + CnZ +
=
=
=
0,
0,
I
t
.
~ ~~
I
~,
!
I
0,
always has the trivial solution x = 0, y = 0,
, as is
obvious if these values are substituted in the equations.
If the number of equations is equal to the number of
unknowns, and if D, the determinant of the coefficients, is
not equal to zero, this is the only solution; if D =
the
equations are dependent and the system has non-trivial
solutions as well.
°
Example.
Solve the system of homogeneous equations:
+
x
2y
2x - y
5x
+ 3z = 0,
+ 2z = 0,
+ 7z = 0.
(1)
(2)
(3)
EXERCISES
SOLUTION.
323
We find
1
2
5
D=
2
-1
3
2
7
0
"" O.
"'....
;.,,~.
Thus, there are non-trivial solutions.
Multiply (2) by 2 and add to (1):
5x
+ 7z
= O.
7
This is the same as (3), and yields x = - "5 z.
y = 2x
(
+ 2z
= 2(-
~ z) + 2z =
From
-
(2~,
~ z.
By giving vRlues to fJ, we CRn find Rny number of corresponding values for x and y. For example, if z = -5, then x = 7,
y = 4. All solutions are in the continued proportion
x:y:z
= 7 :4: -5.
For a complete discussion of systems of linear equations,
the student is referred to Maxime Bocher, Introduction to
Higher Algebra, The Macmillian Company, 1907.
EXERCISES XVIII. 0
Solve the following systems of equations by means of determinants:
+
+ 5y 4z
13,
2. 8x + 6y 5y - 4z + 3w
50,
4y - gz +
7z - 6w
4x = -50,
6w - 5z 3w - 2x - 5y = -12.
2z + 5x +
3. 2x - 2y + 3z + 4w = 33,
4x
3y - 4z - 5w = 3,
5.x + 4y 6z - 2w = 45,
3x - 5y - 7z
3w = 11.
4. 7 x + 2y + llz - 3w = 42,
lOx - 3y
6z
4w = 40,
4x + 5y - 4z + 2w = 28,
5x
2y
9z - 5w = 34.
1. 3x
+
+
+
+
+
+ +
+
3z = 33,
2w = 17,
7x = 1,
2y = 5.
DETERMINANTS
324
5.
+ + +
+ 3v =
x
2y
3z
u
2x - 7y
6z
4u 3x + 4y - z - 3u +
4x - 3y
6z - 7u 2x - 8y
5z
u -
+ +
+
+
[Ch. XVIII
+
- 5,
5v =
3,
5v =
3,
7v =
0,
2v = -1.
I
C;
"
,
Test the following systems of equations for consistency, and
solve when possible:
+
6. 3x - 2y
5z =
1,
x+2y+3z= -1,
5x - 6y
7z =
3.
7.
+
8. 4x
7x
3x
+
+
+
3y
5y
2y
+ 2z
+
+
+
+
x + 2y +
9. x - 3y
2,
3z = - 3,
z =
4.
=
, 11.
+
z = 6,
z = 0,
7z = 1.
+ 3y +
x - 3y + z =
2x + y + 2z =
x
10. 5x + 3y z = 8,
8x + 7y
5z = 5,
3x - 4y - 18z = 2.
12. 3x + 2y - 4z =
1,
2x - y - 5z = -4, ;
x + 4y + 2z =
7.
4x + 3y z = 2,
lOx - 6y - 16z = 5,
6x
7y +
z = 3.
+ 5y + 4z =
7 x + 3y - 5z =
14x + 6y - 10z =
4x
13.
i
6,
2,
-10.
-
21x
+ 9y -
15z
=
4,
8,
12.
14. 5x + 3y =
1,
3x - 4y = 18,
8x + 7y = - 5.
16.
16. 3x - 2y = 8,
5x - 4y = 3,
4x
5y = 14.
17. 11x - 3y = 148,
2x + 7y = 125,
9x - 4y = 101.
18. 17x - 23y = 47,
7x - 11y = 91,
15x + 19y = 179.
19. 3x
lOy =
38,
4x
y = -11,
9x - 7y = 40.
20. 2x + 3y - 6z = 8,
3x
2y
7z = 1,
x - 3y - 5z = 15,
5x + 7y + 4z = 2.
21. 5x + 6y - z = -10,
24,
4x - 7y + 2z =
4,
9x
y - 3z =
5.
8x + 3y - z =
+
+
+
7x - 4y =
7,
3x + 5y =
50,
lOx - 9y = -13.
+
+
+
\
EXERCISES
22. 7 x - 2y + 8z = - 44,
2x + 5y - lOz =
0,
3x
6y
4z =
12,
4x y + 5z = -27.
+
23.
+
3iS
x
4x
7x
+
+
+
lOx +
2y
5y
8y
lly
+
+
+
+
3z
6z
9z
12z
=
=
=
=
13,
25,
37,
49.
Obtain nontrivial solutions, if they exist, of the following
systems of equations:
24. 5x - 6y - 4z = 0,
x + 2y + 4z = 0,
3x + 2y + 6z = 0.
26. 4x - 5y - 8z = 0,
7x - 12y z;'" 0,
3x - lly
13z = 0.
+
y + 4z
28. 5x 2x + 3y
7x - 2y
";-.
= 0,
+ 5z =
+& =
0,
0.
,
"~
\
11'
25. 2x
3x
4x
L':
27.
+
+
+
3y
4y
5y
+ 5z =
+ 7z =
0,
0,
+ 9z = 0.
2y + 4z = 0,
3x 4x
5y + 7z = 0,
14x - 7y - lOz = 0.
+
+ 9y + z
8y
2z t
2z + t - 2x
4t + 5x
2y
29. 3x
+
+
= 0,
= 0,
= 0,
= O.
CHAPTER XIX
Partial Fractions
I'.
I
!
,
~i!,,~;
19.1. Partial Fractions.
An algebraic rational fraction is the quotient of two
polynomials. In elementary algebra the student learns to
combine such fractions into a single fraction whose denominator is the lowest common denominator of the separate
fractions. Thus,
. ",.,
5x - 1
(x - 2)(x + 1)
_3_ +_2_
x-2
x+l
In this chapter it will be shown how to perform the inverse
process of resolving a rational fraction into a sum of fractions, called partial fractions, whose denominators are of
lower degree. This inverse process is often useful, particularly in calculus.
A rational fraction is proper if its numerator is of lower
degree than its denominator. An improper fraction can
always be reduced to the sum of a polynomial and a
proper fraction by dividing the denominator into the
numerator until the remainder is of lower degree than the
denominator; for example,
2x
3
+x +2
2
x
2 -
1
=
2x
+ 1 + 2x + 3.
X2 -
1
We shall, except for several exercises at the end of the
chapter, consider only proper fractions which ha"e been
reduced to lowest terms, that is, having no common factor
in numerator and denominator.
326
:
1. FACTORS OF THE DENOMINATOR LINEAR
§19.21
~--------------------------------------
327
Methods of resolving rational fractions into partial fractions will be given without proofs. For proofs that the
methods are universally valid the reader is referred to more
advanced treatises.
Four different cases will be considered, and will be
explained by means of examples.
19.2. Case 1.
Factors of the denominator linear, none
repeated.
For each factor of the denominator we must have a
partial fraction having that factor as denominator and
having a constant numerator.
Example.
Resolve into partial fractions:
x
+ 21
(x - 5)(2x
SOLUTION.
(1)
+ 3)'
Let
x
(x -
+ 21
5) (2x
+ 3)
== _:!___
x - 5
+
B
2x
+3
(2)
We must now determine the values of the constants A and B so
that both ,members of (2) will be equal for all values of x except
those for which some of the denominators are zero. Clear
(2) of fractions:
x
+ 21 == A(2x + 3) + B(x
-
5).
(3)
Method 1. Since both members of (3) are equal for all values
of x except possibly 5 and -i, by the corollary of § 13.8 they
are equal for all values of x, including 5 and -i. Set x = 5 in
(3) (to make the coefficient of B equal to zero);
26 = 13A,
Set x =
-i in
A = 2.
(3) (to make the coefficient of A equal to zero):
39
2
_ 13 B
2'
B
=
-
3.
PARTIAL FRAnllJNS
328
Thus,
+
x
21
(x - 5) (2x
[C~.
2
3
- 2x
-----,
+ 3) = x---- 5
+3
XIX
(4)
~--~~~~
as can be verified by combining the fractions in the right member of (4).
Two values other than 5 and --£ could have been given to x,
and (wo 8f1uations in A and B obtained and solved. The advantage of assigning these values that cause the factors obtained
frOIr thp oPilllmindtor to vanish is that·,the resulting equations
contain oniy A, and only B, respectively, and can be solved for
these constants directly.
Method 2. Expand (3) and collect terms:
x
+ 21 = (2A + B)x + 3A
- 5B.
!
I
') (5)
Since this is an identity in x, the coefficients of like pOWeni of x
are equal by the corollary of § 13.8. Thus,
2A + B = 1,
3A - 5B = 21.
(6)
Equations (6) have the solution A = 2, B
same result as Method 1.
= -3, yielding the
19.3. Case 2. Factors of the denominator linear, some
repeated.
For each repeated factor such as (x - a)k we must have
a series of partial fractions:
-'.,
_:!!_
X- a
+ (x A2
+
- a)2
•••
+
Ak
(x - a)k'
The constant Ak is not zero ;-the other A's mayor may not
be.
Example.
Resolve into partial fractions:
".! ,
: •.
"
.-
7X2+3X+2~
(x-2)(x+l)2
.
(1)
.
§19.4]
3. f AOORS QUADRATIC; NONE REPEATED
Let
SOLUTION.
7x 2
329
+ 3x + 2
(x - 2)(x
ABC
+ 1)2 == X + 1 + (x + 1)2 + X -
2'
(2)
Clear of fractions:
7x 2
+ 3x + 2 == A(x + 1)(x -
Set x = -1:
Set x = 2:
Set x = 0:
2)
+ B(x
- 2)
6 = -3B,
B = -2.
36 =
9C,
C = 4.
2 = -2A - 2B + C,
2A = -2 - 2B + C = 6,
+ CCx + 1)2.
A = 3.
Substituting these values of A, B, and C in (2) we get
7x 2 +3x+2 _ _
3__
2
(x - 2)(x + 1)2 - X + 1
(x + 1)2
+_4_.
X -
2
We have used Method 1. Note that after setting x = -1
and x = 2, which values cause the factors obtained from the
denominator to vanish, it is necessary to assign a third value to
x since there are three constants (A, B, C) to be obtained and
three equations are necessary. The third value chosen was
x = 0 as this value is extremely easy to substitute.
19.4. Case 3. Factors of the denominator quadratic, none
repeated.
By a quadratic factor we mean here a factor like x 2 + 4
or x 2 - x + 2, which cannot be factored further into real
linear factors. For each such factor, such as ax 2 + bx + c,
we must have a partial fraction whose denominator is that
factor and whose numerator is linear, namely,
Ax+B
ax 2
+ bx + c'
Here A or B (but not both) may be zero.
Example.
Separate into partial fractions:
4x 2
-
5x
+ 11
+ 4)
(x - 1)(x 2
(1)
[Ch. XIX
PARTIAL FRAGIONS
330
SOLUTION.
Let
I t
"
4:1"2 5x + 11 = ~
(x - 1) (X2 + 4) - x - I
+ Bx2 + C.
x +4
(2)
Clear of fractions:
4.1:2 -
4x 2
-
5x
5x
+ 11 == A(x + 4) + (Bx + C)(x - 1),
+ 11 == (A + B)x (B - C)x + 4A 2
2
-
We shall use a combination of Met~ods 1 and 2.
(3)
(4)
C.
Set x = 1 in
(3) :
A = 2.
10 = 5A,
Equating coefficients of like powers of x on the opposite sides of
(4), we get
A + B
= 4,
B - C = 5,
4A
- C = 11.
.
Only two of these equations are needed, since we have already
found A = 2. We readily find B = 2, C = -3. Thus, (2)
becomes
4x 2 - 5x + 11
ex - l)ex 2 + 4)
2
2x - 3
== x - I + x 2 + 4'
.t
(5)
The values of Band C can be very neatly determined by using
the complex roots of the equation x 2
4 = O. In equation
(3) above, viz.,
+
4x 2
-
5x
+ 11
==
A(x 2
+
4)
+ (Bx + C)(x -
1),
substitute x = 2i, getting
or
-5 - lOi = (2iB + C)(2i - 1),
-5 - lOi = -4B - C + 2(-·B
+ C)i.
Equating real parts and imaginary parts respectively, we have
4B
+C
= 5,
-B
+C =
-5,
from which we find B = 2', C = -3.
Substituting x = -2i gives the same result, of course.
4. FACTORS QUADRATIC, SOME REPEATED
§ 19.5]
331
19.5. Case 4. Factors of the denominator quadratic, some
repeated.
For each repeated quadratic factor such as (ax 2
+ C)k we have a series of partial fractions
A 2 x+B 2
- - A1x+B 1
. ax 2 + bx + c + (ax 2 + bx + C)2
+ bx
Akx+B k
+ ... + (ax 2 + bx + C)k'
in which Ak and Bk are not both zero.
Example.
Resolve into partial fractions:
3x 4
SOLUTION.
+
4x 3
13x 2
(x - 3)(x 2
-
llx
-
+ 2)2
+ 23
(1)
Let the fraction (1) be identically equal to
__:i_ + Bx + C
x - 3
x2 + 2
+
Dx + E.
(x 2 + 2) 2
(2)
Clear of fractions:
3x 4
-
==
==
+ 13x
4x 3
A_(x 2
+
2
llx
23
2)2
(Bx + C)(x - 3)(x 2 + 2)
+ (D.r E)(x - 3)
(A + B)x 4 - (3B - C)x 3
(4A
2B - 3C + D)x 2
+ (-6B + 2C - 3D + E)x + 4A - 6C - 3E.
(3)
+
+
+
+
+
Equate coefficients of like powers of x:
A
4A
4A
+
+
B
3B ~ C
2B - 3C
6B - 2C
- 6C
=
=
+
+
3,
4,
D
= 13,
3D - E = 11,
- 3E = 23.
These equations have the solution
A = 2,
B = 1,
C = -1,
D = 0,
E = -3.
Consequently from (1) and (2) we find
3x 4
-
4x 3 + 13x 2 - llx
(x - 3)(x 2 + 2)2
+ 23
2
==
X -
3
x-I
3
2 - (x 2 + 2)2'
+ x2 -
PARTIAL FRACTIONS
[Ch. XIX
In this exercise a combination of Methods 1 and 2 might be
considered preferable: Substitute x = 3 in (3) to find A, then
proceed as above.
:
EXERCISES XIX. A
3x + 4
2. (x - 2) (x - 7)
3x
4. (x + 2)(x + 5)
6
x + 1
x - 10
1
.
· (x - 1) (x - 4)
3
· (x
5.
1
+
l)(x - 3)
3x - 2
4)(2x - 3)
Cx 1
8.
7 . -2--9'
x 9
x + 14 .
• x 2 - 2x - 8
x
11. 2X2 - 7x + 6'
"'-,
19.
1
.
x+1
+ llx + 4'
2
6x - x - 10 .
(x - l)(x - 2)(x +~4)
2x + 3
I~~
16. (2x + 5)2'
18. 3x 2 - 13x 16.
(x - 2)3
2X2 - 3x + 4
20.
.
(x - 3)(x - 2)2
14.
2x + 3
+ 6x + 9'
+
x 2 + 2x - 2
X2(X
4) .
+
X2 + 7x - 2
21. (x _ 1)(x 2 + 1)
x2 + 1
22. x3 + 4x'
23
24.
· (x
+
x(x - 3)
3) (x 2 - 2x
+ 3)
1
25. x-3--8'
x3 + 1
3x 4
-
31.
28.
+9
+ 4)2'
2xa + 19x
x(x2
1
32. x(x2
2
3x
11x - 2
(3x - 2) (3x 2
2)
x-I
+
26. x a +
27. X 4 + X 2'
x3
3x
29. (X2 _ 2x
+
+ 3)2'
- 3)
9x
2x + 3
10. x~ + 3x - 10'
X2 -
12. 6x2
1
.
13. (x + l)(x - 3)(z - .5)"
x+3
15. (x _ 2)2'
17. x2
+ 5)(4x
• (3x
.1
:!
2
-
+ 3)2
lOx
+
30.
27
•
+
(
2x a - 2x + 1
4
x + x3 + x2
xa+ 5
(2X2
+ 5)2'
i
!
EXERCISES
33
+ llx + 17x 2 + llx + 2.
(x - 2)(X2 + + 1)2
- 8x + 17x 2 - 31x + 15
(x + 1)(x 2 - 2x + 3)2
.
3
4X4
•
X
X4
34.
3
5x s + 2x - 3
35. (x
1)(2x2
X
1)2'
1
37. (X2 _ 1)(x 2 1)2'
+
+ +
+
1
+ l)s'
39. x(x2
f.
333
+
l1x 3 - 23x 2
23x - 12
36. (2x - 3)(X2 - X
1)2
X5
X4 - X3
2x 2 x
38.
(x _ 1)2(x2
1)2
+
+ +
+
+
cx+d
40
. (x - a)(x - b)
Reduce each of the following improper fractions to a mixed
expression (see § 4.3) in which the numerator of the fractional
part is of lower degree than the denominator. Then eeparate
into partial fractions.
4x 2 - 7x
41. x2 _ 3x
2·
+
43 2x S
7x 2
-
•
x
x3
2
45. -3--1'
x -
-
llx + 8.
2x - 8
-
3x 2
9x
-
42. x2 - 4x
44.
+x2 3x_
X4 + 1
--a-+
.
x3
46. x
47 2X4 - 9x 3 - 13x 2 + 98x - 93.
•
(x - 3)2(2x - 3)
X4 + 7x 3 + 13x 2 - X - 15
48.
(x + 5) (x 2 + 3x + 3) .
X
2
+4
+3.
-
1
.J'
1
.
•
CHAPTER XX
Infinite Series
,
~;-~;
I
i .
!
i
"
.
I,
~
20.1. Sequences and series.
. t'S
I
A sequence is a set of numbers arranged in a definite'!
order, for example, the terms of a progression, or the set •
of positive integers.
A series is the indicated sum of a sequence. Thus, if
the sequence is ai, a2, aa, . . . ,an . . . ,the series is
al + a2 + aa + . . . + an + . . . .
If a series is the sum of a limited number of terms it is a
finite series, if it is the indicated sum of an unlimited number of terms it is an infinite series. In this chapter
we shall consider infinite series.
A series may be defined by means of its general or nth
term. For example, if its general term is an = n/(n + 1), '
we find by assigning to n the successive values 1, 2, 3, ... ,
1
2
3
.'.>
al = -, a2 = -, aa = -, . . .
2
3
4
/
Consequently the series is
!+~+~+
234
On the other hand, it is impossible to find a unique
general term for a series in which several numerical terms
are gIven. For example, in the series
!+!+!+
...
248
the nth term could equally well be
1
2n
or
_~
lr---
n 2 -_-nc........,.+---=-2·
334
LIMIT
§20.2J
335
In fact, an infinite number of different formulas could be
devised for the general term of this series. Nevertheless,
in a series such as
........ ---.__
it might be permissible to assume that the nth term is
1ln(n + 1).
"
20.2. Limit.
Consider the unending or infinite sequence
8 1, 8 2 , • • • , 8 n , • • • •
(
If there is a number 8 such that, beyond a certain place in
the sequence, the absolute value of the difference between
8 n and 8 is smaller than any positive number named in
advance, then 8 is called the limit of the sequence. We
write
lim 8 n = 8,
which may be read" the limit of 8 n , as n increases without
limit, is 8."
For example, in the infinite geometric progression
111
2+4+8+'" ,
let 8 ... be the sum of the first n terms; we then have
1
8 1 =-,
82
2
1
1
3
= - + - =-,
244
~
+ I4 + ~8
1
+ 4-1 +
8a
=
8 ...
= -
2
2
=
I,8
1
+2
n
1
1- -,
2n
!NFINITE SERIES
336
[Ch. XX
The limiting value of the sequence of 8' s is 8 = 1; for the
absolute value of the difference between 8 n and 8, namely
in' can be made smaller than any positive number named
!
/'
,
in advance, by taking n sufficiently large. Thus, if we
wish to make this diffeI~ence less than 0.001 we have
merely to take n = 10, or any integer greater than 10;
for
2~0 = 1~24'
which is less than 0.001.
Consequently,
any of the terms from 8 10 on differs from 1 by less than
0.001.
20.3. Convergence and divergcmce.
In any series,
al
+
a2
+ . . . +
a n_1
+ an +
... ,
(1)
let 8 n be the sum of the first n terms; that is, let
81 =
82
a1,
= a1
+ a2,
.
;,
,,'
If Sn has a limit 8 the series (1) is said to be convergent.
The limit 8 is called the sum of the series. (It is not a
sum in the ordinary sense, but the limit of the sum of n
terms as n increases without limit.) A series which is not.
convergent is said to be divergent.
An example of a convergent series is an infinite geometric progression in which the ratio is numerically less
than 1.
As an example of divergent series consider an arithmetic
progreSSIOn;
2+5+8+11+"',
or a geometric progression whose ratio is greater than 1)
1+2+4+8+···.
1,
NECESSARY CONDITION t:OR CONVERGENCE
§ 20.4]
337
The fum of n terms of either of these series is seen to
increase \yithout limit as we take more and more terms.
20.4. Necessary condition For convergence.
THEOHE:'.!. If a series is convergent, the nth term
proachcs zero as n increases without limit.
Consider the series
• with the sum S.
Let Sn be the sum of the first n terms.
ap~
Then
(
Thus, lim an = lim (Sn - Sn_l) = S - S = O.
n--+
00
(It is as-
n-+ co
sumed here that the limit of the difference between Sn and
Sn_l is the difference between their limits.)
Thus, a series is divergent if its general or nth term does
not approach zero. On the other hand, the fact that the
general term approaches zero does not imply that the series
is convergent. In other words, the condition lim an = 0
n->O
is necessary but not sufficient for convergence.
