Download MAT 110 Module 1 - West East University

MODULE
FOUNDATION MATHEMATICS- MAT 110
MODULE 1
COPPERBELT UNIVERSITY COLLEGE
Mathematics department
Copyright
© COPPERBELT UNIVERSITY COLLEGE 2009
COPPERBELT UNIVERSITY COLLEGE
BOX 20382
KITWE
ZAMBIA
Fax: +260 21 2239003
E-mail: [email protected]
Website: www.[Add website address]
Acknowledgements
The COPPERBELT UNIVERSITY COLLEGE wishes to thank the following for their
contribution to this module.
ZACODE
DODE
Mr. D Kambilombilo
Lecturer - Foundation Mathematics
Mrs. R. M. Mvula
Lecturer – Analytic Geometry