For example, consider the harmonic series (so called
because its terms form a harmonic progression),
Its general term approaches zero, yet the series is divergent. For the terms may be grouped as follows:
(2)
INFINITE SERIES
338
I
[Ch.XX
t.
But
1
-31 + -41 > 1
-4 +1
-4 =-,
2'
11111
- +- +- +- > 5 6 7 8 8
,,
!
111
+ 8-1 + 8
- +8
- =2-,
J!
l i
and so. on. We can form as many groups of terms as we
wish, the sum of each group being greater than t. By taking enough groups we can make their sum as great as we
please. The series therefore diverges, since Sn increases
without limit.
20.5. Fundamental assumption.
If Sn always increases (or always decreases) as n increases
but always remains less (greater) than a fixed value A, then
as n increases without limit, Sn approaches a limit which is
not greater (less) than A.
20.6. Comparison test.
The following theorems are often useful in establishing
the convergence or divergence of a series by comparing it
with a series which is known to be convergent or divergent:
THEOREM 1. If each term of a series of positive term~ is
less than or equal to the corresponding term of a convergent
series, then the given series is convergent.
Let the given series be
al
+ az + . . . + an + .
(1)
and the known convergent series, with sum S', be
(2)
Let
Sn = al
S~ = b1
Since, by hypothesis,
+ az + . . . + an,
+b + ... +b
ak
n•
2
~
bk for all values of
~
But
(3)
(4)
-
~
we have
~--.-=----~-
(5)
(6)
COMPARISON TEST
UO.6]
339
It follows from (5) and (6) that
Sn < Sf,
(7)
and consequently, by the fundamental assumption, the
___ .~ a series (1) is convergent.
THEOREM II. If each term of a series of positive terms
!
is greater than or equal to the corresponding term of a divergent series of positive terms, then the given series is divergent.
For, if the given series were convergent the second series
would also be, since each of its terms is less than the
corresponding term of the first series.· But this contradicts the hypothesis that the second series is divergent.
In comparing two series it is not sufficient to compare
a few terms at the beginning. The general terms must be
compared.
Example 1.
Show that the following series is convergent:
1
1
1
1
1 + 2 . 2 + 3 . 22 + 4 . 2 3 + . . . + n . 2n 1 + .
.
(1)
SOLUTION. Compare with the convergent geometric series
(see § 20.2),
1
+ 221 + 21 + ... + 2 1 + .
3
n- 1
(2)
Each term of (1) is less than or equal to the corresponding term
of (2). Hence (1) is convergent.
Example .2.
Show that the following series is divergent:
1
1
1
1+y'2+y'3+"'+y-n+
(0
SOLUTION Each term of (1) is greater than or equal to the
corresponding term of the divergent harmonic series (see § ZOA)£
1+~+~+ I
2
Hence (1) is divergent.
3
+~+
n
INFINITE SERIES
340
[Ch. XX
NOTE. The convergence or divergence of a series is unaffected by
omitting (or inserting) a finite nllmber of terms. For this merely
changes Sn by a constant for each value of n beyond the point
of omission (or insertion) of terms. It is often convenient to
neglect several terms at the beginning of a series.
20.7. UseFul comparison series.
The following series will be found useful
comparison tests:
:'" The geometric series,
III
making
+ ar + ar2 + arB + . . . + ar n - +
~(>nvergent for IrI < 1, divergent for IrI ~ 1.
a
1
(1)
. The p series,
1
+ 2p1
+ 3p 1
+ ..1
. +n + ..
P
HI.
(2)
convergent for p > 1, divergent for p ~ 1.
; ii, '
If p = 1 the series is the divergent harmonic series.
", If p < 1 the series is greater, term by term, than the
harmonic series and is consequently divergent.
If p > 1 the convergence of the series can be established
as follows:
(3)
But the expression on the right side of (3) is a convergent
geometric series, since its ratio, 1/2 p -1, is less than 1 when
p > 1. Consequently the p series converges when p > 1.
RATIO TEST
§20.8]
341
20.8. Ratio test.
THEOREM. If, in a series of positive terms, the ratio of the
(n + l)th term to the nth term approaches a limit R as n
inc1 eases without limit, then the series is convergent 'if R < 1
_-___ and divergent if R > 1. * If R = 1 the series may be convergent or may be divergent.
Let the series of positive terms be
(1)
R
If R < 1, choose a number r between Rand 1 (i.e.,
< r < 1). Since, according to hypothesis,
(
then, by the definition of a limit there is a term in the
series, say ak, beyond which an+t/a n is less than r. Thus,
('
i\t,j
ak+l
< r,
ak
ak+2
< r,
ak+l
ak+3
< r,
ak+2
or
ak+l < ak r ;
or
ak+2 < ak+l r < akr2;
or
ak+3 < ak+2 r < ak r3 ;
Let us now compare the series
ak+l
+ ak+2 + ak+3 +
(2)
with the series
(3)
Each term of (2) is less than the corresponding term of (3).
But (3) is a convergent geometric series, since r < 1.
Therefore (2) is convergent. However, (2) is the original
* If the ratio becomes larger without limit (which may be written, symbolically, R ---> 00) the series is obviously divergent.
under the case R > 1.
This may be included
INFINITE SEidES
342
ICh. XX
series (1) with the first k tenIlS omitted, and consequently
(1) is convergent.
If R > 1, choose a number r between Rand 1 (i.e.,
R > r > 1). As before there is a term, say ak, beyond
which we ha'Te
ak+l
> r,
ak
ak+2
> r,
ak+l
ak+3
~- >r,
ak+2
or
ak+l > akr ;
or
ak+2 > ak+l r
::> akr2;
, 1
or,
ak+3
>
ak+2r
>
akr3;
Comparing the series
ak+l
+ ak+2 + aH3 +
(4)
with the geometric series
akr
+ a/,r 2 + akr3 +
(5)
which, since r > 1, is divergent, we establish the divergence of (4) and consequently of the tested series.
If R = 1 we can draw no conclusion from the ratio test,
since, as the following illustrations show, there are both
convergent and divergent series for which R = 1.
Consider the series
1
+-+.
n
(6)
For this deries,
R
=
lim
n--+"
(_1_
!)
n +1
n
-7-
=
1.
The ratio test fails to give any information concerning the
series (6). Jt is, however, the divp"!I~nt harmonic series.
§20.8J
RATIO TEST
343
----------------------------------------
Consider next the series
(7)
Here
R
.
=
~~ [(n ~ 1)2 + ~2]
=
lim ( _ 1
n---+O)
1
)2
+!
=
lim ( -nn+1
)2
n-.....
1.
n
The ratio test gives us no information concerning the series
(7). However, (7) is the p series with p > 1 and is
, convergent.
(
Example 1.
Apply the ratiu test to the series "
SOLUTION.
R
=
·
11m
n-+
00
(n + 1 + n)
2 n +1
1+!
= lim _ _
n
n---+
Since R
0)
2
2n
=,
n
1.un - 2
+ -1
00
n
n~
=!.
2
< 1, the series is convergent.
Example 2.
Apply the ratio test to the series
2
12
22
22
23
+ +3 +
2
SOLUTION.
R =
!~rr; Ln2~+~)2 + ~: ] = !~m.. 2 (n ~ 1)' = 2.
Therefore the series is divergent.
344
SE~IES
INFINITE
[Ch. XX
EXERCISES XX. A
Determine whether the following series are convergent or
divergent:
l,~+~+~+
... +~+
03
G"
;)
2
2. ;;
8
3. I)
...
;)2
2
2
2
+ 10 + 15 + ... + 5n +
'2
3,,-3
1
+ 3" + 2 + ... + :2
~.
t',
+
2 ,,-5
111
.4
5 . (j
111
6. 1
3" "5
2n - 1
4.
<
1
1-2 + 3
+
+ ... + (2n - 1)2n \,
+ + + ...+
+ . . ..
1
1
1
1
+ -+ . G + 6 . 8 + ... + 2n(2n + 2) + . ~j';,''.'''~
1
1
1
1 + _r + ~ r + ... + _/
- + ..
v:3
v 5
v 2n - 1
6. 2. 4
7.
1
8. ;;
+ 522 + 533 + . . . + 5"n + . . . .
__!_Q__ + __!_Q__ + . . . + -;-!2__ + . ". :;
20 3V3
nvn
1
1
'1
1+
+
+ ... + n + ....
v"2
v3
v
+ 1000 + .
1000 + 1000 + 1000 +.
.
9. 10 +
10.
11
.
_3;-
1!
12. 100
+
_3;;)
2!
2!
(100)2
q'"
r,
_3;-
n!
3!
3!
(100)3 +
n!
+
l_ + . . . + l_ +
3
n"
13 1 + l_ + 3
•
22
1
2
3
14. 22 + :32
,-102
~
. + (100)" +
. . .
.
n
+ + ... + (n + 1)2 + .
1
1
16.
1
1
+ 0.01 + 1 + 0.02 + 1 + o.m~ + ... + 1 + O.Oln
+ ....
1
1
1
(l + 0.01)2 + (1 + 0.02)2 + (1 +-0.-m~)2 + ..
15. 1
+
1
(1
+ O.Oln)2 +
'¥,
SERIES OF NEGATIVE TERMS
UO.9]
17.
1
1'2+
+
_ _ _18.
VI
+
1
2'3+ y'2
1
n(n
+ 1) + vn
+
1
3'4+
+
0
+ ....
1
+
1
+
1
1·2+0
2'3+0
3·4+0
+
D
52
53
111
20. 1 + _;;:;
_31')
+ nv 2
v 3
Vn
1
1
21. loge 3
2" loge 4 :3 loge 5
+
+
+ ....
1
:3 log, 3
1
23. loge 3
+
+ ...
+
1
+ \") loge 9
+ ....
+
+ ....
1
n(n + 1) + vn + 1
19. 1 ~ 2 + 2 . 3 + 3 . 4 + . . . + n(n + 1)
22.
345
+
5"
+ ....
1
+n
loge (n + 2)
+
1
27 loge 27
+
1
+ . . . + 3" loge 3 n
1
1
1
loge 9 + loge 27 + . . . + loge 3" +
24. (loge 3)-1 + (loge 4)-2 + (loge 5)-3
+ [loge (n + 2)]-n + . . . .
+
20.9. Series of negative terms.
A series with all of its terms negative is not (:'':)sentially
different from a series with all of its terms positive. Thus,
.
if the series
(1)
in which all of the a's are positive, converges to the limit S,
then the series
(2)
obviously converges to the limit - S Furthermore, if
(1) diwrges so also does (2).
The case in which some of the terms of a series are positive and some negative presents a different situation,
which will be discussed in the next few sections.
0
346
INFINITE SERIES
[Ch. XX
,
I
20.10. Alternating series.
An alternating series is one whose terms are alternately
positive and negative.
THEOREM. An alternating series is convergent if the absolute value of each term is less than that of the preceeding, and
if the limit of the nth term is zero as n increases without limit.
Let the given series be
.-
i
(1)
in which the a's are positive. The sum of an even number
of terms, say 2k terms, may be written in either of the
following two forms:
S2k = (al - a2)
+ (aa -
S2k = al - (al - aa) -
a4)
+ ... + (a2k-1
- aZk),
(2)
. . . - (a2k-2 - a2k-l) - a2k'
(3)
Since an > an+l, each difference in parentheses in (2) and
(3) is positive. Then (2) shows that S2k is positive and
increases as 2k increases, while (3) shows that S2k is never
greater than al. Therefore, by the fundamental as sumption of § 20.5, Sn approaches a limit S, which is less
than or equal to al. Equation (3) shows that S is actually
less than al.
N ow consider the sum of an odd number of terms,
S2k+1
=
S2k
+ a2HI.
We have already shown that S2k approaches a limit S
and by hypothesis a2k+1 approaches zero. Therefore S2k+l
approaches S. Thus, whether n is odd or even Sn converges to a limit S.
Example 1.
Show that the following series is convergent: "
1
1- 2
1
1
+ -3 - -4 +
1
-~--_/
+ (-1 )n+l -n + . . . .
I
)
347
ALlERNAliNG SERIES
§20.10]
SOLUTION. The series is alternating, and the absolute values
of the nth and (n + l)th terms are, respectively,
1
an =-,
n
~-
so that an+l
< an.
Also,
lim an = lim
n-t
00
n---+
00
!n
=
o.
Therefore the series is convergent.
If the absolute value of each term of an alter·
nating series is less than that of the preceding term, and if the
limit of the nth term is zero as n increases without limit, then
the error made in taking the sum of the first n terms for the sum
oj the series is less in absolute value than the \'f~ -1- 1)1.h i~rff£.
Given the alternating series,
COROLLARY.
(
with the a's positive, an+l < an, and lim an
n-""
=
o.
If 8 is
the sum of the series, and 8 n the sum of the first n terms,
we may write
IS -
8 n!
a n +l - a n+2 + an+3 - an+4 + ...
a n+l - (a n+2 - an+3) - (a n-+-4 - a n +6)
=
=
Since the differences in parentheses are positive,
18 - 8 n l < a n +l.
This corollary is useful, if we are using an alternating
series for purposes of computation, ill determining the
accuracy of the approximation.
Exampie 2.
Consider the alternating series,
1-
1
1
2 + 22
-
. . .
+ (-1)n-1 2 n1- + (_l)n 2"1 +
1
which satisfies the conditions of the ecrpllary.
'"
INFINITE SERIES
348
[Ch. XX
The series is an infinite geometric progression whose sum is
1/(1 + i) = i. The sum of n terms of the progression is
1
1
1--+-2
22
The error made in taking the sum of n terms for the sum of the
senes IS
This error is less in absolute value than the (n + l)th term,
(-1)n/2n.
To illustrate with a definite numerical value of n, let us take
n = 4. Then the sum of four terms of the series is
1
2
1 - -
+ -41 - -81 = -85.
,"
!
'j
Subtracting the true sum of the series, we have for the ,~r~;;:
!
521
8 - :3 = -
24'
This is less in absolute value than the 5th term, which is 1/16.
20.11. Absolute and conditional convergence.
THEORE:\L A series with both positive and negative terms
is convergent if the series of absolute values of its terms is
conver(Jcnt.
Let the given series be
(1)
in which some of the a's are positive and some n~gative.
The convergent series of absolute values is, th~
~'_--F"----.-
jail
+ la21 + . . . + lan.!"'+
(2)
I
§20.11J
ABSOLUTE AND CONDITIONAL CONVERGENCE
349
Denote by Sn the sum of the first n terms of (1), by P n
the sum of the positive terms among the first n terms, and
by N n the sum of the absolute values of the negative terms
among the first n. Denote by S~ the sum of the first n
terms of (2). Then
(3)
By hypothesis S~ converges to a limit, say S'.
is a series of positive terms,
S~ = P n
for all finite values of n.
P n < S',
+ Nn
< S'
ri
Hence
Nn
Since (2)
< S'.
Moreover, since P nand N n continually increase with n,
by the fundamental assumption of § 20.5 they approach
limits, say P and N respectively. Thus,
lim Sn = lim (P n - N n) = P - N,
n-+
n---+ co
00
assuming that tb8 limit of a difference is equal to the
difference of the limits. That is, series (1) is convergent.
A series is said to be absolutely convergent if the series
of absolute values of its terms is convergent. If a series
with both positive and negative terms is convergent, but
the series of absolute values of its terms is divergent, the
given series is said to be conditionally convergent.
For example, the series
1
1 - 22
+ 21
3 -
1
24
+ ... + (_1)n+l 21n +
is absolutely convergent, sinc0 the series
is convergent.
. ..
[Ch. XX
INFINITE SERIES
350
The series
I
I
1-!+!-!+
234
is conditionally convergent. We proved in § 20.10 that
it is convergent, but the series of absolute values of its
terms is
111
1+-+-+-+"
234
the divergent harmonic series.
20.12. Ratio test extended.
If in a series of real terms the absolute value
of the ratio of the (n + 1)th term to the nth term approaches
a limit R as n increases without limit, then the series is
absolutely convergent if R < 1 and divergent if R > 1. If
R = 1 the series may be convergent or may be divergent.
Since R is the limit of the ratio of the (n + 1)th to the
nth term of the series of absolute values, by the Theorem
of § 20.8, this series of absolute values is convergent if
R < 1. That is to say, the given series is absolutely convergen t if R < 1.
If R > 1, each term, after a certain place in the series,
will be numerically greater than the preceding, and consequently the general term does not approach zero. Therefore the series diverges. (See § 20.4.)
If R = 1 no conclusion can be drawn.
THEOREM.
EXERCISES XX. B
Write out several terms of the series whose general term is
given below, starting with the smallest value of n for which the
term is defined and different from zero. Determine whether the
series is convergent or diverg~nt.
1. (_I)n+l
1
•
v'2n - 1
3. (_I)n+l _1_.
.
loglO n
II
2. (_I)n+l n ~ (
4. (_I)n+l
1
yn'
POWER SERIES
UO.13]
1
5. (-l)n+1 1 _ O.ln·
7. (-l)n+1
1
Vr/:
1
n-
1
vn
.
n~
10. (-l)n+1 (lOoo)n'
2n
(-l)n+I~.
12. (-l)n+l n-"'-(n-+-:--:1")
V n !
13. (-l)n+1 loglo n
6. (-l)n+l(l - O.ln).
8. (-l)n+1
9. (-1)n+1(2 - !n).
11.
351
+1 1"
15. (_1)n+1 l0ge (n + 1).
n
e
17. (_l)n+llog (n,+ O.
n.
1
14. (-l)n+1 loge (n
+
16. (_1)n+1loglo (~
n
1)
+ 1).
18. (-1)n+1 _2_.
loglo n
20.13 Power series.
A series of the form
in which the c's are constants, is called a power series in x.
A series such as
+ cn(x - a)n
+ .. "
(2)
is a power series in x-a.
The set of values of x for which a power series converges
is called the interval of convergence. This interval of
convergence can be determined by means of the extended
form of the ratio test.
Example.
Find the interva! of convergence of the following series:
x2
x3
X4
x+-+-+-+
2
3
4
INFINITE SERIES
352
(Ch. XX
SOLUTION.
.
Xn+l
R = n-+
hm -+'
n
1
00
Xn
-;- -n
n
.
= n---+
11m -+1 [x[ = [x[.
n
00
The series is convergent if R < 1, that is, if [x[ < 1.
The end values, x = ± 1, must be investigated separately.
For x = 1, the series is the diyergent harmonic series, for x =
-1 it is a convergent alternating series. Thus, the interval of
convergence IS
-1~x<1.
'
, '.,1, EXERCISES XX. C
Determine .the"b;lWpral of convergenC?6,;qfl,~t~, ~h~
general term IS: ',i
•
,
I
.I" ,
I
t
I
X,,-1
4.
xn -
9. n(n
n
6. (_1)n-1x2n.
xn
8. (-1),,-1 n (
1
7. (-1)n- 1 ""F"'
xn -
-3-'
x n-
1
+
11. xnyn.
.,,~ 12. (-1)n
13.
(~yn-1.
14. ( _
16.
(x - 2)n
2n
.
16.
17. (-1)n (j-x 5n
19.
(x - a)n-1
bn
1
10. (_1)n-1 2n(2n _ 1)'
1)'
•
3)~' :,'
x"
-v1i:
~)2n+l.
(_1)n-l
"
(x -2:)"-1
(3x - 7)n+1.
18. -'-------'.5n
20. (_1)n-1 (x ~" a)n.
TABLES
INDEX
ANSWERS
f
TABLE I.-POWERS AND ROOTS
No. Square
-1 --12
4
3
9
Cube Square Cube No. Square
root root
--r
51 260T
2704
2809
132 651
140 608
148877
7.141 3.708
7.211 3.733
7.280 3.756
1.587
1.710
1.817
54
55
56
2 916
3 025
3 136
157 464
166 375
175 615
7.348 3.780
7.416 3.803
7.483 3.826
57
58
59
1.000
1.260
1.442
64 2.000
125 2.236
216 2.449
4
5
6
16
25
36
7
8
9
10
IT
12
13
49
64
81
100
121
144
169
1
1
1
2
2.646
2.828
3.000
3.162
3.317
3.464
3.606
1.913
2.000
2.080
2.154
2.224
2.289
2.351
14
15
16
196
225
256
2 744 3.742
3 375 3.873
4 096 4.000
2.410
2.466
2.520
64
65
66
4096
4 225
4356
2.571
2.621
2.668
2.714
2.759
2.8(12
2.844
67
68
69
4 489
4 624
4 761
-4-!JOO
-
4.123
4.243
4.359
4.472·
4.583
4.690
4.796
---so
22
23
484
529
24
25
26
576
625
676
13 824 4.899
15 625 5.000
17 576 5.099
27
28
29
32
33
729
784
841
900
961
1 024
1 089
19
21
24
27
29
32
35
34
35
36
1 156
1 225
1 296
39304 5.831 3.240
42875 5.916 3.271
46 656 6.000 3.302
84
85
86
37
38
39
50
54
59
64
68
74
79
3.332
3.362
3.391
3.420
3.448
3.476
3.503
87
88
89
42
43
1
1
1
1
1
1
1
44
45
46
1 936
2 025
2 116
47
48
49
2 209
2 304
2 401
289
;)24
361
20 ~
21 ---:«1
30
31"
To
41
369
444
521
600
681
764
849
60 2500
913
832
859
000
261
648
167
3 249
185 193
3364
195 112
3 481
205 379
3600 -216000
6f"" 372T 226 981
62
3 844
238 328
63
3 969
250 047
4
5
6
8
9
10
12
17
18
19
683
952
389
000
791
768
937
653
872
319
000
921
088
507
5.196
5.292
5.385
5.477
5.568
5.657
5.745
6.083
6.164
6.245
6.325
6.403
6.481
6.557
2.884
2.924
2.962
6.856
6.928
7.000
7.071
3.849
3.871
3.893
3.915
3.936
3.958
3.979
8.000 4.000
8.062 4.021
8.124 4.041
763
432
509
000
911
248
017
8.185 4.062
8.246 4.082
8.307 4.102
8.367 4.121
8.426 4.141
8.485 4.160
8.544 4.179
72
73
5 184
5 329
74
75
76
5 476
5625
5 776
405 224
421 875
438 976
8.602 4.198
8.660 4.217
8.718 4.236
456
474
493
512
8.775
8.832
8.888
8.944
9.000
9.055
9.110
7 056
7 225
7 396
592 704
614 125
636 056
9.165 4.380
9.220 4.397
9.274 4.414
92
93
7 569
7744
7 921
8 100
8 281
8464
8 649
658
681
704
729
753
778
804
9.327
9.381
9.434
9.487
9.539
9.592
9.644
94
95
96
8 836
9 025
9 216
830 584
857 375
884 736
---oro
71 504T
----so
3.609
3.634
3.659
3.684
262 144
274 625
287 496
7.550
7.616
7.681
7.746
7.810
7.874
7.937
300
314
328
343
357
373
389
77
3.000
5929
3.037
78
6084
3.072
6 241
79
3.107
6400
3.141 8 1 656l
3.175
82
6 724
3.208
83
6 889
85 184 6.633 3.530
91 125 6.708 3.557
97 336 6.782 p.583
103 823
110 592
117 649
125000
Square Cube
root root
52
53
1.000
8 1.414
27 1.732
343
512
729
000
331
728
197
Cube
90
91
97
98
99
100
355
533
552
039
000
----:53T 441
551 368
571 787
_--
9 409
9 604
9 801
10000 1
503
472
969
000
571
688
357
4.254
4.273
4.291
4.309
4.327
4.344
4.362
4.431
4.448
4.465
4.481
4.498
4.514
4.531
9.695 4.547
9.747 4.563
9.798 4.579
912 673 9.849
941 192 9.899
970 299 9.950
000000 10.000
4.595
4.610
4.626
4.642
iTABLE
N
-55
II.-LoGARtTlnoIS (Ccm:linUEd)
2
4
Ii
1
3
0
-- - - - ------7404 -7412 7419 7427 7435 7443 -7451 7459 7466 7474
6
7482 7490
57 7559 7566
58 7634 7642
59 7709 7716
50
7497
7574
7649
7723
7505
7582
7657
7731
7513 7520
7589 7597
7664 7672
7738 7745
7528
7604
7679
7752
7
8
9
7536 7543
7612 7619
7686 7694
7760 7767
7551
7627
7701
7774
7839
7910
7980
804.8
8116
7846
7917
7987
8055
8122
- - - - - - - - - - - - - - - - - ---7796 7803 7810 7818 7825
7868 7875 7882 7889 7896
7938 7945 7952 7959 7966
8007 8014 8021 8028 8035
8075 8082 8089 8096 8102
7782
7853
7924
7993
64 8062
7789
7860
7931
8000
8069
65
66
67
68
69
8129
8195
8261
8325
8388
8136 8142 8149 8156 8162 8169 8176 8182 8189
8202 8209 8215 8222 8228 8235 8241 8248 8254
8267 8274 8280 8287 8293 8299 8306 8312 8319
8331 8338 8344 8351 8357 8363 8370 8376 8382
8395 8401 8407 8414 8420 8426 8432 8439 8445
70
71
72
73
74
8451
8513
8573
8633
8692
8457 8463
8519 8525
8579 8585
863!) 8645
8698 8704
8751
8808
8865
8921
8976
8756 8762 8768 8774
8814 8820 8825 8831
8871 8876 8882 8887
8927 8932 8938 8943
8982 8987 8993 8998
60
61
62
63
7832
7903
7973
8041
8109
_ - - -_ - - - - - - - -- - - - -- - -
(
_ - - - - - - - - - - - - - - - - ---8470
8531
8591
8651
8710
8476
8537
8597
8657
8716
8482 8488
8543 8549
8603 8609
8663 8669
8722 8727
8494 8500
8555 8561
8615 8621
8675 8681
8733 8739
8506
8567
8627
8686
8745
_- - - - - -- - - ----
-
------
-80
_- - - - - _- - - - - -- - - - - - -
75
76
77
78
79
8779
8837
8893
8949
9004
8785 8791 8797 8802
8842 8848 8854 8859
8899 8904 8910 8915
8954 8960 8965 8971
9009 9015 9020 9025
9031 9036 9042 9047 9053 9058 9063 9069
81 9085 9090 9096 9101 9106 9112 9117 9122
82 IH38 9143 9149 9154 9159 9165 9170 9175
83 9191 9196 9201 9206 9212 9217 9222 9227
84 9243 9248 9253 9258 9263 9269 9274 9279
-
85
86
87
88
89
9074 9079
9128 9133
9180 9186
9232 9238
9284 9289
- - - - - - - - - - - -- - - - - - 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340
9345
9395
9445
9494
9350 9355 9360 9365 9370 9375 9380 9385 9390
9400 9405 9410 9415 9420 9425 9430 9435 9440
9450 9455 9460 9465 9469 9474 9479 9484 9489
9499 9504 9509 9513 9518 9523 9528 9533 9538
- - - _- - - - - - - - - --- - "
-90 -9542- 9547 9552 9557 9562 9566 9571 9576 9581 ()S86
91
92
93
94
-95
96
97
98
99
9590 9595 96GO 9605 960<;1 <;1614 ';)01,;) 0024 OO2S 91'>33
9638 9643 9647 9652 9657 9661 9666 9671 9675 9680
9685 9689 9694 9699 9703 9708 9713 9717 9722 9727
9731 9736 9741 9745 9750 9754 9759 9763 9768 9773
_- -9782- - - - - -9795- -9800- -9805- -9809- -9814- -98189777
9823
9868
9912
9956
9786 9791
9827 9832 9836
9872 9877 9881
9917 9921 9926
9961 9965 9969
9841 9845 9850 9854 9859 9863
9886 9890 9894 9899 9903 9908
9930 9934 9939 9943 9948 9952
9974 9978 9983 9987 9991 9996
TABLE II.-LOGARITU:'lS
N
0
1
a
3
-10 -0000
- -0043
- -0086 0128
11 04H 0453 0492 0531
12 0792 0828 0864 0899
13 1139 1173 1206 1239
14 1161 1492 1523 1553
6
1
8
9
----0170 0212 0253 0294 0334 0374
6
4
0569
0934
1271
1584
0607 0645
0969 1004
1303 1335
1614 1644
0682 0719 0755
1038 1072 1106
1367 1399 1430
1673 1703 1732
-
- - - - - - -- - - - - -- - - - - --
-
- - - ---- - - - - -- - - - - - 3010 3032 3054 3075 3096 3118 3139 3160 3181
-
- - - - ---- - - - - -- - - - - - -
-
-4771- -4786
- -- - - - -- - - - - -- -4800
4814 4829 4843 4857 4871 4886 4900
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
1761
2041
2304
2553
2788
1790 1818
2068 2095
2330 2355
2577 2601
2810 2833
1847 1875 1903 1931 1959
2122 2148 2175 2201 2227
2380 2405 2430 2455 2480
2625 2648 2672 2695 2718
2856 2878 2900 2923 2945
1987 2014
2253 2279
2504 2529
2742 2765
2967 2989
3201
3222 3243 3263 3284 3304 3324 3345 3365 3385 3404
3424 3444 3464 3483 3502 3522 3541 3560 3579 3598
3617 3636 3655 3674 3692 3711 3729 3747 3766 3784
3802 3820 3838 3856 3874 3892 3909 3927 3945 3962
3979 3997 4014 4031 4048 4065
4150 4i66 4183 4200 4216 4232
4314 4330 4346 4362 4378 4393
4472 4487 4502 4518 4533 4548
4624 4639 4654 4669 4683 4698
4082 4099
4249 4265
4409 4425
4564 4579
4713 4728
4116 4133
4281 4298
4440 4456
4594 4609
4742 4757
4914 4928 4942 4955 4969 4983 4997 5011 5024 5038
5051 5065 5079 5092 5105 5119 5132 5145 5159 5172
5185 5198 5211 5224 5237 5250 5263 5276 5289 5302
5315 5328 5340 5353 5366 5378 5391 54.03 5416 M28
- - - - - - - - - - - - - -- - - - - -35
36
37
3S
39
-
40
41
42
43
44
-
45
46
47
48
49
-
5441
5563
5682
579S
5911
5453
5575
5694
5465
5587
5705
5SI)~
58'21
5922 5933
5478
5599
5717
5832
5944
5490
5611
5729
5843
5955
5502 5514 5527
5623 5635 5647
5740 5752 5763
5855 5866 5877
5966 5977 5988
5539 5551
5658 5670
5775 5786
5888 5899
5999 6010
6180 6191
6284 6294
6385 6395
6484 6493
6212
6314
6415
6513
6222
6325
6425
6522
6693 6702
6785 6794
6875 6884
6964 6!J72
6712
6803
6893
6981
- - - - - -- - - - - -- - - - - - 6021 6031 6042 6053 6064 6075 6085 6096 6107 6Il7
6128 6138
6232 6243
6335 6345
6435 6444
6149 6160 6170
6253 6263 6274
6355 6365 6375
6454 6464 6474
6201
6304
6405
6503
-- - - - -- - - - -- - - - - -6532 6542 6551 6561
6580 6590 6599 6609 6618
6571
6628 6637 6646 6656 6665 6675 6684
6721 6730 6739 6749 6758 6767 6776
6812 6821 6830 6839 6848 6857 6866
6902 6911 6920 6928 6937 6946 6955
- - - - - - ----7016 1024 7033
50 6990 6998 7007
51 7076 7084 7093
52 7160 7168 7177
53 7243 7251 7259
54 7324 7332 7340
7101 7110
7185 7193
7267 7275
7348 7356
-
7042
7118 7126
7202 7210
7284 7292
7364 7372
----7050 7059 7067
7135 7143 7152
7218 7226 7235
7300 7308 7316
7380 7388 7396
TABLE
+ r)"
1i%
2%
2t%
3%
4%
5%
6%
7%
2 1.0201
3 1.0303
1.0302
1.0457
1.0404
1.0612
1.0506
1.0769
1.0609
1.0927
1.0816
1.1249
1.1025
1.1576
1.1236
1.1\HO
1.1449
1.2250
4 1.0406
5 1.0510
6 1.0615
1.0614
1.0773
1.0934
1.0824
1.1041
1.1262
1.1038
1.1314
1.1597
1.125.'5
1.1593
1.1941
1.1699
1.2167
1.2653
1.2155
1.2763
1.3401
1.2625
1.3382
1.4185
1.3108
1.4026
1.5007
1.0721
1.0829
1.0937
],1046
IT 1.1157
12 1.1268
13 1.1381
1.1098
1.1265
1.1434
1.1605
1.1779
1.1956
1.2136
1.1487
1.1717
1.1951
1.2190
1.2434
1.2682
1.2936
1.1887
1.2184
1.2489
],2801
1.3121
1.3449
1.3785
1.2299
1.2668
1.3048
1.3439
1.:';842
1.4258
1.4685
1.3159
1.3686
1.4233
],4802
1.5395
1.4071
1.4775
1.6651
1.6289
1.7103
1.7959
1.8856
l.5W6 1.605S
1.5938 1.7182
1.6895 1.8385
1. 7908 11.9672
1.8983 2.1049
2.0122 2.2522
2.1329 2.40\)8
14 1.1495
15 1.1610
16 1.1726
1.2318
1.2502
1.2690
1.3195
1.3459
1.3728
1.4130
1.4483
1.4845
1.5126
1.5580
1.6047
1.7317
1.8009
1.8730
1.9799
2.0789
2.1829
2.2609
2.3966
2.5404
2.5785
2.7590
2.9522
1.1843 1.2880
1.1961 1.3073
1.2081 1.3270
1.2202 1.3459
2f 1.2324 1.3671
22 1.2447 1.3875
23 1.2572 1.4084
1.4002 1.5216 1.6528
1.4282 1.5597 1.7024
1.4568 1.5987 1.7535
1.4859 1.6385 1.8061
1.5157 1.5796 1.8603
1.5460 1.7216 1.9161
1.5769 1.7646 1.9736
1.9479
2.0258
2.1058
2.1911
2.2788
2.3599
2.4647
2.2920
2.4066
2.5270
2.6533
2.7850
2.9253
3.0715
2.6928
2.8543
3.0255
3.2071
3.3996
3.60.15
3.8197
3.1588
3.3799
3.6165
Z.il697
4.1406
4.4304
4.7405
24 1.2697
25 1.2824
26 1.2953
1.4295
1.4509
1.4727
1.6084
1.6406
1.6734
1.8087
1.8539
1.9003
2.0328
2.0938
2.1566
2.5633
2.6658
2.7725
3.2251
3.3864
3.5557
4.0-189
4.2919
4.5494
5.0724
5.4274
5.8074
27 1.3082
28 1.3213
29 1.3345
30 1.3478
IT 1.3613
32 1.3749
33 1.3887
1.4948 1.7059
1.5172 1.7410
1.5400 1.7758
1.5631 1.8114
1.5865 1.8476
1.6103 1.8845
1.6345 1.9222
1.9478
1.9965
2.0454
2.0976
2.1500
2.2038
2.2589
2.2213
2.2879
2.3566
2.4273
2.5001
2.5751
2.6523
2.8834
2.9987
3.1187
:i.2434
3.3731
3.5081
3.6484
3.7335
3.9201
4.1161
4.3219
4.5380
4.7649
5.0032
4.8223
5.1117
5.4184
5.7435
G.0881
6.4534
6.8405
5.2139
6.6488
7.1143
7.6123
8.1451
8.7153
9.3253
34 1.4026
35 1.4166
36 1.4308
U\590
1.6839
1.7091
2.3153
2.3732
2.4325
2.7319
2.8139
2.8983
3.7943
3.9461
4.1039
5.2533
5.5150
5.7918
7.2510 9.9781
7.6861 10.6765
8.1473 11.4239
n
1%
III.-CoMPOUND AMOUNT: (1
- - --- --- --- --- --- --- - -1.0100
--1.0150 1.0200 1.0250 1.0300 1.0400 1.0500 1.0600 1.0700
1
7
8
9
10
17
18
19
110
1.9607
1.9999
2.0399
1.6010
1.551~
37 1.4451 1.7348 2.0807 2.4933
38 1.4595 1.7608 2.1223 2.5557
39 1.4741 1.7872 2.1G47 2.6196
40 1.4889 1.8140 2.2080 2.6851
IT 1.5038 1.8412 2.2522 2.7522
42 1.5188 1.8588 2.2972 2.8210
43 1.5340 1.8969 2.3432 2.8915
2.9852 4.2681
3.0748 4.4388
3.1670 4.6164
3.2520 4.80io
a.3599 4.9931
3.4607 5.1928
3.5645 5.4005
44 1.5493
45 1.5648
46 1.5805
2.3901 2.9638
2.4379 3.0379
2.4865 3.1139
3.6715
3.7816
3.8950
5.6165
5.8412
6.0748
2.5363
2.5871
2.6388
2.5915
4.0119
4.1323
4.2562
4.3839
6.3178 9.9060 15.4659 24.0457
6.5705 10.4013 16.3939 25.7289
6.8333 10.9213 17.3775 27.5299
7.1067 1l.4()74 18.4202 29.4570
47
48
4.9
50
1.9253
1.9542
1.9835
1.5963 2.0133
1.612212.0435
1.6283 2.0741
1.6446 2.1052
3.1917
3.2715
3.3533
3.4371
358
f
/
/
6.0814
6.3855
6.7048
7.0400
7.3920
7.7516
8.1497
8.6361
9.1543
9.7035
10.2857
10.9029
11.5570
12.2505
12.2236
13.0793
13.9948
14.9745
15.0227
17.1443
18.3444
8.5572 12.9855 19.6285
8.9850 13.7545 21.0025
9.4343 14.5905 22.4726
TABLE
n
1%
1!%
N.-PRESENT
2%
21%
--- ------ I --.99010 ,98522 .98039 .97561
(1
-3% - 4%
2 98030
3 .97059
97066
.95632
.96117
94232
.97087
.95181 .94260
.92860 .91514
4 .96098
.95147
6 .94205
.94218
.92826
.91454
,92385
.90573
.88797
.90595 .88849
.88385 .86261
.86230 .83748
7 .93272
8 .92348
9 91434
10 90529
IT .89632
12 .8,,745
13 .87866
.90103
.88771
.87459
.86167
.84893
.83639
.82403
14 .86996
15 .86135
16 .85282
17
18
19
20
21
22
23
.96154
.92456
.88900
.85480
.82193
.79031
+ r)-n
5% 6% 7%
--------.95238 .94340 .93458
.90703
.86384
.89000
.83962
.87344
.81630
.82270
.78353
.74622
.79209
.74726
.70496
.76290
.71299
.66634
.87056
.85349
.83676
.82035
.80426
.78849
.77303
.84127 .81309 .75992 .71068
.82075 .78941 .73069 .67684
.80073 .76642 .70259 .64461
.78120 .74409 .67556 .61391
.76214 .72242 ~ .58468
.74356 .70138 .62460 .55684
.72542 .68095 .60057 .53032
.66506
.62741
.59190
.55839
.52679
.49697
.40884
.62275
.58201
,54393
.50835
.47509
.44401
.41490
.81185
.79985
.78803
.75788
.74301
.72845
.70773
.69047
.67362
.50507
.48102
.45811
.44230
.41727
.39365
.38782
.36245
.33873
.84438
.83602
.82774
.8HI54
.81143
.80340
.79544
.77639
.76491
.75361
.74247
.73150
.72069
.71004
.71416 .65720
.70016 .64117
.68043 .62553
.67297 .61027
.65978 .59539
.64684 .58086
.63416 .56670
.60502 .51337
.58739 .49363
.57029 .47404
.55368 A5639
.53755 .43883
.52189 .42196
.50669 .40573
.43630
.41552
.39573
.37689
.35894
.34185
.32557
.371313 .31657
.35034 .29586
.33051 .27651
.31180 .25842
.29416 .24151
.27751 .22571
.26180 .21095
24 .78757
25 .77977
26 .77205
.69954
.68921
.67902
.62172
.60953
.59758
.49193
.47761
.46369
.31007
.29530
.28124
.24698
.23300
.21981
.19715
.18425
.17220
27
28
29
30
31
32
33
.76440
.75684
.74934
.74192
.,'3458
.72730
.72010
.66899
.65910
.64936
.63976
.63031
.62099
.61182
.58586 .51340 .45019 .34682 .26785 .20737
.57437 .50088 .43708 .33348 .25509 .19563
.56311 .48866 .42435 .32065 .24295 .18456
.55207 .47674 .41199 .30832 .23138 .17411
.54125 .46511 .39999 ~ .22036 .16425
.53063 .45377 .38834 .28506 .20987 .15496
.52023 .44270 .37703 .27409 .19987 .14619
.16093
.15040
.14056
.13137
.12277
.11474
.10723
34 .71297
35 .70591
36 .69892
.60277
.593R7
.58509
.51003
.50003
.49022
.43191
.42137
.41109
.36604
.35538
.34503
.26355
.25342
.24367
.19035
.18129
.17266
.13791
.13011
.12274
.10022
.09366
.08754
37
38
39
40
41
42
.69200
.68515
.67837
.67165
.66500
.65842
43 .65190
.57644
.56792
.55953
.55126
.54312
.53509
.52718
.48061 .40107
.47119 .39128
.46195 .38174
.45289 .37243
.44401 .36335
.43530 .35448
.42677 .34584
.33498
.32523
.31575
.30656
.29763
.28896
.28054
.23430
.22529
.21662
.11579
.10924
.10306
.20028
.19257
.18517
.16444
.15661
.14915
.14205
.13528
.12884
.12270
.09172
.08653
.08163
.08181
.07646
.07146
.06678
.06241
.05833
.05451
44 .64545
45 .63905
46 .63273
.51939
.51171
.50415
.41840 .33740 .27237
.41020 .32917 .26444
.40215 .32115 .25674
.17805
.17120
.16461
.11686 .07701
.11130 .07265
.10600 .06854
.05095
.04761
.04450
47 .62646
48 .62026
49 .61412
&Q .60804
.49670 .39427
.48936 .38654
.48213 .37896
.47500 .37153
.15828
.15219
.14634
.14071
.10095
.09614
.09156
.08720
.06466
.06100
.05755
.04159
I)
(
VALUE:
.55288
.53939
.52623
.31331
.30567
.29822
.29094
.66112
.64186
.62317
.24926
.24200
.23495
.22811
359'
.57748
.55526
.53391
.39012
.37512
.36069
-:20829
-:o972z
.03887
.03632
--:o5-i29 -:03395
TABLE V.-AMOUNT 011' AN ANNUITY: [(1
n 1%
-,
11%
+ r)" -
I]/r
3%
4%
5%
6%
"1%
--- -2!%- ----------1.0000 1.0000 1.0000 1.0000 1.0000
2%
I1.0000 1.0000 1.0000 1.0010
2 2.0100 2.0150 2.0200 2.0250
3 3.0301 3.0452 3.0604 3.0756
2.0300
3.0909
2.0400
3.1216
2.0500
3.1525
2.0600
3.1836
2.0700
3.2149
4 4.0604 4.0909 4.1216 4.1525
5 5.1010 5.11)23 5.2040 5.2563
6 6.1520 ~.2296 6.3081 6.3877
4.1836
5.3091
6.4684
4.2465
5.4163
6.6330
4.3101
5.5256
6.8019
4.3746
5.6371
6.9753
4.4399
5.7507
7.1533
7 7.2135 7.3230 7.4343 7.5474
8 8.2857 8.4328 8.5830 8.7361
9 9.3685 9.5593 9.7546 9.9545
10 10.4622 10.7027 10.9497 11.2034
IT 11.5668 11.8633 12.1687 12.4835
12 12.6825 13.0412 13.4121 13.7956
13 13.8093 14.2368 14.6803 15.1404
7.6625
8.8923
10.1591
11.4639
12.8078
14.1920
15.6178
7.8983
9.2142
10.582B
12.0061
13.4864
15.0258
16.6268
8.1420
9.5491
11.0266
12.5779
14.2068
15.9171
17.7130
8.3938
9.8975
11.4913
13.1808
14.9716
16.8699
18.8821
8.6540
10.2598
11.9780
13.8164
15.7&36
17.8885
20.1406
-~-
14 14.9474 15.4504 15.9739 16.5190 17.0863 18.2919 19.5986 21.0151 22.5505
1516.0963 16.6821 17.2934 17.9319 18.5989 20.0236 21.5786 23.2760 25.1290
16 17.2579 17.9324 18.6393 19.3802 20.1569 21.8245 23.6575 25.6725 27.8881
17 18.4304 19.2014 20.0121 20.8647
1819.6147 20.4894 21.4123 22.3863
19 20.8109 21.7967 22.8406 23.9460
20 22.0HlO 23.1237 24.2974 25.5447
21 23.2392 24.4705 25.7833 27.1833
22 24.4716 25.8376 27.2990 28.8629
23 25.7163 27.2251 28.8450 30.5844
21.7616
23.4144
25.1169
26.8704
28.6765
30.5368
32.4529
23.6975
25.6454
27.6712
29.7781
31.9692
34.2480
36.6179
25.8404
28.1324
30.5390
33.0660
35.7193
38.5052
41.4305
28.2129
30.9057
33.7600
36.7856
39.9927
43.3923
46.9958
30.8402
33.9990
37.3790
40.9955
44.8652
49.0057
03.4361
24 26.9735 28.6335 30.4219 32.3490 34.4265 39.0826 44.5020 50.8156 58.1767
25 28.2432 30.0630 32.0303 34.1578 36.4593 41.6459 47.7271 54.8645 63.2490
26 29.5256 31.5140 33.6709 36.0117 38.5530 44.3117 51.1135 59.1564 68.6765
27 30.8209 32.9867 35.3443 37.9120
28 32.1291 34.4815 37.0512 39.8598
29 33.4504 35.9987 38.7922 41.8563
34.7849 37.5387 40.5681 43.9027
31 36.1327 39.1018 42.3794 46.0003
32 37.4941 40.6883 44.2270 48.1503
33 38.8690 42.2986 46.1116 50.3540
30
40.7096
42.9309
45.2189
47.5754
M.0027
52.5028
55.0778
47.0842
49.9676
52.9663
56.0849
59.3283
62.7015
66.2095
54.6691
58.4026
,62.3227
66.4388
70.76~
75.2988
80.0638
63.7058
68.5281
73.6398
79.0582
84.8017
90.8898
97.3432
74.4838
80.6977
87.3465
94.4608
lO2.0730
110.2182
118.9334
34 40.2577 43.9331 48.0338 52.6129 57.7302 69.8579 85.0670 104.1838 128.2588
35 41.6603 45.5921 49.9945 54.9282 60.4621 73.6522 90.3203 111.4348 138.2369
36 43.0769 47.2760 51.9944 57.3014 63.2759 77.5983 95.8363 119.1209 148.9135
37 44.5076 48.9851 54.0343 59.7339
38 45.9527 50.7199 56.1149 62.2273
47.4123 52.4807 58.2372 64.7830
~
40 48.8864 54.2679 60.4020 67.4026
tIT 50.3752 56.0819 62.6100 70.0876
42 51.8790 57.9231 64.8622 72.8398
43 53.3978 59.7920 67.1595 75.6608
66.1742
69.1594
72.2342
75.4013
78.6633
82.0232
85.4839
81.7022 101.6281
85.9703 107.7095
90.4091 114.0950
95.0255 120.7998
99.8265 127.8398
104.8196 135.2318
110.0124 142.9933
127.2681 160.3374
135.9042 172.5610
145.0585 185.6403
154.7620 199.6351
165.0477 214.6096
175.9505 230.6322
187.5076 247.7765
44 54.9318 61.6889 69.5027 78.5523 89.0484 115.4129 151.1430 199.7580 266.1209
285.7493
63.6142 71.8927 81.5161 92.7199 121.0294 159.7002 212.7435
58.0459 65.5684 74.3306 84.5540 96.5015 126.8706 168.6852 226.5081 306.7518
!~ 56.4811
59.6263 67.5519 76.8172 87.6679 100.3965 132.9454 178.1194 241.0986 329.2244
61.2226 69.5652 79.3535 90.8596 104.4084 139.2632 188.0254 256.5645 353.2701
~~
49 62.834871.6087 81.9406 94.1311 108.5406 145.8337 198.4287 272.9584 378.9990
liD 64.4632 73.6828 84.5794 97.4843 112.7969 152.6671 209.3480 290.3359 406.5289
1
360'
-
/
TABLE VI.-Pro:SENT VALUE OF AN
1!%
2%
ANNUITY: [1- (1
3% 4%
5% 6% 7%
-----------.9346
.9756
.9615
.9709
.9524
.9434
21%
n
1%
2
3
1.9704
2.9410
.9852
.9804
1.9559 1.9416 1.9274 1.9135 1.8861
2.9122 2.8839 2.8560 2.8286 2.7751
4
5
6
3.9020
4.8534
5.7955
3.8544 3.8077 3.7620 3.7171
4.7826 4.7135 4.6458 4.5797
5.6972 5.6014 5.5081 5.4172
7
8
9
6.7282
7.6517
8.5660
-1 - .9901
6.5982
7.4859
8.3OO5
10 9.47l3 9.2222
IT 10.3676 10.0711
12 11.2551 10.9075
13 12.1337 11.7315
6.4720
7.3255
8.1622
8.9826
9.7868
10.5753
11.3484
+ r)....]/r
6.3494
7.1701
7.9709
8.7521
9.5142
10.2578
10.9832
6.2303
7.0197
7.7861
8.5302
9.2526
9.9540
10.6350
1.8594 1.8334
2.7232 2.6730
1.8080
2.6243
3.6299 3.5460 3.4651
4.4518 4.3295 4.2124
5.2421 5.0757 4.9173
3.3872
4.1002
4.7665
6.0021
6.7327
7.4353
8.1109
8.7605
9.3851
9.9856
5.3893
5.9713
6.5152
7.0236
7.4987
7.9427
8.3577
5.7864
6.46327.1078
7.7217
8.3064
8.8633
9.3936
5.5824
6.2098
6.8017
7.3601
7.8869
8.3838
8.8527
14 13.0037 12.5434 12.1062 11.6909 11.2961 10.5631 9.8986 9.2950 8.7455
15 13.8651 13.3432 12.8493 12.3814 11.9379 11.1184 10.3797 9.7122 9.1079
16 14.7119 14.1313 13.5717 1;3.0550 12.5611 11.6523 10.8378 10.1059 9.4466
I
17 15.5623 14.9076 14.2919
18 16.3983 15.6726 14.9920
19 17.2260 16.4262 15.6785
iO 18.0456 17.1686 16.3514
21 18.8570 17.9001 17.U112
22 19.6604 18.6208 17.6580
23 20.4558 19.3309 18.2922
13.7122
14.3534
14.9789
15.5892
16.1845
16.7654
11.3321
13.1661
13.7535
14.3238
14.8775
15.4150
15.9369
16.4436
12.1657
12.6593
13.1339
13.5903
14.0292
14.4511
14.8568
11.2741
11.6896
12.0853
12.4622
12.8212
13.1630
13.4886
10.4773
10.8276
11.1581
11.4699
11.7641
12.0416
12.3034
9.7632
10.0591
10.3356
10.5940
10.8355
11.0612
11.2722
24 21.2434 20.0304 18.9139 17.8850 16.9355 15.2470 13.7986 12.5504 11.4693
25 22.0232 20.7196 19.5235 18.4244 17.4131 15.6221 14.0939 12.7834 11.6536
26 22.7952 21.3986 20.1210 18.9506 17.8768 15.9828 14.3752 13.0032 11.8258
27 23.5596 22.0676 20.7069
28 24.3164 22.7267 21.2813
29 25.0658 23.3761 21.8444
30 25.8077 24.0158 22.3965
31 26.5423 24.6461 22.9377
32 27.2696 25.2671 23.4683
33 27.9897 25.8790 23.9886
19.4640
19.9649
20.4535
20.9303
21.3954
21.8492
22.2919
18.3270
18.7641
19.1885
19.6004
20.0004
20.3888
20.7658
16.3296
16.6631
16.9837
17.2920
17.5885
17.8736
18.1476
14.6430
14.8981
15.1411
15.3725
15.5928
15.8027
16.0025
13.2105
13.4062
13.5907
13.7648
13.9291
14.0840
14.2302
11.9867
12.1371
12.2777
12.4090
12.5318
12.6466
12.7538
34 28.7027 26.4817 24.4986 22.7238 21.1318 18.4112 16.1929 14.3681 12.8540
35 29.4086 27.0756 24.9986 23.1452 21.4872 18.6646 16.3742 14.4982 12.9477
,36 30.1075 27.6607 25.4888 23.5563 21.8323 18.9083 16.5469 14.6210 13.0352
37
38
39
40
41
42
43
30.7995
31.4847
32.1630
32.8347
33.4997
34.1581
34.8100
28.2371
28.8051
29.3646
29.9158
30.4590
30.9941
31.5212
25.9695
26.4406
26.9026
27.3555
27.7995
28.2348
28.6616
23.9573
24.3486
24.7303
25.1028
25.4661
25.8206
26.1664
22.1672
22.4925
22.8082
23.1148
23.4124
23.7014
23.9819
19.1426
19.3679
19.5845
19.7928
19.9931
20.1856
20.3708
16.7113
16.8679
l'7.0170
17.1591
17.2944
17.4232
17.5459
14.7368
14.8460
149491
15.0463
15.1380
15.2245
15.3062
13.1170
13.1935
13.2649
13.3317
13.3941
13.4524
13.5070
44 35.4555 32.0406 29.0800 26.5038 24.2543 20.5488 17.6628 15.3832 13.5579
45 36.0945 32.5523 29.4902 26.8330 24.5187 20.7200 17.7741 15.4558 13.6055
46 36.7272 33.0565 29.8923 27.1542 24.7754 20.8847 17.8801 15.5244 13.6500
47
48
49
50
37.3537
37.9740
38.5881
39.1961
33.5532
34.0426
34.5247
34.9997
30.2866
30.6731
31.0521
31.4236
27.4675
27.7732
28.0714
28.3623
25.0247
25.26G7
25.5017
25.7298
361
21.0429
21.1931
21.3415
21.4822
17.9810
180772
18.1687
18.2559
15.5890
156500
15.7076
15.7619
13.6916
13.7305
13.7668
13.8007
TABLE VII.-COMMISSIONERS'
1941
STAl'<DARD ORDINARY MORTALITY
TABLE
Number Number
Number Number
Number Number
dying
dying Age living
dying Age living
living
Age
x
Ix
x
dX
Ix
x
dX
Ix
dx
--
I---
0 1023102 23102 35
1 1000000 5770 36
994230 4116 37
2
990114 3347 38
3
986767 2950 I 39
4
906554
902393
898007
893382
888504
4161
4386
4625
4878
5162
70
71
72
73
74
454548
427593
400112
372240
344 136
26955
27481
27872
28104
28154
28009
27651
27071
26262
i
983817
981 102
978541
976 124
973869
2715
2561
2417
2255
2065 ,
40
41
42
43
44
883342
877883
872098
865967
859464
5459
5785
6 131
6503
6910
75
76
77
78
79
315982
287973
260322
233251
206989
252241
10
11
12
13
14
971 804
\)69890
968038
966179
\)64266
1914 45
1852 . 46
1859 47
1913 ; 48
1996 49
852554
845214
837413
829114
820292
7340
7801
8299
8822
9392
80
81
82
83
84
1817651
157799
135297
114440
95378
23966
22502
20857
19062
1
17 157
15
1&
17
18
19
962270
960201
958098
955942
953743
2069 I 50
2103 51
2156 52
2199 : 53
2260 54
810900
800910
790282
778981
7eG 961
9990
10628
11 301
12020
12770
85
86
87
88
89
78221
63036
49838
38593
29215
15185
13 198
11 245
9378
7638
20
21
22
23
24
951 483
949 171
946789
944337
941806
2312
2382
2452
2531
? 609
55
56
57
58
59
754 191
740631
726241
710990
694843
13560 90
14390 91
15251 ,92
16 147 i 93
17072 94
21577
15514
10833
7327
4787
6063
4681
3506
2540
1776
939197
25
26 i 936492
933692
27
28
930788
927763
29
2705
2800
2904
3025
3154
60
61
62
63
64
677 771
659749
640761
620782
599824
18022 I 95
18988 96
19979 : 97
20958 . 98
21942 . 99
3011
1818
1005
454
125
1193
813
551
329
125
924609
921317
917880
914282
910515
3292
3437
3598
3767
3961
65
66
67
68
69
577 882
554975
531 133
506403
480850
22907
23842
24730
2555:3
26302
5
6
7
8
9
I
I
!
30
31
32
33
34
362
i
I
I
-
TABLE
angle sin
cot
VIII. -TRIGONOMETRIC
cos
angle sin
FUNCTIONS
cot
cos
-- -- - - - - - - -9° ---- - - ---00' .1564 .1584 6.3138 .9877 81° 00'
10 .1593 .1614 6.1970 .9872 -w
---w 20
.1622 .1644 6.0844 .9868
40
tan
1.0000 90' 00'
0°00' .0000 .0000 10 .0029 .0029 343.77 1.0000
40
20 .0058 .0058 171.89 1.0000
3a
30 ,0087 .0087 114.59 1.0000
20
40 .0116 .OU6 85.940 .9999
10
50 ,0145 .0145 68.750 .9999
1°00' .0175 .0175 57.290 .9998 89° 00'
10 .0204 .0204 49.104 .9998
20 ,0233 ,0233 42,964 ,9997
40
30
30 .0262 ,0262 38,188 .9997
40 .0291 .0291 34.368 .9996
20
31.242
.0320
.0320
.9995
50
10
2°00' .0349 .0349 28.636 .9994 88° 00'
10 .0378 .0378 26.432 .9993
40
20 .0407 .04(}7 24.542 .9992
30 ,0436 .0437 22.904 .9990
30
40 .0465 .0466 21.470 ,9989
20
,0494
,0495
20.206
.9988
10
50
3° 00' ,0523 ,0524 19,081 .9986 87" 00'
10 .0552 .0553 18.075 .9985 --go
40
20 .0581 .0582 17,169 .9983
30
30 .0610 .0612 16.350 ,9981
20
40 .0640 .0641 15.605 ,9980
14.924
.9978
50 .0669 .0670
10
~ooo' .0698 ,0699 14,301 .9976 86° 00'
10 .0727 .0729 13.727 .9974
40
20 .0756 .0758 13.197 ,9971
30
30 .0785 .0787 12.706 .9969
20
40 .0814 .0816 12.251 ,9967
10
50 .0843 .0846 11.826 .9964
lio 00' .0872 .0875 11.430 .9962 85· 00'
10 .0901 .0904 11.059 .9959
40
20 .0929 .0934 10.712 .9957
30 .0958 .0963 10.385 .9954
30
40 .0987 .0992 10.078 .9951
20
10
50 .1016 .1022 9.7882 .9948
6° 00' .1045 .1051 9.5144 .9945 84'00'
10 .1074 .1080 9.2553 .9942
40
20 .U03 .1UO 9.0098 .9939
30
30 .1132 .1139 8.7769 .9936
20
40 .1161 .1169 8.5555 .9932
10
50 .1190 .1198 8.3450 .9929
7°00' .1219 .1228 8.1443 .9925 83° 00'
10 .1248 .1257 7.9530 .9922
40
20 .1276 .1287 7.7704 .9918
30
30 .1305 .1317 7.5958 .9914
20
40 .1334 .1346 7.4287 .9911
10
50 .1363 .1376 7.2687 .9907
8° 00' .1392 .1405 7.1154 .9903 82° 00'
10 .1421 .1435 6.9682 .9899 --go
20 .1449 .1465 6.8269 .9894
40
30
30 .1478 .1495 6.6912 .9890
20
40 .1507 .1524 6.5606 .9886
10
50 .1536 .1554 6.4348 .9881
9°00' .1564 .1584 6.3138 .9877 81°00'
---w
---w
-w
-w
-w
-w
tan
30 .1650 .1673 5.9758
40 .1679 .1703 5.8708
50 .1708 .1733 5.7694
100 00' .1736 .1763 5,6n3
10 .1765 .1793 5.5764
20 .1794 .1823 5.4845
30 .1822 .1853 5.3955
40 .1851 .1883 5.3093
50 .1880 .1914 5.2257
11°00' .1908 .1944 5.1446
10 .1937 .1974 5.0658
20 .1965 .2004 4.9894
30 .1994 .2035 4.9152
40 ,2022 .2065 4,8430
50 .2051 ,2095 4.7729
12°00' .2079 .2126 4.7046
10 ,2108 .2156 4,6382
20 .2136 .2186 4.5736
30 ,2164 .2217 4,5107
40 .2193 ,2247 4.4494
50 .2221 .2278 4.3897
13° 00' .2250 .2309 4,3315
10 .2278 .2339 4.2747
20 .2306 .2370 4,2193
30 .2334 ,2401 4.1653
40 .2363 .2432 4.1126
50 .2391 .2462 4.0611
14°00' .2419 .2493 4.0108
10 .2447 .2524 3.9617
20 .2476 .2555 3.9136
30 .2504 .2586 3.8667
40 .2532 .2617 3.8208
50 .2560 .2648 3.7760
15° 00' .2588 .2679 3.7321
10 .2616 .2711 3.6891
20 .2644 .2742 3.6470
30 .2672 .2773 3.6059
40 .2700 .2805 3.5656
50 .2728 .2836 3.5261
lSo 00' .2756 .2867 3.4874
10 .2784 .2899 3.4495
20 .2812 .2931 3.4124
30 .2840 .2962 3.3759
40 .2868 .2994 3.3402
50 .2896 .3026 3.3052
170 00' .2924 .3057 3.2709
10 .2952 .3089 3.2371
20 .2979 .3121 3.2041
30 .3007 .3153 3.1716
40 .3035 .3185 3.1397
50 .3062 .3217 3.1084
18° 00' .3090 .3249 3.0777
30
.9863
20
.9858
10
.9853
.9848 SOoOO
.9843
40
.9838
,9833
30
.9827
20
.9822
10
.9816 79° 00'
.9811
.9805
40
.9799
30
20
.9793
.9787
10
.9781 78° 00'
,9775
.9769
40
.9763
30
.9757
20
.9750
10
.9744 77°00'
.9737
.9730
40
30
.9724
.9717
20
.9710
10
.9703 76° 00'
.9696
.9689
40
.9681
30
.9674
20
.9667
10
.9659 7&·00'
.9652
.9644
40
.9636
30
20
.9628
.9621
10
.9613 74° 00'
.9605
40
.9596
.9588
30
20
.9580
10
.9572
.9563 73· 00'
.9555
.9546
40
.9537
30
.9528
20
.9520
10
.9511 71° 00'
-w
-so
----w
-w
-w
-w
-w
----w
- --- --- - - - - -- -- -cos ----cot tan sin angle
cos cot tan sin angle
363
TABLE
angle sin
VIII. -
tan
cot
TRIGONOMETRIC FUNCTIONS
cos
- -- -- ---- 72° 00'
18° 00' .3090 .3249 3.0777 .9511 -10 .3118 .3281 3.0475 .9502 -w
20 .3145 .3314 3.0178 .9492
40
angle sin 1 tan
(Continued)
cot
,
cos
I
27°00 ' .4540 .5095 1.9626 .891063° 00'
10 .4566 1.5132 1.941'6 .8897
20 .4592 .5169 1.9a47 .8884
40
30 .3173 .3346 2.9887 .9483
30
30 .4617 .5206 1.9210 .8870
30
40 .3201 .3378 2.9600 .9474
20
.5243
1.9074
.4643
.8857
40
20
50 .3228 .3411 2.9319 .9465
10
50 .4669 .5280 1.8940 .8843
10
0
19 00' .3256 .3443 2.9042 .9455 71°00' 28° 00' .4695 .5317 1.8807 .8829 62° 00'
10 .3283 .3476 2.8770 .9446
10 .4720 .5354 1.8676 .8816
20 .3311 .3508 2.8502 .9436
40
20 .4746 .5392 1.8546 .8802
40
30 .3338 .3541 2.8239 .9426
30
30 .4772 .5430 1.8418 .8788
30
40 .3365 .3574 2.7980 .9417
20
.5467
40 .4797
1.8291 .8774
20
50 .3393 .3607 2.7725 .9407
10
10
50 .4823 .5505 1.8165 .8760
20" 00' 3420 .3640 2.7475 .9397 70° 00' 29°00' .4848 .5543 1.8040 .8746 61°00'
10 .3448 .3673 2.7228 .9387
10 .4874 .5581 1.7917 .8732
20 .3475 .3706 2.6985 .9377
40
20 .4899 .5619 1.7796 .8718
40
30 .3502 .3739 2.6746 .9367
30
30 .4924 .5658 1.7675 .8704
30
40 .3529 .3772 2.6511 .9356
20
40 .4950 .5696 1.7556 .8689
20
50 .3557 .3805 2.6279 .9346
10
10
50 .4975 .5735 1.7437 .8675
21°00' .3584 .3839 2.6051 .9336 69° 00' 30°00' .5000 .5774 1.7321 .8660 60° 00'
10 .3611 .3872 2.5826 .9325
10 .5025 .5812 1.7205 .8646
20 .3638 .3906 2.5605 .9315
40
20 .5050 .5851 1.7090 .8631
40
30 .3665 .3939 2.5386 .9304
30
30 .5075 .5890 1.6977 .8616
30
40 .3692 .3973 2.5172 .9293
.5930
1.6864 .8601
20
40 .5100
20
50 .3719 .4006 2.4960 .9283
10
10
50 .5125 .5969 1.6753 .8587
22°00' .3746 .4040 2.4751 .9272 6S000' 31°00' .5150 .6009 1.6643 .8572 59° 00'
10 .3773 .4074 2.4545 .9261
10 .5175 .6048 1.6534 .8557 --W
20 .3800 .4108 2.4342 .9250
40
20 .5200 .6088 1.6426 .8542
40
30 .3827 .4142 2.4142 .9239
30
30
30 .5225 .6128 1.6319 .8526
40 .3854 .4176 2.3945 .9228
20
40 .5250 .6168 1.6212 .8511
20
50 .3881 .4210 2.3750 .9216
10
50 .5275 .6208 1.6107 .8496
10
23° 00' .3907 .4245 2.3559 .9205 670 00' 32°00' .5299 .6249 1.6003 .8480 58° 00'
10 .3934 .4279 2.3369 .9194
10 .5324 .6289 1.5900 .8465
20 .3961 .4314 2.3183 .9182
40
40
20 .5348 .6330 1.5798 .8450
30 .3987 .4348 2.2998 .9171
30
30
30 .5373 .6371 1.5697 .8434
40 .4014 .4383 2.2817 .9159
20
20
40 .5398 .6412 1.5597 .8418
50 .4041 .4417 2.2637 .9147
10
10
50 .5422 .6453 1.5497 .8403
24°00' .4067 .4452 2.2460 .9135 66°00' 33° 00' .5446 .6494 1.5399 .8387 57° 00'
10 .4094 .4487 2.2286 .9124
10 .5471 .6536 1.5301 .8371
20 .4120 .4522 2.2113 .9112
40
20 .5495 .6577 1.5204 .8355
40
30 .4147 .4557 2.1943 .9100
30
30
30 .5519 .6619 1.5108 .8339
40 .4173 .4592 2.1775 .9088
20
20
40 .5544 .6661 1.5013 .8323
50 .4200 .4628 2.1609 .9075
10
10
50 .5568 .6703 1.4919 .8307
25°00' .4226 .4663 2.1445 .9063 65" 00' 34° 00 .5592 .6745 1.4826 .8290 56° 00'
10 .4253 .4699 2.1283 .9051
10 .5616 .6787 1.4733 .8274
20 .4279 .4734 2.1123 .9038
40
40
20 .5040 .6830 1.4641 .8258
30 .4305 .4770 2.0965 .9026
30
30 .5664 .6873 1.4550 .8241
30
40 .4331 .4806 2.0809 .9013
20
20
40 .5688 .6916 1.4460 .8225
50 .4358 .4841 2.0655 .9001
10
10
50 .5712 .6959 1.4370 .8208
26°00' .4384 .4877 2.0503 .8988 64" 00' 35°00' .5736 .7002 1.4281 .8192 55° 00'
10 .4410 .4913 2.0353 .8975
10 .5760 .7046 1.4193 .8175
20 .4436 .4950 2.0204 .8962
40
40
20 .5783 .7089 1.4106 .8158
30 .4462 .4986 2.0057 .8949
30
30 .5807 .7133 1.4019 .8141
30
40 .4488 .50?,} 1.9912 .8936
20
40 .5831 .7177 1.3934 .8124
20
50 .4514 .5059 1.9768 .8923
10
50 .5854 .7221 1.3848 .8107
10
27° 00' .4540 .5095 1.9626 .8910 63°00' 36'00' .5878 .7265 1.3764 .8090 54° 00'
--w
-w
-w
-w
-w
-w
-w
-w
----so
-w
----so
----so
-w
--:so
----so
-w
-- ----- - - -- - -- ---- - cos cot tan I sin angle -- cos cot tan sin angle
364
TABLE
VIII -
TRIGONOMETRIC FUNCTIONS
angle sin
tan
cot
(Continued)
cos
Il36° 00' .5878 .7265 1.3764 .8090 64°00'
10 .5901 .7310 1.3680 .8073 ~
20 .5925 .7355 1.359-7 .8056
40
30 .5948 .7400 1.3514 .8039
30
40 .5972 .7445 1.3432 .8021
20
10
50 .5995 .7490 1.3351 .8004
37° 00' .6018 .7536 1.3270 .7986 63° 00'
10 .6041 .7581 1.311l0 .7969
20 .6065 .7627 1.3111 .7951
40
30 .6088 .7673 1.3032 .7934
30
40 .6111 .7720 1.2954 .7916
20
50 .6134 .7766 1.2876 .7898
10
38°00' .6157 .7813 1.2799 .7880 52°00'
10 .6180 .7860 1.2723 .7862 ~
20 .6202 .7907 1.2647 .7844
40
30 .6225 .7954 1.2572 .7826
30
40 .6248 .8002 1.2497 .7808
20
50 .6271 .8050 1.2423 .7790
10
39'00' .6293 .8098 1.2349 .7771 61"00'
10 .6316 .8146 1.2276 .7753
20 .6338 .8195 1.2203 .7735
40
30 .6361 .8243 1.2131 .7716
30
40 .6383 .8292 1.2059 .7698
20
50 .6406 .8342 1.1988 .7679
10
iOoOO' .6428 .8391 1.:i918 .7660 60°00'
10 .6450 .8441 1.1847 .7642 ~
40
20 .6472 .8491 1.1778 .7623
30
30 .6494 .8541 1.1708 .7604
20
40 .6517 .8591 1.1640 .7585
10
50 .6539 .8642 1.1571 .7566
41°00' .6561 .8693 1.1504 .7547 49° 00'
10 .6583 .8744 1.1436 .7528
20 .6604 .8796 1.1369 .7509
40
30 .6626 .8847 1.1303 .7490
30
20
40 .6648 .8899 1.1237 .7470
50 .6670 .8952 1.1171 .7451
10
42,°00' .6691 .9004 1.1106 .7431 48° 00'
10 .6713 .9057 1.1041 .7412
40
20 .6734 .9110 1.0977 .7392
30
30 .6756 .9163 1.0913 .7373
20
40 .6777 .9217 1.0850 .7353
50 .6799 .9271 1.0786 .7333
10
43° 00' .6820 .9325 1.0724 .7314 47" 00'
10 .6841 .9380 1.0661 .7294
20 .6862 .9435 1.0599 .7274
40
30 .6884 .9490 1.0538 .7254
30
20
40 .6905 .9545 1.0477 .7234
50 .6926 .9601 1.0416 .7214
10
44°00 ' .6947 .9657 1.0355 .7193 46° 00'
10 .6967 .9713 1.0295 .7173
20 .6988 .9770 1.0235 .7153
40
30 .7009 .9827 1.0176 .7133
30
40 .7030 .9884 1.0117 .7112
20
50 .7050 .9942 1.0058 .7092
10
45°00 ' .7071 1.0000 1.0000 .7071 45°00'
-w
-w
-w
---w-
---w-
-w
- - --------cos cot tan sin angle
365
/
Index
-
(
i
(Numbers, refer to pages)
Abscissa, 25.
Absolute convergence, 348, 349.
Absolute inequality, 145.
Absolute value, 4; of complex number, 194.
Addition; associative law for, 2;
~ommutative law for, 2; of complex numbers, 98, 190, geometric,
192; of fractions, 66; of polynomials, 15; of radicals, 93; of
signed numbers, 5; of similar
terms, 7.
Algebra; fundamental theorem of,
209; number cystem of, 189.
Algebraic expression, 6.
Algebraic number, 189.
Algebraic solution, 236.
Alternating series, 346.
Amount; of annuity, 268; compound,263.
Amplitude of complex number, 194.
Annuity, 267; amount of, 268; present value of, 269.
Antilogarithm, 242; finding the, 246.
Approximate numbers, 249.
Approximations, successive, to irrational root, 221.
Argument of complex number, 194.
Arithmetic means, 175.
Arithmetic progression, 173.
Associative law; for addition, 2; for
multiplication, 3.
Assumptions, fundamental, 2, 3.
Axis; of imaginary numbers, 192;
of real numbers, 24, 192; x, 25;
Binomial coefficients and combina·
tions, 280.
Binomial formula, 164, 165; generai
term of, 167, 168.
Binomial series, 170.
Binomial theorem, 168.
Bounds for roots, 207.
Braces, 11.
Brackets, 11.
Briggsian logarithms, 258.
Cardan's formulas for roots of
cubic, 232.
Change of base of logarithms, 259.
Character of roots of quadratic
equation, 114.
Characteristic of logarithm, 241.
Clearing of fractions, 77.
Coefficient, 7
Coefficients; binomial, 280; in termF
of roots, 246.
Cologarithm, 253.
Combination, 279.
Combinations, number of, 279, 280,
total, 280.
Commissioners' Standard Mortality Table, 286.
Common difference of arithmetic
progression, 173.
Common logarithms, 258.
Common ratio of geometric progression, 176.
Commutative law; for addition, 2;
for multiplication, 2.
Comparison series, 340
Comparison test, 338.
y,25.
Completing the square, 102.
Base, 7; logarithmic, 239, change of, Complex fraction, 72.
259.
Complex number, 98, 188; absolute
value of, 194; amplitude of. 194:
Bases, logarithmic, other than
argument of, 194; geometric rep"10," 257.
resentation of, 192; imaginary
Binomial, 14.
367
3611
INDEX
part of 98, 188; mixed, 98, 188;
modulus of, 194; powers of, 196;
pure, 98, 188; real part of, 98,
188; roots of, 197; trigonometric
or polar form of, 194.
Complex numbers; addition of, 98,
99, 190, geometric, 192; conjugate, 98; division of, 99, 190, in
trigonometric form, 195; multiplication of, 99, 190, in trigonometric form, 195; subtraction of,
98, 99, 190, geometric, 192.
Components of complex number,
98.
Compound amount, 263.
Computation with logarithms, 249.
Conditional convergence, 34S, 349.
Conditional equation, 30.
Conditional inequality, 145.
Conjugate complex numbers, 98.
Conjugate imaginary numbers as
roots of equation, 211.
Conjugate surds as roots of equation, 212, 213.
Consistent equations, 319, 320, 322;
condition for, 322.
Constant, 20; of proportionality or
variation, 154.
Continuation symbol, 25S.
Continuous function, 206.
Convergence, 336; absolute, 348,
3·19; comparison test for, 338;
conditional, 348, 349; interval
of, 351; necessary condition for,
337; ratio test for, 341, extended,
350.
Convergent series, 336.
Conversion period, 263.
Coordinate paper, 26.
Coordinates, rectangular, 25.
Cube, perfect, 87.
Cube root, 81.
Cubic; general, 230; reduced, 230;
resolvent for quartic, 234.
C\lbic surd, 87, 212.
Decimal, repeating, 181.
Decimal point, standard position of,
241.
Degree; of equation or expression,
126; of integral rational equation
or function, 53, 201; of poly-
nomial, 19, 53; of term, 5:1, 126.
De Moivre's theorem, HJ6.
Denominator; lowest common, 66;
rationalizing the, 89.
Dependent equations, 42, 31S, 319.
Dependent events, 292.
Dependent variable, 21.
Depreciation fund, 273.
Depressed equation, 219.
Descartes' rule of signs, 214.
Determinant, 301, 302, 303, 304,
30S; expansion of, 309, by minors,
312.
Determinants; properties of, 309;
solution of linear equations by,
316.
Development of determinant, 313.
Diagonal, principal, of determinant,
302,308.
Difference of arithmetic progression,173.
Discriminant of quadratic equation, 114.
Distributive law, 3.
Divergence, 336; comparison test
for, 338; ratio test for, 341, extended, 350.
Divergent series, 336.
Dividend, 19.
Division, 1; of complex numbers, 99,
190, in trigonometric form, 195;
of fractions, 71; of mixed expressions, 71; of polynomials, 18; of
radicals, 96; of signed numbers,
5; synthetic, 203; of terms, 10.
Divisor, 19.
Double root, 114, 210.
e, base of natural logarithms, 25S.
Elements of determinant, 302;
negative pairs of, 308; positive
pairs of, 308.
Equation, 30; conditional, 30; cubic,
230; degree of, 126; depressed,
219; exponential, 256; fractional,
77; homogeneous, 135; identical,
30; integral rational, 201; linear,
32; quadratic (see Quadratic
equation); in quadratic form,
109; quartic, 233; radical, 110;
with given roots, 118.
Equations; algebraic solution of,
j
INDEX
(
236; consistent, 319, 320, 322;
dependent, 42, 318, 319; equivalent, 31; homogeneous, 322; homogeneous linear, 322; inconsistent, 41, 318, 319; linear, 40, 42,
320, solved by determinants, 316;
non-homogeneous, 322; simultaneous, 40; symmetric, 131;
systems of quadratic, 126.
Equivalent equations, 31.
Expansion of determinant, 309, 312,
313.
Expectation, 287.
Expected number, 287.
Exponent, 7; fractional, 82; negative, 84; zero, 83.
Exponential equation, 256.
Exponents, laws of, 80.
Expression, 6; algebraic 6; degree
of, 126; integral rational, 53;
mixed, 67; prime, 53; radical,
simplest form of, 92.
Extended ratio test, 350.
Extraneous roots, 76.
Extremes of proportion, 154.
Factor, 7, 53; highest common, 62;
prime 53.
Factor theorem, 202.
Factor types; elementary 55; more
complicated, 57.
Factorial, 167.
Factoring; by grouping, 57; by
solving quadratic, 120; suggestions for, 60.
Ferrari's method of solving quartic,
235.
Finite series, 334.
Fourth proportional, 154.
Fraction; complex, 72; improper,
68; in lowest terms, 64; proper,
326; rational, 189; signs of, 64.
Fractional equations, 77.
Fractional exponents, 82.
Fractions, 64; addition of, 66; clearing of, 77; division of, 71; fundamental principles of, 64; lowest
common denominator of, 66;
multiplication of, 70; partial, 326;
powers of, 70; subtraction of, 66.
Frequency, relative, 285.
Fulcrum, 38.
369
Function, 21; continuous, 206;
graph of, 26; integral rational,
201; quadratic (see Quadratic
function); zero of, 201.
Fundamental assumption for series,
338.
Fundamental assumptions, 2, 3.
Fundamental operations, 1.
Fundamental principle; for fractions, 64; for permutations, 274.
Fundamental properties of inequalities, 146.
Fundamental theorem of algebra,
209.
General number, 1.
General quadratic function, 149.
General term; of arithmetic progression, 173; of binomial formula, 167, 168; of geometric
progression, 176.
Geometric addition and subtraction of complex numbers, 192.
Geometric means, 178.
Geometric progression, 176; infinite, 180.
Geometric representation of complex numbers, 192.
Geometric series, 180, 340.
Graph, 26; of factOled polynomial,
213.
Graph paper, 26.
Graphic representation; of linear
equations, 41; of quadratic equations, 136; of quadratic function,
122.
Grouping; factoring by, 57; symbols
of 11.
Harmonic means, 183.
Harmonic progression, 183.
Harmonic series, 337.
Highest common factor, 62.
Homogeneous equation, 135.
Homogeneous equations, 322.
Horner's method, 225.
Hyperbolic logarithms, 258.
i, imaginary unit, 98, 188.
Identity, 30.
Imaginary cube roots of unity, 231.
Imaeinary number, 97, 98, 188;
370
INDEX
mixed, 98, 188; pure, 98, 188.
Imaginary numbers, axis of, 192.
Imaginary part of complex number,
98, 188.
Imaginary roots, 211.
Imaginary unit, 98, 188.
Improper fraction, 68.
Inconsistent equations, 41, 318, 319.
Independent events, 291.
Independent variable, 21.
Index of radical, 81, reducing, 88.
Induction, mathematical, 160.
Inequality, 145; absolute, 145; conditional, 145; sen~e or order of,
145.
Infinite geometric progression or
series, 179, 180.
Infinite series, 334.
Integer, 189.
Integral rational equation, 201.
Integral rational expression, 53.
Integral rational function, 201.
Integral root, 218.
Interest, 263.
Interpolation, 244.
Interval of convergence, 351.
Inverse variation, 155.
Irrational number, 87, 189.
Irrational roots, 221.
Joint variation, 155.
Latin square, 308.
Law; associative, for addition, 2,
for multiplication, 3; commutative, for addition, 2, for multiplication, 2; distributive, 3.
Laws; of exponents, 80; of logarithms, 247; of radicals, 87.
Lever, 38.
Light-year, 86.
Like terms, 7.
Limit, 335; of infinite geometric
progression, 180.
Linear equation, 32.
Linear equations, 40, 42, 320;
graphic solution of, 41; homogeneous, 322; solved by determinants, 316.
Literal number, 1.
Logarithm, 239; characteristic of,
241; common or Briggsian, 258;
mantissa of, 240, 243; natural,
hyperbolic, or Napierian, 258;
of a power, 248; of a product,
247; of a quotient, 248; of a root,
248.
Lower bounds for roots, 207.
Lowest common denominator, 66.
Lowest common multiple, 62.
Lowest terms, fraction in, 64.
Magic square, 307.
Mantissa, 240, 243.
Mathematical induction, 160.
Maximum value of quadratic func- ,
tion, 123.
Mean proportional, 154.
Means; arithmetic, 175; geometric,
178; harmonic, 183; of a proportion, 154.
Members of equation, 30.
Minimum value of quadratic function, 123.
Minor of determinant, 312.
Mixed expression, 67.
Mixed expressions, division of, 71.
Mixed imaginary number, 98, 188.
Modulus of complex number, 194.
Moment of force, 38.
Monomial, 14.
Mortality table, 286; Commis..;;
sioners' Standard, 286.
Multinomial, 14.
Multiple, lowest common, 62.
Multiple root, 210.
Multiplication; associative law for,
3; commutative law for, 2; of
complex numbers, 99, 190, in
trigonometric form, 195; of frac_/' tions, 70; of polynomials, 16; of
radicals, 94; of signed numbers,
5; of terms, 8.
Multiplicity of a root, 210.
Mutually exclusive events, 290.
N apierian logarithms, 258.
Natural logarithms, 258.
Negative exponent, 84.
Negative number, 3.
Negative pair of elements in a
determinant, 308.
Negative-term series, 345.
Nominal rate of interest, 263.
'"
/
INDEX
'(
Non-homogeneous equations, 322.
Non-trivial solution, 322.
Number; algebraic, 189; complex,
98, 188; expected, 287; general,
1; imaginary, 97, 98, 188; irrational, 87, 189; literal, 1; negative, 3; positive, 3; rational, 87,
189, 217; signed, 3; transcendental, 189.
Number of combinations, 279;
total, 280.
Number of permutations, 276; of
things, some alike, 276.
Number of roots of an equation,
209.
Number system of algebra, 189.
Numbers; approximate, 249; conjugate complex, 98.
Operations; with complex numbers,
190; fundamental, 1; yielding
equivalent equations, 31.
Order; of inequality, 145; of radical,
81.
Ordinate, 25.
Origin, 24, 25.
p-series, 340.
I,
Parabola, 122.
'\
Parentheses, 11.
Partial fractions, 326.
Pascal's triangle, 166.
Perfect cube, 87.
Perfect power, 87.
Perfect square, 87.
Period, conversion or interest, 263.
Permutation, 275.
Plotting, 26.
Polar fOTm of complex number, 194.
Polynomial, 14, 53, 201; degree of
19, 53; graph of factored, 213;
loca tion of real zeros of, 205.
Polynomials; addition of, 15; division of, 18; multiplication of, 16;
subtraction of, 15.
Positive number, 3.
Positive pair of elements in a
determinant, 308.
Power, 7; of complex numbe:-, 196;
of fraction, 70; perfect, 87; of
term, 9.
Power series, 351.
371
Present value, 264; of annuity, 269
Prime expression, 53.
Principal, 263.
Principal diagonal of determinant,
302,308.
Principal root, 81.
Probability, 285; of dependent
events,
292; of independent
events, 291; of mutually exclusive events, 290; in repeated
trials, 295.
Product, logarithm of, 247.
Product of roots, 117,237.
Products, type, 54.
Progression; arithmetic, 173; geometric, 176, infinite, 179, 180;
harmonic, 183.
Proper fraction, 326.
Properties of determinants, 309,
of inequalities, 146.
Proportion, 154.
Proportional; fourth, 154; mean,
154; third, 154.
Proportionality, constant of, 154.
Pure imaginary number, 98, 188.
Quadratic equation, 101; character
of roots of, 114; discriminant of,
114; product of roots of, 117;
solution by completing square,
102, by factoring, 101, by formula,
103; sum of roots of, 117.
Quadratic equations, simultaneous,
126; graphic representation of,
136.
Quadratic function,
122, 149;
graphic representation of, 122;
maximum or minimum value of,
123.
Quadratic surd, 87, 212.
Quartic, 233.
Quotient, 1, 19; logarithm of, 248.
Radical, 81; reducing index of, 88.
Radical expression, simplest form
of,92.
Radicals; addition of, 93; division
of, 96; equations involving, 110;
laws of, 87; multiplication of, 94;
subtraction of, 93.
Radicand, 81.
Rate of interest, 263.
'372
INDEX
Ratio, 154; of geometric progression, 176.
Ratio test, 341, extended, 350. .';
Rational fraction, 189.
Rational number, 87, 189, 217.
,:
Rational roots, 217.
Rationalizing denominator, 89.
Real numbers, axis of, 24, 192.
Real part of complex number, 98,
188.
Real roots, suggestions for finding,
228.
Real zeros of a polynomial, location of, 205.
Rectangular coordinates, 25.
Reduced cubic, 230.
Relative frequency, 285.
Remainder, 19.
Remainder theorem, 201.
Repeated trials, probability in, 295.
Repeating decimal, 181.
Resolvent cubic, 234.
Root; cube, 81; double, 114,210; of
equation, 30; extraneous, 76;
integral, 218; irrational, 221;
logarithm of, 248; multiple, 210;
nth, 81; of number, 80; of order
or llluitiplicity n, 210; principal,
81; rational, 217; simple, 210;
square, 81; triple, 210.
Roots; bounds for, 207; coefficients
in terms of, 236; of complex number, 197; imaginary, 211; number of, 209; of quadratic equation, 104, character of, 114;
quadratic surd, 212; transformation to diminish, 224.
Rounding off, 245.
Rule of signs, Descartes', 214.
Scientific notation, 85.
Sense of an inequality, 145.
Sequence, 334.
Series, 334; absolutely convergent,
349; alternating, 346; binomial,
170; comparison, 340; comparison
test for, 338; conditionally convergent, 349; convergent, 336;
divergent, 336; finite, 334; fundamental assumption for, 338;
geometric, 180, 340; harmonic,
337; infinite, 334; interval of
convergence of, 351; negativeterm, 345; p, 3·10; power, 351;
ratio test for, 341, extended, 350;
sum of, 336.
Set of equations, 40.
Signed numbers, 3.
Significant digits, 249.
Signs; Descartes' rule of, 214; variation in, 214.
Similar terms, 7.
Simple root, 21(1.
Simplest form of radical expression,
92.
Simultaneous equations; linear, 40,
42, 316, 318, 320, 322; quadratic,
126.
Sinking fund, 272.
Solution; of equation, 30; of inequality, 148; non-trivial, 322;
trivial, 322.
Square, completing the, 102.
Square, perfect, 87.
Square root, 81.
Standard position of decimal point,
241.
Subtraction, 1; of complex numbers, 98, 190, geometric, 192; of
fractions, 66; of polynomials, 15;
of radicals, 93; of signed numbers, 5; of similar terms, 7.
Successive approximations to irrational root, 221.
Sum; of arithmetic progression,
173; of geometric progression,
177; of infinite geometric progression, 180; of roots of equation,
237; of roots of quadratic equation, 117; of series, 336.
Surd, 87, 212.
Symbol of continuation, 258.
Symbols of grouping, 11.
Symmetric equations, 131.
Synthetic division, 203.
Systems of equations (see Simultaneous equations).
Term, 7; of arithmetic progression,
173; degree of, 53, 126; of geometric progression, 176.
Terms; addition of, 7; division of,
10; like or similar, 7; multiplication of, 8; powers of, 9; subtrac.
INDEX
tion of, 7.
Third proportional, 154.
Transcendental number, 189.
Transformation to diminish roots,
224.
Trigonometric form of complex
number, 194.
Trinomial, 14.
Triple root, 210.
Trivial solution, 322.
Unknown, 30.
Upper bound for roots, 207.
Value; absolute, 4, 194, preI8Ilt,
264, of annuity, 269;
I
373
Variable, 21; dependent, 21; independent, 21.
Variation, 154; constant of, 154;
direct, 154; inverse, 155; joint,
155; in signs, 214.
Vector, 192.
Vinculum, 11.
x-axis, 25.
y-axis, 25.
Zero; division by impossible, 2;
exponent, 83; of function, 201; of
quadratic function, 122.
Answers to
Odd. numbered Exercis~s
Exercises I. A, page 6 .( ,;,
!
1.
11.
21.
31.
4.1.
51.
61.
71.
13.
0.
3.
-90.
40.
-2025.
3.
13.
23.
33.
43.
53.
63.
73.
5.
O.
-2.
8.
-12.
8.
-35.
O.
-3.
6.
15.
25.
36.
45.
55.
66.
~:;,:
-20.
-148.
10.
16.
75.
5412.
7.
17.
27.
37.
47.
57.
67.
75.
7.
Impossible
16.
9.
-20.
19.
-18.
29.
20.
39.
-17.
49.
O.
59.
-1.
69.
Impossible.
-4.
1.7.
-16.
3.3.
-60.
-2.00.
1.
Exercises I. B, page 8
1.
9.
17.
25.
22a.
-13u 3v2•
22a.
-r 2s.
I',.
3.
11.
19.
27.
5.
13.
21.
29.
llx 3 •
O.
-12:ry. :
6abe.
-49z 2.
-57t 2w S•
45abe.
-2.5z b •
7. 13mn'.
16. 56a 3.
23. O.
,.
,;;
Exercises I. C, p.ges 10-11
1. 16ax.
3. -63pq.
5. 21a 2.
9. 50X2y3.
11. 27x 4t b• 13. 30a 4b4e.
16.
21.
27.
35.
4.1.
47.
55.
49x 2y 4 Z 5.
-54u 4V 7•
27a 9.
729m 27 n 9•
441x 12 y 12Z 12. /'
as.
27a 6bge 12 .
17.
23.
29.
37.
43.
49.
67.
7. -12a7 •
462a&x1'l.y1.
19. 256abx 2 y 2.
72m 3n 4s 3t 4 •
25. XiS.
a 8b12 .
31. '31r12s20.
33. lOO,OOOx 50•
40a sb8e9.
39. -4000U I9 V I8 .
40,OOOU 19V22 .
45. -324,000x22yt9.
-7x3•
51. 4t. 53: 4x byz z.
-3m 4n 2.
59. 375x Sy 13Z 13.
Exercises I. D, pages 13-14
1.
7.
11.
15.
19.
23.
-7.
-a - 1Sb
a -
3. 7a - 2b
+ 10.
(2& + 3e + 4d).
+ 6e + 6.
- (y2 - 2yz + Z2).
(1 + a
b - c)x.
-46.
25. -390.
x2
+
9.
13.
17.
21.
375
5. ac - ad + be - bd.
17x 2 - 80xy
52x.
5p - (-6q
7r - 8s).
(a - b
3)x.
(-3y - 5)x + yZ + 7y + 4.
27. 255,724.
+
+
+
376
ANSWERS
to
ODD-NUMBERED EXERCISES
\ '
Exercises I. E, pages 15-16
1. 24a + lob. 3. ox 3
7. 10x 3 • 9. ab + 2be
+
+
2x~ - 3x + 20.
+ ae + bd + cd.
-
3a2 - 5b 2
4e 2 2X2
O.
19. 43m 2n -4x 3 - 4x 2
17x -12x 2 - ox
8.
8p2
16pq
15q2.
-4x 3 - 73x 2 - 99x
39. -m 2n - 49mn 2.
43. -28a - 24b
1I5e.
13.
17.
23.
27.
31.
35.
+
+
+
+
5. 19a~b - 15ab~ - a Sb2 • '
11. _2p2 - 20pq
19q2. !
- 7y2 - 3z 2. 15. 10x 3 - 59x 2 - 174x - 2.
13mn 2
m2
n 2.
21. 8a
2b.
22.
25. -5a 2b
21ab 2
a 3 b2.
29. ab
cd - ac - bd.
33. 11a2 11b2 - 14e2 - 3y2 - 9z 2 •
32. 37. p - 2q
T.
41. -5x 2 17xy - 22y2.
."
j" •
45. -x 3 - 98x 2
155x
57.
: i
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Exercises I. F, pages 17-18
1.
5.
'1.
9.
11.
13.
15.
1'1.
19.
21.
23.
25.
27.
29.
31.
+
+
+
3x 3
27x 2
37x - 35.
3. 18x 3 - 33x 2 41x - 60.
x·
lOx 3
23x 2
46x
40.
5x· - 18x 3y
33x 2y 2 - 28xy 3
6y4.
2
2
8a - 21b
24c 2
2ab - 38ae - 47bc.
7x· - 30x'
21x 3
37x 2 - 68x
33.
6xsy - llx'y2
14x 3y 3 - 2x 2 y' - 4xy 5
5y 6.
<~ 1
-3x 6
16x 5 - 24x'
20x 3 - 44x2
20x
15.
3
2
2x' - 15x
l3x
36x - 36.
36a 6 - 97a 5b - 17a'b 2
101a 3b 3
9a 2b' - 26ab 5 - 66',""
a2
b2
e2
2ab
2ac
2bc.
Xl - 4X3y - 2x2y2
12xy 3
9y4.
x3
6x2y
12xy2
8 y 3.
8x 6
60x4y2
150x2y4
125y 6.
36x' - 60x 3y - 47x 2y 2
60xy 3
36y'.
/
4a' - 12a 3b
21a 2b2 - 14ab 3
3b'. / / /
+
+
+ +
+
+
-t+
+
+
+
+
+
+
+
+ +
+
+
+
+
+ + + + +
+
+
+
+
+
+
+
+
+
+
+
+
,
"
Exercises I. G, page 20
The first expression given is the quotient, the second is the remainder. If the latter is zero it is, not given.
1. x 3 + 7x + 6.
3. 4x 3 - 3x~ + 2x + 5.
5. a 2 + ab + b2.
7. a 2 - ab + b2 , -2b 3•
9. X2 - 2x - 8, -36.
11. x.2 - 2x - 2, 30x - 2.
13. 5x 2 + 2xy, -6xy 3 - 7y4.
15. as - a 2b + ab 2 - b 3•
19. a 3 - a 2b + ab 2 - b 3, 2b'.
23. -x 3 + 8x 2 - 55x + 384.
27. 4X2 - 9x - 2, 02x 2 + 9x
31. 1:' + x2y2 + y', 2y ••
+
17. a 2 - b 2 •
21. a 4 - a 3b
a 2b2 - ab 3
b'.
3
2
25. x
5x
lOx
12, 14x - 6.
11.
29. X2
y2.
+
+
+ +
+
+
\
.
.
ANSWERS TO ODD-NUMBERED EXERCISES
371
Exercises I. H, pages 23-24
1. 3,5,7,1, -1,4, 2a + 3, -2a + 3, 2a 2 + 3, 5 - 2y.
3. 2,0,6, -7,3 - X2, 3 + X3.
5. 13, 110,3,4, -504, -30a.
7. 9a 2 + 6a - 8.
11.
X4 -
6X 3
+ 10x2 -
3x.
+ 3x + 7, 3x 2 + 2xy 6x - 3y + 7, 3x 2 - 2xy -
+ + + 7,
+
+ 7.
13. 3y2 - 2yx - 4x 2 - 6y
4y2
6x
3y
4y2 4y2
6x - 3y
19. V = f(a) = a 3, S = f(a) = 6a 2.
21. g = f(m) = m/18.
23. V = f(r,h) = 7rr2h, T = f(r,h) = 27rrh
27rr2.
25. L = f(r,h) = 7rrVr2 h2, T = f(r,h) = 7rr 2 7rrVr2 h2.
3x 2
+ 2xy -
+
+
+
+
Exercises I. I, pages 27-29
43. 4x(5 - x)(8 - x), 2.
41. 2x(25 - x), 12.5.
(
Exercises II. A, pages 33-34
1. -6.
3. 5.
5. 2.
7. _2J-.
,. 9. 9/(3a - 5b).
11. O.
13. Iden.
15. 3a/2.
17. Impossible.
19. (A - P)/Pn.
21. (2S - na)/n.
23. 2(c -p)/(2gh + V2).
Exercises II. 8, pages 35-40
1.
9.
15.
19.
23.
31.
35.
71,54.
3. 17~ X 30~ in.
6. 25 ft.
7. 12.
372 at $1.25, 219 at $1.75.
11. 200.
13. 225 lb.
17. Ii mi.
$7500 at 3~%, $8500 at 4%.
27~ mi./hr., 32~ mi./hr.
21. 81 ft./sec.
32-tT min.
26. 2l-fr min.
29. 4j ft. from 65-lb. child.
1025 lb.
33. 6 in. from one end.
(a) 3 ft. j in. from 100-lb. wt., (b) 3 ft. 3i in. from 100-lb. wt.
Exercises II. (, pages 43-45
1. (13,5).
9. (i,¥).
3. (6~,-lj).
11. Inconsistent.
17. (3a,-2a).
5a + 3b 3a - 2b)
19. ( ~,
95
.
25. (4,-11,6).
33. (3,-2,6,7).
37. riCa
b
c
i( -a b
27. (-1,2,3).
29. (13,-9,20).
35. (3,-1,8,5).
- d), i(a
b - c
d), i(a - b
c
d)].
+ +
+ + +
+
c-th
5. (7~,-2).
13. Dependent.
+
Exercises II. D, pages 48-52
1. 120 lb (75c.), 30 lb. (90c.).
3. Lemons 60 c./doz., oranges 75 c./doz.
7.
15. (12,13).
21. (-}J).
23. (j,2).
31. (2,-4,5).
+ c + d),
378
ANSWERS TO ODD-NUMBERED EXERCISES
5.
11.
15.
21.
25.
29.
175 lb. tin, 0 lb. lead.
7. 24 X 13 ft.
9. 120 X 185 ft.
41°,82°,57°. 13. $6500 at 4%, $3300 at 3}%, $1350 at 3%.
10 yr., 14 yr.
37. 13 in., 8 in., 6 in.
19. 1 hr.
A 30 days, B 20 days, C 60 days.
23. 673.
325 mi./hr., 25 mi./hr.
27. 1056 ft., 36 mi./hr.
1500 mi., 132 mi.
31. 12 mi., 48 mi./hr.
;
I
I
.n :
•i
Exercises III. A, page 55
1.
9.
11.
15.
19.
23.
27.
29.
35.
41.
45.
47.
49.
m2 - n 2 •
3. x 2 _ 9y2,
5. 4a 2 - 81.
36x 2 y 2 - U 3V 2 •
(a
b)2 - e2.
13. 9(x
y)2 - 25.
17. (a - 2b)2 - ge2.
16 - 9(m
n)2.
21. u 2
2uv
v2 •
_,
<!} :'11' j'_,
25(p - q)2 - 36(r + S)2
2
2
25.
25p4q2
40p2qrs2
l6r2s4.
4a
28ab
49b .
16a 2b6 - 88ab 3d2e
121d4e2.
X2
18x
77.
31. x 2 - 18x
45.
33. x 2
5bx
6b 2.
15x 2
29x - 14. 31. 63x 2 - 82x
24. 39. 15x 2
4Ix
14.
15x 2
txy - 2y2. 43. X2
y2
Z2 - 2xy
2xz - 2yz.
9x 2
4y2
Z2
12xy
6xz
4yz.
a2
b2 e2 d 2
2ab
2ae
2ad
2be
2bd
2ed.
r2
48 2
9t 2
16u2
25v 2 - 4rs
6rt - 8ru
IOrv
16ru - 20sv - 24tu
30tv - 40uv.
- 12st
. ,,', .i>
+
+
+ +
+
+
+
+
+
+
+
+
+
+
+
+ +
+ + +
+ +
+ + + + + + +
+ + +
+
+
+
+
+
+
+
+
+
+
+
57.
+
+
3x(2yz
3ab).
3. 17x 2y 2(3x 3 - 7y 3).
5. (7
3x)(a - b).
(p
3q)(5x - 8y).
9. (6p
7q)(6p - 7q).
3(3x
2y)(3x - 2y).
13. (m
n
5)(m
n - 5), .IS
(9x - y
4u
3v)(9x - y - 4u - 3v).
.,it'
2a(7a
6)(7a - 6).
19. (x - 2y)2.
(3m
8n)2.
23. (x - 6)(x - 4). 25. (a
18b)(a - 2b).
(x
3) (x - 3)(x
2)(x - 2). 29. (3x
2)(2x - 5).
3(p
3q)(2p - 9q).
33. (5x
y)(x
5y).
(ax
b)(bx - e).
37. (x
2)(x 2
1).
(a
3b)(x
l)(x - 1).
41. (3x - 2a)(2y
3a).
(2x
y)(4x 2 - 2xy
y2).
(7a - 5b)(49a 2
35ab
25b 2).
"
(a
b - c)[(a
b)2
(a
b)c
c2].
(x
y)(x 2 - xy
y2
3).
(x
2y)(x 4 - 2x 3y
4x2y2 - 8xy 3
l6y.).
(x 4 y4)
y2) (x
y) (x - y).
~
(x
y)(x 6 - x 5y
x4y2 - X 3y 3
X 3y 4 xyi
y6).
(x 2
3y2)(X4 - 3X 2y2 9 y 4).
59. (a - b
c)(a - b - c).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+
+
+
+
+
+
+ + + +
+ +
+
+
ex2 +
+
+
+
+
'"J
q_
+
,
,i
+
Exercises III. 8, pages 61-62
1.
7.
11.
15.
17.
21.
27.
31.
35.
39.
43.
45.
47.
49.
51.
53.
~o5.
,
+
-,I
ANSWERS TO
ODD-NUMBE~ED
37~
EXERCISES
61. (a - b + c + 2d)(a - b - c - 2d).
63. (3x 2 + y2)2.
65. (5x 2 - 2y2 + 2xy)(5x2 - 2y2 - 2xy).
67. (x - 1)(x 2 + X
69. (x + 3)(X2 - X + 3).
71. (x -I- 3) (x 2 + X
73. (x - 2)(x - 4)(X2 + X + 1).
75. (a + b)(a 2 - ab + b2)(a 6 - aBbB + b6).
-
4).
+ 2).
Exercises III. (, page 63
3.
7.
11.
13.
15.
17.
29x 2y 2z8, l044x 5y 7z lo w B.
6. x - 6, (x - 6)(x + 2)(x - 3).
1, (a B+ bB)(a + b).
9. 6x - 7y, (6x - 7y)2(7x + 4y).
2a - b, a(2a + b)(2a - b)3.
a2 + b2, (a B - bB)(a 4 - a 2b2 + b4).
x - 3, (x - 3)(x 2 + 3x + 4)(x 2 - 2x - 6).
1, (16x 4 + 81 y 4)(4x 2 + 9y2).
Exercises IV. A, pages 65-66
B
x-I
6. a+~.
3.--·
x +4
a + 2b
9. (x 2 + y2)(X + Y), 11. y - x,
X2 + xy + y~
y+x
2 2
4
4
a
a
b
+
b
17.
X2 + 2x + 10,
15.
.
a 2 - b2
l ..l
X2 + 5x + 22
1. I3bc •
63a 2
21. a - b + c.
a+b-c
+ 2b.
3a - 1
2x + 2
7. a
13.
X2
.
X2
"
x+y
19. -_'-"--_
x 2 + xy + y2
,
~".
"H
Exercises IV. B, pa"s 68-70
1. _!l_.
3. a
2
I2x
7. 182: - 20x
21y.
24xyz
11. 30a 2
30ab
15b 2 •
(2a - 3b)(3a
4b)
+b
Z
+
+
13.
+ y)(x
9.
.
17.
- y)2
lIx - Y
(X2 _ y2)(X _ y)
1
-,
5a - 4b
- x + 8x + 11
.
(3x + 5)(2x + 3)(3x - 2)
27. ~.
29. 2X2 + 12x
3x - 5
(x - 2) (x
19.
2
~
33. - 35. 4x
3
3(3x + 8)
39. 2x _ 76 + 246x - 307.
X2 + 3x - 4
+3 -
25. 12ab - 2c.
3b
3
31.1+--'
x+2
23. O.
+ 13.
+ 7)
8
_--'
3x - 4
+ 28x .
+ 3)(2x + 5)
12x2
(x
2
21.
+ 56x",
144x 3y 4
+
+
-2y
(x
5. 81y2z
•
abc
37. 3
+
9x - 14 .
5x 2 - X - '2
ANSWERS TO ODD-NUMBERED EXERCISES
380
Exercises IV. (, pages 71-72
+
+
+
+
(x
8)(x
3)
6. (a
2)(a2 - 2a
2).
2
10z2
(x - 2)(x - 3x
9)
a - 2
7. (2x - 1)(5x
4)(7x - 2).
9. (a 2
b2 )(x
y)
2
(8x
9)(x
2)(x - 1)
x
xy
y2
11. (x - 12y)(x - 8y).
13. (3x - 4y)(4x
9y) •
16. a2(a
b).
(x - 9y)(x
6y)
(2x - 5y)(2x
lly)
b- a
1. 9a 2cx2y.
3.
+
+
+
+
+
+
+
+
+
+
+
+
Exercises IV. D, page 75
1.
3. ~.
i.
i:;(
5. -1.
C
. x
+
X2 - 4y2
1
13.
.
X2 - 4y2
11. 2 _. x.
\t
7 _x_.;
+ 3."
~4'
4x-5
9.--·
x-2
1
15. 1.
Exercises IV. E, pages 78-79
1. 28.
-)-J-.
3.
5. 4.
7. No solution.
15. 825H
13. _!]!_.
x+y
19. 240 mi., 50 mi./hr.
11. (4 ,4.
)
+ wV 2•
v
17. 7.
g
,
,
I
Exercises V. A, page 82
13. 2x.
Exercises V. 8, pages 85-87
1. 7.
11. 13.
x
9. 1.
7.2x.
r_
1
13. _.
17.
8x 3
q4
23. -_.
16p2
31.
+ x.
y-x
1!_
37. 5x 2
-=:..
27
27. 27 y 6.
64x 48
33. 6x 3
+ 18x -
-
17x 2
9 - 4x- 1•
+ 22x -
15.
I
19.
29.
35. 3X 5/ 2
41. 1.7 X 10- 7 •
-H.
+ 5X 3/2
-
8Xl/2.
43. 5.8 X 10- 6•
ANSWERS TO ODD-NUMBERED EXERCISES
47. 0.0000000215.
63. 6.7 X 10- 19 •
46. 3.361 X 10 6 •
61. 5.0 X 1011.
381
49. 9.0 X 10- 12 •
Exercises V. (, page 89
1. 12Va.
9. O.3u 4 Vu.
17.
3. llxyV2xz.
11. -Zx-~~.
50a3b5
~- _e& .
7. 3a4V~.
13. 2Ya'l.
19. ~.
.3;:::;:27. v'L7a.
26. V9x 3 •
29.
,
V x2n+1.
21.
~3x
_.
7. 2xyzV"4xy 2Z'.
2b'l..
15. Y16ab.
-
23. Vxn+2.
31. {Ia3 b2e.
25y
33 .
.,yp2 q3.
Exercises V. D, pages 90-92
1.
V77.
3.
7
13. 3(V6
19. 3 -
2~.
6.
iv'15.
+ V2).
7.
j{l6. 9.
23. aex -
16. a(Vb - Vc).
b- c
1
r.:.6 - ~V
L r.:.
+ ~V
2.
bdy + (be - ad) v;jj.
3L
+ 11 v'5 + 7V7 _
6V35)/131.
1 - a 2/3
33.
1 - a
36. (x + h)2/'
37. 0.816.
1O~~.
a
17. a(Vb - e).
b - c2
21. 183 - 23v65.
128
c2x - d2 y
25. (9
11.
3y
. r.:.3
V
~12by2.
a 2b2/3
_
2
27. x(x + 4) .
(x2 + 3)3/2
abl/ 3e l/3 + C2/3
.
a 3b + c
.~
1
+ (x + h)1/3X 1l3 + X 2/3
41. -0.115.
39. 2.129.
Exercises V. E, page 93
1. 5pq2V2pq.
~. 3~3.
V2x
17. 18x'
+y -
11. -
aV~
5b
~675
19. -5-'
yV;)j y(x - y)
31. a3b 5 Va 2b3 •
33.
2~
xy
3. 2abc2~5ab2.
.y;y.
Vn.
6.n~n.
~3ab2
13.~·
~18
21. -2-'
27.3+
7.~.
54 - 17V10
16. ---13--'
aVb -
bV~
23. --a---b-
Vlo.
3Si
ANSWERS 10 ODD-NUMBERED EXERCISES
"
Exercises V. F, page 94
1. -9v17.
17~675
11.
9.5~.
3. l8V3.
13•
5
V-
~.
2
"
'
!
Exercises V. G, pages 95-96
1. V133.
3. ~20.
5. ~3.
,t+
7. 35V35· x if x ~ 0, -35v35' x if x < O.
,9. 2aV"2a.
6/,3;~5/-,3;:::
11. V 108.
13. l6abva.
15. 2v 30375. 17. 2 - 2v 2.
19. 2{!6.
21. 72 - l2{i2 + 2{i54.
28~l x 2 - 3x 2
25. x - 4x - ¥. 27. 0
29. O.
'
4.;
Exercises
1.
I
i
i
V. H, page 97
0.
3.
V(17)'
VB.
7. -1-7-
9.
-\77.
2V 12'
V a2bc 3
1 6,
4
13. -3-' ' 15. - - .
17. - (V2;5 - ~).
c
x
-27 + 31V6
19. 11 + 2V30.
21. -----'-7-3--
/
-b
+' eV~
eVb
- -~
23. -- - -+
:_
_-c2
-
b
acx - b y + (be - ab)V;Y
.
c2 x - b2 y
_3/-
2
25.
27.
1 - V x2
I-x' /
Exercises V. /, pages 99-100
1. 5
9.
+ 6i.
3. 11
+ 11i.
a-bVc·i,c~O.
+
7. x
5. 7 - 2v2 . i.
V2·
11.8-i
S
+ ivY, y ~ O.
13. -2+2v2·i.
+
- {15
iv2.
17. 81 - 35i, 61
13i.
9 - (V3
v6)i, 3
(V6 - V3)i.
27 + (VIS - v17)i, -1 - (V18
v17)i.
23. -v15.
-6V5' i. 27. -11 VU' i. 29. 49.
31. -4
33i.
8 - l2li.
35. 23 37. 40 - 42i. 39. -46
9i.
47 - 32i
41. 1.
43.2.
45. i.
47.
53
15.
19.
21.
25.
33.
+
+
+
iva.
+
+
ANSWERS TO ODD-NUMBERED EXERCISES
49.
58.
~O -
v21 -
+ 7v6 + (2.J2i -
3VU)i
102 - 53i
181
6
51. •
27
(2V3
11
+ 5V7)i
.
9 - 7i
55.~·
.
383
59. O.
57. O.
Exercises VI. A, pages 105-108
1. 5, 7.
8. 7, -8.
5.
± 11.
7.
± 17i.
_ r.::
.r
3 + iV31
1
9. -2 ± v 3. 11. -1 ± 2v 2. 18.
- 2
15. 1, -"2".
(
17.
i, -i.
19.
25.
-4, -4.
27. -2,
-i, -i.
-l
-2
21. r,8.
28.
29. 0.3, 004.
31.
± iv'll
3
15
+
5V3· i
.
12
33. i(k - 3 ± v'k
6k - 23).
67. 8, -2.
37. 4, -1.
39. 1, 1.
41. 2, -5.
43. 7 in. X 10 in. 45. 36,37.
47. 27, 29. 49. 3 in.
51. 10 in. X 20 in. 53. 60 mi./hr.
55. 60.
57. 20 min., 30 min. 59. 9 ft.
61. 32 mi.
63. (x - 5)2
(y - 3)2 = 16, (5, 3), 4.
2 -
+
3 + (y + "2")2
5 =1~, (-"2",-~),
35v'2
+ "2")2
2'
(x - 4)2 + (y + 4)2 = 9, (4,-4),3.
V3V(x - 3)2 - 5.
71. V5V(x + 4)2 - 18.
V7V (x - 3)2 + 2.
75. 2V% - (x - 1)2.
v'2v¥ - (x - %)2.
79. V2V4 - (x - 2)2.
65. (x
67.
69.
73.
77.
Exercises VI. B, pages 109-110
_r:..
1. ±3, ± v 2.
.r
3. ±iv 3, ±4i.
9. 216, -27.
11. 22 ±
15
17. 81.
19. 15,4.
3
5. ±"2",
V3
± 3'
1OV7.
13. "2"fl.;, _L~.!!..
21. ±3.
23. 9.
7. 9.
15. ±8, ±27.
25. 0, 1, 1,2.
Exercises VI. (, pages 112-114
1. 11.
15.4.
3. %.
17.4.
5. 1.
19. 17 ft.
7. 9.
9. 8.
21. 25 ft., 30 ft.
11. 2.
13. 4, 5.
23. 15 mi., Ii mi.
Exercises VI. D, pages 115-117
1. Real, unequal, irrational.
5. Imaginary.
3. Real, equal, rational.
7. Real, unequal, rational.
384
ANSWERS TO ODD-NUMBERED EXERCISES
9. Real, unequal, irrational.
11. Real, equal, rational.
15. 2.
17. -·~l-.
19. 28.
21. ± 1.
23. -3.
25. 11.
27. 1,2.
29. 4.
31. -1.
33. ±5.
35. --frs.
37. (a) t, (b) 0, -4, (c) 2.
Exercises VI. E, pages 119-120
1.
9.
19.
27.
23.
27.
41.
45.
J-i-,
-7, 12.
3.
-¥. 5. -J,
~, O.
11. 5, ¥-.
13. -¥. 15.
± 16.
21. ±¥-.
23. -a.
x 2 - 35x = O.
29. 81x 2 - 25 = O.
x2
45 = O.
35. x 2 '36x 2 - 204x
301 = O.
39. x 3 x 3 - Dx 2 - 25x
150 = O.
43. x$ x 3 - 8x 2 + 29x - 34 = O.
47. ~3 -
+
+
+
t.
-~.
7. -i-,
17. ¥-, -.!f!. 25. X2 - 4x - 32 = O.
31. X2 + 169 = O.
lOx + 28 = O.
3x 2 - 28x + 60 = O.
llx 2 + 37x - 35 = O.
8x 2 + 68x + 176 = O.
¥.
Exercises VI. F, page 121
1. (12x - 77)(14x
7. (x
+ V6)(x -
+ 73)(50x + 49).
!_
!+
3. (lOx
9
VB).
(x _ iV3)(x _ iv'3).
v's9) .
+ 10
•
2
2
2
2
v's9)(x iV2)(
i'V'2)
x--- x--+-'
3
W
4
3
3
11. 5 ( x - 10 133 (
.
+ 99).
3
10
4
3
3
-
+ 3 - -VS).
_3V5 _ v'57)(2X _ 3V5 +
15. (x - 3 - V5)(x
17
.
(2X
2
2
2
v'57).
2
Exercises VI. G, pages 124-125
15. -14, min., 4.
21. -1, min., 2.
27.
-is, min., --1.
33. 77c. or 78c.
,
17. 7, min., -2.
23. ¥, max., -£.
29.
3
2' min.,
v'2
2
19. 19, max., 3.
25. -5, min., -2.
31. 20,000 sq. ft.
35. $6.50.
_/
Exercises VII. A, pages 127-128
1. (3,2), (¥,!).
3. (4,2), (-j,j).
6. (-7,-3), (¥>H).
7. (2,3), (-¥,-¥-). 9. (4,-2), (-W,W).
_7'
f
3V7 9
3V7) (9
3V7 9
0 )
9
2 - -2- i, 2 + 3
-2- i .
11. (2 + -2- i, 2 - -2i ,
"
/
ANSWERS TO ODD-NUMBERED EXERCISES
13. (3,4), (-t-H).
19. (3,-1),(-1,-4).
. 25. 4 in., 6 in.
385
15. (-t,6),( -t,6).
17. (0,5)(-7,4).
21. (-2,2), (HJ-~-h 23.5 in.
Exercises VII. B, page 129
1.
3.
5.
7.
(3,2),(3,-2), (-3,2), (-3,-2).
(v7, V2), (v'7,-V2), (-v7,V2),(-V7,-V2).
(1l,2V2), (1l,-2V2), (-1l,V2), (-1l,-2V2).
(2V2' i, Va), (2V2· i,-Va), (-2V2' i, Va),
(-2V2 . i, - Va).
2i 4) (2i 4) (
9. ( 3' 3""' 3'-3'
2i 4) ( 2i 4)
-3'
3"" ' -3'-3 .
,
11. (2,3), (2, -3), (-2,3), (-2, -3). 13. 20Va, 1OV'33:'
f
Exercises VII. (, page 131
Va
Va) ' ( - -9-'
7Va
3. (1,-2), (-1,2), ( 7-9-'-""3
v'a)
3"""" .
6. (2, -4), (-2,4), (3v6, - V6), (-3V6, V6).
7. (2, -6),( -2,6),(4,2), (-4, -2).
9. (2V5,3VS)
,
I
,}if
•
x
.ooU:
,
'
(- 2V5, -3V5).
11. G,2} ( - ~t 2} (2~, _7~). ( _2~, 7~3}
_
13. (3, -6), (-3,6), (tVS, 5V3), (~4V3,
16. (6,2), (-6, -2), (3,7), (-3, -7).
-5V3).
17. 12 in., 18 in.
Exercises VII. D, pages 133-134
1. (3,4), (4,3), (-1 + iV2,-l-iV2), (-1 - iV2, -1 + iV2).
3. (3, -2), (-2,3), (3, -2), (-2,3).
6. (2,0), (0,2), (-2 + iV2,-2-iV2), (-2 - iV2,-2 + iV2).
ivfU)
1"" + v'ai 1"" - V31)
+ v'31)
77· 7
7'
(3 ivfU 3 iv0U)
3 ivfU 3
7. (2,1), (1,2), ( 8 + -8-' 8 - -8- , 8 - -8-' 8 + -8- .
9. (3,2), (2,3), ( -
20
_ 20 _ V31,_ 20
(
11. 12 in. X 16 in.
-7-' -
20
-7- ,
-,
~86
ANSWERS TO ODD-NUMBERED EXERCISES
Exercises VII. E, page 136
!
1. (5,6), (-5,-6), (6,5), (-6,-5).
3 (1
.
(
1
,3), (6,-~),
-11 -
(-n +2 iV47' -n-iV47)
4
'
iV47,
2
5. (3,4), (4,3).
-11
+ iV47)
4'
,
i
I
.
7. (4,6), (6,4). 9. (5,3), (5-3), (-5,3), ,(-,,-3).
I
,
11. (2,1), (-2, -1),
(v2, vi), (-v2, -vi).
13. (4,6), ( _ 1, __
~}
( - 1
Vi7,
-1
+
(
17.
21.
-1 -
2
~ vT7, - 1 ~
-
v'i7).
. 1
v17)
,
4'
VB+ vi
vi33
I
"
19. V 15 - ViD.
23. V21 + V13.
v22.
Exercises VII. F, pages 139-140
3. ±4.
9. ±12.
16. ±5.
I
. I
7. j.
13. ±V46.
5. ±25.
11. - 1,3.
17. p/2m.
, ! 19.
± V ma':
2
I
ti
1";r. u
Exercises VII. G, pages 140-144
1. (~1,10), (-2,3).
O.
7.
11.
13.
15.
17.
19.
(-V6,':"81.
3. (V5,3) (V5,-3), (-V5,3),
(7i,3i), (-7i, -3i), (2i,6i), (-2i, -6i).
(2,6), (6,2), (3,8), (8,3).
9. (3,3), (H,ll).
(3,7), (7,3), (1,-3), (-3,1).
,
CV7,O), (-v7,0), (2V7· i, -5V7' i), (-2v7' i, 5~ •. ;).
(2, -1), (-iJ).
(0,0), (3
V3,3 - V3), (3 - V3, 3 V3),
+
+
~ + V21, _ ~ _ V21), (_ ~ _ v21, _~ + V21).
(_2
222
2222
+ V13
+
(3,4), -_]' - '2'
_ 6 ; - - - 4 _.' '
. ,
_/.....
. r ~.,.\;. ::.J
1 - 6V1r; _1 + 4V13).' ....
(
( 8 9) (1
-1 V13)
~
I.
I
,li.
ANSWERS TO ODD-NUMBERED EXERCISES
21. (1, -1), (
-1
+ iva
2
'
387
1 - iVS) (-1 - iVa 1 + iV3)
2
'
2
'
2
.
Last two solutions are double.
23. (7,7), (7,-7), (-7,7), (-7,-7).
6
3
25. (2,1), (T, -T),
[
27.
29.
31.
37.
43.
[18(15 - 4i) -6(4 + 15i)]
241
'
241
'
18 (15 + 4i) 5(4 - 15i)].
241
'
241
(2,2), (2 + 2v'6 . i, 2 - 2v'6 . i), (2 - 2v'6 . i, 2 + 2v'6 . i).
(±2Y2, ±3, ±iV5) with all possible combinations of signs.
(2, - 5,3) (- 4,3, - 5).
33. (3,2, - 2), (- 3, - 2,2).
35. 8i in.
5 in. X 8 in., 5* in. X 6~ in.
39. 8.
41. lW hr., 14 hr.
A 1000, B 600, 2i hr.; A 975, B 650, 2rt hr.
45 20,
. 3
svi3,
4V73.
3
3
47. 38.
49.
ii.
51. 8 mi./hr., 2 mi./hr .
. 63. 7 hr., 5t hr.
Exercises VIII. 8, pages 151-153
1. x < !.
3. x ~ -1.
5. x < 1.
7. x ~ j.
9. x < -¥-.
11. x < -2, x > 8.
13. 1 ;;:; x ;;:; 7.
15. x < -i, x > 4.
17. x < -i, x > i.
19. All values of x.
21. No solution.
23. x < -1 - v'6, x > -1 + VB. 25. - 3 < x < 2, x > 5.
27. x ;;:; -1, x = 2, x ~ 4.
29. x ~ -1.
31. x < -1, x > i.
35. -i < x < i, x > ¥.
37. -4 < x < -2.
39. -1 < x < 3, x > 5.
41. k > ¥-.
43. k < 4 - 4 Va, k > 4 + 4. Va.
45. k < -llf.
47. -6 - 12V5 < k < -6 + 12V5.
49. -2 < k < --ft.
51. k < - Va, k > Va.
53. k < 1.
55. k < 1.
67. -7 < k < 7.
69. k < -6, k > 6.
Exercises IX. A, pages 155-159
3.
·13.
15.
21.
16, J-f, 10.
5.!, f, t.
7. (a) 4, (b) i.
9. ¥.
11. 30 lb.
(a) 25 lb./sq. in., (b) 33t lb./sq. in., (c) 12i lb./sq. in.
19. (a) 200 f.p.m., (b) 0.864 H.P.
120 amp.
17. 85j lb.
4 lb.
23. 75 lb.
25. 3125 lb.
27. (a) 1200, (b) 52i lb.
-
6v'i5
29. 3V6 in., - 5 - in.
388
ANSWERS TO ODD-NUMBERED EXERCISES
Exercises
X.
, it
A, pages 163-164
27. False.
29. True.
31. False.
33. True.
I
35. False.
X. 8, pages 165-166
1. X S + 5x 4y + 1Ox 3y2 + 10x2y3 + 5xy4 + yS.
3. a 6 + 6a sb + 15a 4b2 + 20a 3b3 + 15a 2b4 + 6ab s + b6 •
5. x B + 8x 7 + 28x 6 + 56x 5 + 70X4 + 56x 3 + 28x 2 + 8x + 1.
7. a9 + 9a Bb + 36a 7b2 + 84a 6b3 + 126a 5b4 + 126a 4b5 + 84a 3b6
+ 36aW + 9ab B+ b9 •
9. 1 - 10x + 45x 2 -. 120x3 + 21Ox 4 - 252x· + 210x 6 - 120x7
+ 45x B - 1Ox9 + x 10 .
11. 27x 3 + 108x 2y + 144xy2 + 64y3.
13. 16x8 + 32x 6yl/2 + 24x4y + 8X2y3/2 + y2.
15. -ha 3 + .feaS!2b1!3 + -fsa 2b2l3 + -Aa 3/ 2b + rlsab 4!3
+ira 1!2b 6!3 + -ri.rb 2.
17. a 14b7 - 14a 12b6c2 + 84a 10b6c4 - 280a Bb4c6 + 560a 6b3cB
- 672a 4b2c10 + 448a 2bc 12 - 128c 14 •
,,"(" I
19. 1,000,000,000x9 - 1,800,000,OOOxBy
1,440,000,000x7y2
- 672,000,000X6y3 + 201,600,000x6y 4 - 40,320,000X4y 5
+ 5,376,000X 3y 6 - 460,800X2y7 + 23,040xy 8 - 512y 9.
21. x So + 50X49y + 1225x 48 y2 + 19,600x 47 y3.
23. x 3 + 3x 2y + 3xy 2 + y3 + 3x 2z + 6xyz + 3y 2Z + 3xz 2 + 3y,. ~.
25. 23.4256.
27. 0.885,292,81.
29. 9,509,900,499.1: ~,
31. 1.012,048,064.
33. 1,01 7,054,108,081.
.,' ,:J:;
35. 1.4859.
37. 0.9044.
,,'./
Exercises
+
+
Exercises
X. (,
page 168
4
3. 1,088,640x y 6. 5. 3" 210 • lla 5/ 2b7/ 3•
9. 3003a 10b 6.
11. -2 10 • 37 • llx 21y lO. -f
15. 278,460X 2y 4z12.
Exercises
1.
5.
9.
13.
15.
17.
23.
29.
35.
1 - x
X.
+ h - -ir;x + rhx
2
3. 1
3
•
1 - 0.9x - 0.945x 2 - 1.6065x 3.
1 + 0.2X2
1.0149.
2.234.
0.993289.
4.020725.
..'(ol"
D, page 172
+ X2
- xa.
1 + ix - h2 + s\x 3 •
1
7. -2 7 • 3 7 • 7 6xil':;&
13. -9240X 3y 5Z 3. ,~'J<
+ 0.22x 4 + 0.308x
19.
25.
31.
37.
+ jx -
}x2 + -/sx 3 •
7. 1 - jx + h2 - .fsx 3 •
11. 1 - x + iX2 - tx 3 •
6.
10.247.
7.8103.
9.9933289,
1.007872.
21.
27.
33.
39.
5.291.
9.055384.
2.22.
4.9594.
., .
ANSWERS TO ODD-NUMBERED EXERCISES
389
Exercises XI. A, pages 175-176
1.
9.
15.
17.
25.
37.
47, 245.
3. 402, 4050.
5. -43, -225. 7. 109.5, 5505.
11. 15, 442j..
13. i, 273.
5{-, 7j., gi, 12, 14{-, 16j., 18t 21, 23{, 25j., 27t.
76,5720.
19. -30, -60.
21. - ni, 28j.
23. 5,34.
7, 11. 27. 9, 3i. 29. 62. 31. 31. 33. 3, -1. 35. n(n + 1).
al = an - (n - l)d, S" = in[2a" - (n - l)dJ.
2S"
2(na" - S,,)
39. al = - n - a", d = nn(
1)'
2S"
a! - a~
.!,
a" - al
41. n = - - , d = ----"----=-43. d = - - ·
n-1
al + a"
2S" - al - a"
Ji,
io, H.
Exercises XI. 8, pages 178-179
(
1.
7.
11.
15.
19.
21.
29.
3. 2916,2188.
5. j, 1534j.
9. 0.000,000,000,3, 3.333,333,333,3.
1.082,432,16, 5.204,040,16. 13. -192, -189(2 - v2).
5,48%.
17. i, 422; -i, 110.
42,294,2058; -42,294, -2058.
625, 103l.
23. 3, 768.
25. 6, 0.016.
27. 9, 5621.
!,6.
31. 52,5.
33. 5,225; -6,324.
625J), 7812.
'2"i-h, HH.
n
35. 374,556. 37. 7,
41. al
=
a"r l -", S..
(an)_l_
(a,,!aa)n-l - 1
39. r = - n-l, Sn = al
l'
al
(a,,!al)n-l - 1
a,,(r" - 1)
43. al = ra" + (1 - r)S".
r,,-l(r _ 1)'
1
-If.
=
Exercises XI. C, page 181
1. 2f.
11. -1-#-(2
6. Ii.
13. 16
3. 41i.
+ 3V7).
7. 16j.
+ 9vi
9. t(5
16.
i.
+ Vs).
17. 26.
Exercises XI. D, pages 182-183
*.-M.
1.~. 3. 1.
15. -Nir;.
5.
17.
7. ~.
9.
19.
&.
ill.
11.
m.
13. f.
21.~.
Exercises XI. E, page 184
1.
lrJ.
9. (a) 5, (b) 4.
3.
i.
5.
11. 3,
A.
-t.
7.
¥, i, ¥, ¥-, 4, 5, ¥'.
13. 2ab!(a
+ b).
390
ANSWERS TO ODD-NUMBERED EXERCISES
Exercises XI. F, jJages 184-187
1. 54, 236.
3. 21 j, 4431
6. 3f, 56i-.
9.
13.
23.
29.
7. 6.
(a) 0, (b) 4, (c) 5j.
11. (a) 26 ft., (b) 98 ft., (c) 8 sec.
100 m.
15. 247,500. 17. 71,071. 19. j in.
21. $512. , I '
3~! gal.
25. 26,956.8 ft. lb.
27. 11 in.
4,11,18; 22,11, O.
31. 4,36.
,i
'-''i.~~ I
I
I
Exercises XII. B, pages 191-192
1. 13 + 9i.
3. 14 - 12i.
9. -6 - iVa. 11. 4 + 12i.
17. -182 - 61 V10 . i. 19. 16
23.
29.
35.
H - -Hi.
if - Hi.
H - -Hi.
25.
(:
31.
5. 9 - 2i.
'1. -13 + 16i.
13. 111 + 2i.
16. 29 - 31i.
+ 30i.
21. -i!r; + ffi.
V2l
1
-"5 - 15 i.
-ih + H-3i.
5V15
+ --w- i.
ill + i-Hi.
28
27. - 19
33.
Exercises XII. D, pages 194-195
cos A + i sin A is abbreviated cis A.
7. 12 cis 240°.
3. 7v2 cis 45°.
5. 2Vs cis 26.6°.
13. 9v2 cis 225°.
11. 17 cis 118.1°.
9. 8 cis 90°,
15. v'7 cis 32.3°.
17. ViO cis 63.4°.
19. H- cis 337.4°.
21. 5 + 5VS . i.
23. 3v2 + 3V2' i. 25. -2V2 + 2"'2·,.
27. 1 - i.
29.
33. -17.36
+ 98.48i.
'I
VS
2- 2
35. -5.541
i.
+ 7.092i.
31. -5i.
37. -20.
Exercises XII. E, page 196
1. 10 cis 100° = -1.736 + 9.848i.
+ 30Va . i.
6.
7.
9.
11.
-3V2' i.
3. 60 cis 60° = 30
84V2 cis 75° = 30.75 + 114.75i.
2 cis 40° = 1.532 + 1.286i.
2 cis 110° = -0.6840 + 1.8794i.
cis 270 = -i.
13. 3V2 cis 270 0 =
0
Exercises XII. F, pages 199-200
1. 64 cis 36 = 51.78 + 39.62i.
3. 27 cis 150° =
0
5. 32 cis 270 = -32i.
7. 65,536 cis 120°
9. k cis 296 = 0.0007014 - 0.001438i.
11. cis 300 0 = j(1 - iVS).
0
0
J-i-( - VS
= 32,768( -1
+ i).
+ iVa).
I
,i
392
ANSWERS TO ODD-NUMBERED EXERCISES
1. X 2 + 8x + 9, 10.
3. x 2 - 6x - 4, 0.
5. x 2 - 7x + 3, o.
7. 5x 2 + 29x + 186, 1293.
9. x 3 + 7x 2 - 2x + 4, 0.
11. x 3 - llx 2 + 46x - 199, 792. 13. 3x 3 - llx 2 + 55x - 282, 1432.
15. X4 - x 3 + 2x 2 - 2x + 3, -2. 17. 4x2 - lOx + 4, 0.
19. x 2 - (8 + Va)x + 7 + Va, 0. fI,',
21. 29, 1835.
Exercises XIII. B, page 209
1.
5.
9,
15.
°
2,4, -3.
3. -1, -4, (0,1), i.e. between and 1.
-4, (2,3), (-3,-2).
7. (0,1), (6,7), (-1,0).
2, (0,1), (-3, -2). 11. (-1,0).
13. 2, -4.
No real zeros. 17. 2,-3. 19. 1, -6. 21. 4, -1. 23. 3,0.
Exercises XIII. C, page 213
3. j, -i, 3 ± iV2.
7. 3 ± i, 3 ± i, 2, -1.
5.
Exercises XIII. E, page 217 '"
NOTE.
imaginary.
1. (1,0,2).
9. (2,1,0),
15. (1,2,0),
21. (2,2,0),
23. (2,2,0),
25. (2,1,2),
± V7, ± 0,
?', ':
(1,0,2), for example, means 1 positive,
3. (0,1,2).
(0,1,2).
11.
(1,0,2).
17.
(2,0,2), (0,2,2),
(2,0,2), (0,2,2),
(0,1,4).
27.
1,
5. (2,1,0), (0,1,2).
(1,2,0), (1,0,2).
13.
(3,1,0), (1,1,2).
19.
(0,0,4).
(0,0,4).
(1,1,4).
! 29.
°
negative, 2
7. (1,2,0), (1,0,2).
(3,0,0), (1,0,2).
(3,1,0), (1,1,2).
j.;tq
(1,0,4).
Exercises XIII. F, pages 220-221
1. 2.
3. -2.
5. i.
7. fl'
9. 3, i, -j.
13. i, -j, -.fr.
15. ±2,3, -4. ,// 17. -1, i, j,
19. 2, -3.
21. ~, -i.
./ .j:. 23. i in.
Enrcises XIII.
- j..
~,::-
11.
-i.
2:
G, page 223
1. 0.682.
3.
1.175.
5. 0.347, 1.532, -1.879.
9. 0.379.
11. 0.555, 2.247, -0.802.
15. 1.262, -0.660, -3.602.
17. 0.414, -2.414.
7. -4.703.
13. -1.761 •.
Exercises XIII. H, page 225
1. x 3 + 6x 2 + llx - 1 = 0.
3. 2x 3
2
3
5. x + 10x + 33x + 16 = 0.
7. 4x 3
9. 8x 3 + 14.4x2 + 16.64x - 0.472 = o.
11. X4 - 16x 3 + 84x 2 - 136x - 37 = O.
+ 17x + 38x+~'-';;' 0.
l3x! + 8x + 6 = O.
2
-
-h
-j .
ANSWERS TO ODD.NUMBERED EXERCISES
393
Exercises XIII. /, pages 228-229
1.
9.
16.
21.
26.
27.
1.328.
3. 1.060, -1.536, - 5.524.
6. 0.870.
7. -1.485.
1.843, -0.306, -3.537.
11. 1.638.
13. 1.202, 3.J28, -1.330.
±1.414.
17. 0.414, -2.414, ±3.464.
19. 0.838.
2.924.
23. 2.187.
(2,0), (-1.463,1.861), (-2.473, -2.115), (1.935,0.254).
29. 6.33 ft.
1.45 X 1.45 in.
Exercises XIII. J, pages 235-236
I
1. ~ + ~, w~2 + W2~, W2~ + w~.
3. 2 cos 40°, - cos 40° - Va sin 40°, - cos 40° + Va sin 40°.
_3/_3;;::
_3;;::
_3;;:;
_3;;:;
_3;;:;
6. - v 9 + v 3, -wv 9 + w2 v 3, -w 2 v 9 + wv 3.
1 _3/_3;;::
1 _3r:
_3;;::
1 _3r:
_3/7. 02"V 4 + v 2, w . "2"V 4 +- w2 v 2, w2 • "2"V 4 + wV 2.
_3)_3r;
_3;;:;
_3)_3/:::
_3r;
9. -2 + v2 + v4, -2 + wv2 + w2 v 4, -2 + w2v2 + wv4.
11.2+ V5 + ~25,2 + wV5 + w2~,2 + w2 V5 + w~.
13. -2 cos 15° = -j(V6 + V2), 2 cos 45° = V2,
2 cos 75° = jeV6 - V2).
16. -2 ± 2V2, -1 ± iVa.
17. 1 ± V2, -2 ± Va.
19. ± v5, j( -5 ± v5).
21. 3 ± V2, ±3i.
Exercises XIII. K, pages 237-238
1. -4, -9.
9. i, o.
15. ±j, j; -4.
1
5
3
3. 4, 7.
6. -"W, -"2".
7. -"2", -4.
11. 6, -1, -2; -16.
13. -i, -i, 4; -46.
17. 2 ± 3i, 2 ± 3i; c = 42, d = -104.
Exercises XIV. A, page 243
1.
13.
21.
29.
37.
1.
3. 1.
6. -2, or 8 - 10.
7. 4.
9. 1.
11. O.
-5, or 5 - 10.
16. -7, or 3 - 10.
17. O.
19. 6.
-2, or 8 - 10.
23. 627.
25. 0.0627.
27. 62,700.
0.0000627.
31. 3690.
33. 0.369.
36. 0.000369.
0.00000369.
Exercises XIV. 8, pages
3. 0.8451.
11. 1.4508.
19. 2.0013.
24~246
6. 3.4771.
13. 4.6506.
21. 0.7398.
7. 8.1106 - 10.
15. 9.5791 - 10.
23. 1.0052.
9. 0.5072.
17. 7.8752 - 10.
26. 8.9546 - 10,
Exercises XIV. C, page 247
9. 100.
1. 1.630.
3. 85.70.
6. 0.09050. 7. 1.726.
17. 10,980.
11. 352.4.
13. 22.42.
16. 1.001.
19. 0.001090.
21. 0.00004143.
23. 2214-
ANSWERS TO ODD-NUMBERED EXERCISES
3••
13. rfu cis 60° = 2-A-s(1 + iV3).
15. 6 cis 25° = 5.438 + 2.536i, 6 cis 205° = -5.438 - 2.536i.
17. 5 cis 50° = 3.214 + 3.830i, 5 cis 230° = -3.214 - 3.830i.
19. 6 cis 9° = 5.926 + 0.9386i, 6 cis 129° = -3.776 + 4.663i,
6 cis 249° = -2.150 - 5.601i.
21.
cis 100° = -0.2188 + 1.241i,
~ cis 220° = - 0.9652 - 0.8099i,
~ cis 340° = 1.184 - 0.430I:li.
~3. 1, cis 120° = ·j(-1
iVa), cis 240° = ·j(-l-iVa).
25. cis 180° = -1, cis 300° = .fr(1 - iVa), cis 60° = .fr(1
iVa).
27. V5 cis 17.7° = 1.629 + 0.5199i,
V5 cis 137.7° = -1.265 + 1.151i,
cis 257.7° = -0:3643 - 1.671i.
29. 1, cis 72° = 0.3090
0.9511i, cis 144° = -0.8090
0.5878i,
cis 216° = -0.8090 - 0.5878i, cis 288° = 0.3090 - 0.9511i.
31. 2 cis 27° = 1.782 + 0.9080i, 2 cis 99° = -0.3129 + 1.975i.
~
2 cis 171° = -1.975
0.3129i, 2 cis 243° = -0.9080 - 1.782i,
2 cis 315° = V2(1 - i).
33. 7, 7 cis 120° = i( -1
iVa), 7 cis 240° = i( -1 - iva).
35. cis 180° = -1, cis 36° = 0.8090 + 0.5878i,
cis 108° = -0.3090 + 0.9511i, cis 252° = -0.3090 - 0.9511i,
cis 324° = 0.8090 - 0.5878i.
37. 1, cis 51.4° = 0.6239 + 0.7815i, cis 102.9° = -0.2232 + 0.9748i,
cis 1M.3° = -0.9011 + 0.4337i, cis 205.7° = -0.9011 - 0.4337i,
cis 257.1° = -0.2232 - 0.9748i, cis 308.5° = 0.6225 - 0.7826i.
-«2
+
-«5
+
+
+
+
+
39. 0, 1, cis 45° =
V2
=
2
cis 225° =
2
cis 135°
V2 (1 + i), cis 90° = 1,
2
(-1
V2 (-1
+ i), cis 180° =
- i), cis 270°
+
=
-1,
-i, cis :Y15° =
V2
2
(1 - i).
+
41. cis 72° ~ 0.3090
0.9511i, cis 144° = -0.8090
0.5878i,
cis 216 = -0.8090 - 0.5878i, cis 288° = 0.3090 - 0.951li.
43. 2 cis 72° = 0.6180
1.902i, 2 cis 144° = -1.618
1.176i,
2 cis 216° = -1.618 - 1.176i, 2 cis 288° = 0.6180 - 1.902i.
0
+
+
Exercises XIII. A, pages 204-205
In exercises 1-19 the first expression given is the quotient, the second
is the remainder.
394
ANSWERS TO ODD. NUMBERED EXERCISES
Exercises XIV. D, pages 254-256
,")_
Answers were computed with four-place logarithms. Where correct
result (to proper number of significant digits) differs, it is given in ( ).
1. 5049.
3. 2.299.
5. 0.0002284 (0.0002285).
7. 2.487.
9. 0.1154.
11. 24.42 (24.43). 13. 121.8(121.9).
15. 0.05764. 17. 0.02650. 19. 3.175 (3.176). 21. 0.04303(0.04302).
23. 2.312 (2.311).
25. 0.05088(0.05086).
27. 24.26(24.25).
29. 4.618 (4.617).
31. 1.745.
33. 0.7810(0.7811).
35. 3.449.
37. 3.405 (3.404).
39. -7.273 (-7.272). 41. 0.9750(0.9754).
43. 0.9990.
45. 0.8058 (0.8059).
47. 1.276 (1.275).
49. 0.6314 (0.6312).
51. 0.9206 (0.9208).
53. 0.05066 (0.05067).
55. -2.465. 67. 1.103. 59. 70.03 sq. in. 61. 3.854 (3.852) cu. in.
63. 256.8 ft/sec.
65. 41.34 (41.35).
Exercises XIV. E, pages 256-257
1. 5.644.
3. 3.455.
5. -0.3903. 7. 1.213.
9. 2.140.
11. 9.427.
13. 0.6181.
15. 10.
17. 7.08.
b log a + d log c
19.
.
21. x = -0.481, Y = 2.10.
log c - log a
25. x = .0.540, y = 2.92.
23. x = -0.251, Y = -1.20.
(log a)2 + log b log c
(log a)2 + (log C)2
27. x = log a(log c - log b)' Y = ll)g a(log c - log b) .
/
Exercises XIV. F, pages 257-258
1. 4.
3. 3.
5. 64.
7. 2.
9. -j.
11. -t
13. 2.
23. Va.
25. -\I-;S / a.
15. 2h.
17. 5.
19. 1.
21. 1.
_br
27. vb. 29. x = any number if a = 1, x = 0 if a ~ 1 and ~O.
Exercises XIV. G, page 260
3. 1.465.
13. 2.219.
6. -2.561.
15. 2.012.
7. 4.248.
9. 3.900.
11. 2.296.
17. 1.215.
19. 0.6990.
Exercises XIV. H, page 262
1. 10, 1.371.
9. -0.7667.
3. 3.453.
11. 0.04767.
6. 1,2.
7. 0.5671.
Exercises XV. A, pages 266-267
Answers given first are from tables in book. Those in ( ) are ~. rect to nearest cent; they are omitted when book tables are adlMlliatil.
1. (a) $1268.30 ($1265.32), (b) $1268.20 ($1268.24),
(c) $1269.70 ($169.73).
3. $1109.64.
•
ANSWERS TO ODD-NUMBERED EXERCISES
5. $5()3.40 ($563.41). 7. $1856.52.
9. 12.
11. 3t.
13. 2}%.
15. 3%.
17. 3.7%.
21. (a) 4.7%, (b) 3.5%, (c) 2.3%. 23. 5446.49.
'95
19. 3.02%.
Exercises XV. B, pages 271-273
Answers given first are from tables in book. Those in ( ) are correct to nearest cent; they are omitted when book answers are adequate.
1. $6468.40 ($6468.41).
3. $4451.80 ($4451.82).
5. (a) $13,486.75 ($13,486.73), (b) $10,621.70 ($10,621.69).
7. $23,589.20 ($23,589.16).
9. $1204.09.
11. $1598.28.
13. $6605.70. 15. $108.18. 17. $555.92. 19. $9259.80 ($9259.79).
21. $916.16.
23. 2.63%.
Exercises XVI. A, pages 277-279
1. (a) 120, (b) 24.
9. (a) 729, (b) 504.
17. 12.
3. (a) 720, (b) 120.
11. (a) 360, (b) 3888.
19. 36.
5. 96.
13. 264.
7. 1.,689.
16•••
Exercises XVI. B, pages 281-282
1. 70.
9. (a) 20,475, (b) 37,128.
17. 14,700.
3. 56.
5. 300.
11. 1120.
13. 2970.
19. 1,010,142.
7. 2,598,960.
15. 450,450.
Exercises XVI. C, pages 282-284
1. (a) 5040, (b) 720. 3. (a) 94,050, (b) 110,814. 6. 325. 7. 105.
9. im(m - l)n(n - 1). 11. 144. 13. 10. 15. (a) 1440, (b) 3600.
17. 56.
19. 1080.
21. 350.
Exercises XVII. A, pages 288-290
1. b. 3. -{o. 5. (a) -io, (b) io, (c) io, (d) {-, (e) {, (f) {-, (g) -fo.
7. (a) $0.25, (b) $0.66i.
11. riih. 13. (a) i, (b)
15. -i-p;;
17. (a)!, (b) H, (c) N-ri.
19. (a) 0.1318, (b) 0.8682.
23. (a) 0.8945, (b) 0.1055.
t.
Exercises XVII. B, pages 293-295
1. i.
3. j.
6. 0.09316.
9.
-dss.
11.
-Jr;.
7. (a)
rh,
(b) }, (c)
-rho
13. }.
Exercises XVII. C, pages 297-298
1. (a) -l-t-rts, (b) 2fu.
3. (a) 0.1536, (b) 0.7373.
5. 0.008847.
7. (a) 0.9039, (b) 0.09224, (c) (UH7Si.
9. (a) 0.001295, (b) 0.0004317, (c) 0.9987.
•
396
ANSWERS TO ODD-NUMBERED EXERCISES
i
Exercises XVII. D, pages 298-300
1.
7.
11.
15.
17.
(a) ~, (b) A, (c) i.
3. $0.45.
6. (a)~, (b)
(a)~, (b) ...J-,jNp;, (c) 1f..f,fn.
9. 0.2036.
A $60/11, B $50/11.
13. $5.005.
(a) A, (b) 0, (c) {-, (d)
(c) i.
(a)
(b) ill, (c) ~.
19. (a)~, (b) -rh.
&.
i,
+H,
Exercises XVIII. 8, pages 306-308
9. 85,66~.
15. (4,-5,2).
23. Ct.-i.!).
1. 18.
3. -39.
5. -434.
7. -320.
11. a3
b3
c3 - 3abc.
13. (x - a)2(x
2a).
17. (-4,3,3).
19. (2,-1,-1).
21. (i-,i-,2).
+ +
+
Exercises XVIII. (, pages 315-316
1. 7.
11. 9.
3. 270.
13. 1080.
+ 3a).
6. (1 - a)3(1
9. -10.
7. 708.
"
Exercises XVIII. D, pages 323-325
1. x = -3, y = 6, z = -2, w = 4.
C; :I'r,
.*1 3. x = 7, y = -1, z = 3, W = 2.
5. x = 3, y = 0, z = -2, u = 1, v = -1.
',',} ,;-, ",';"7. x = c + j, y = -c, z = c, c arbitrary.
9. x = H, y = --!, z = H.
11. x = -c, y = -2, z = c, c arbitrary.
1.;1I
13.
17.
28.
27.
y and z arbitrary, x = -}( -3y + 5z + 4).
15. x = 5, y
x = 17, y = 13.
19. Inconsistent.
21. Inconsistent.
x= -3,y=5,z=2.
26. x:y:z= -1:-1:1.
No nontrivial solution.
29. x:y:z:t = 2: -1 :3: -2.
,
= 7.
--~
Exercises XIX. A_, pages 332-333
1. _3_ _ _
2_.
x-I
x - 4
7.
1
6(x
+ 3)
1
24(x
1
4Cx - 1)
1
6(x - 3)
13.
3.
+ 1)
1
8(x - 3)
1
4(x
+ 1)
9. _3_ _ _
2_. 11. _2_ _ _5_.
x- - 4 x + 2
x - 2
2x - 3
+
1
16. _1_
x - 2
1
3
12(x - 5)
2
3
1'1. - - .
x + 3
(x
3)2
19.
21. __3__ 2x - 5.
x-I
X2 + 1
23. _1_ _
x +3
+
5. _2_ _ _I__
x - 4
2x - 3
+
3
(x - 2)2
5
--+---
8x
2X2
8(x
X2 -
+ 4)
1
2x
, -. + 3
.
I
I
i
ANSWERS TO ODD-NUMBERED EXERCISES
x+4
1
25.
397
12(x 2
12(x - 2)
+ 2x + 4)
x +2
+ 3x + 1 .
29.
+ x2 - I .
x +1
x 2 - 2x + 4
(x 2 - 2x + 4)2
31. ~ _ _2_ + x - 4 .
X
X2 + 3
(X2 + 3)2
33. _4_ _
3x
+ 2x + 1
x - 2
x2 + X + 1
(x 2 + X + 1)2
35.5
+ 5x
+ 5x-1
2(x + 1)
2X2 + X + 1
(2x + X + 1)2
27 .
..!_
x2
2
37. __1_ _ _
8(x·
1
1)
8(x
1
+ 1)
4(x 2
1
+ 1)
+ 1)2
3
2
41. 4 + - - + -_'
x-I
x - 2
2(x 2
1
X
x
1
39. - - _ - x
.r~ + 1
(x 2 + 1)2
(X~ + 1)3
7
10
43. 2x - 3
3(x + 2)
3(x - 4)
+ ---
x+2
3(x 2 + X + 1)
2
3
1
47. x + 3 + - - - - - +
.
2x - 3
x - 3
(x - 3)2
45. 1
+
1
3(x -
1)
Exercises XX. A, pages 344-345
C means convergent, D means divergent.
1. C.
3. C.
5. D.
7. D.
9. C.
15. D. 17. C. 19. C.
21. D.
23. D.
11. C.
13. C.
11. C.
13. D.
Exercises XX. B, pages 350-351
1. C.
15. C.
3. C.
17. C.
5. C.
7. D.
9. D.
Exerci;es XX. C, page 352
1. -1 ~ x < 1.
7. -2 < x < 2.
13. -3 < x < 3.
19. a - I ~ x < a
3. -1
9. -1
~
x ~ 1.
~
x
~
x
< 3.
15. 1
+ 1.
~
1.
5. -1
11. -1
17. -4
< x < 1.
<x <1
< x < 16.
![m1-] 63 NOTE ON THE NUMBER OF LINEARLY INDEPEND](http://s1.studyres.com/store/data/012997862_1-d2f38d28dd32ef9691afaab3955d0a2a-150x150.png)
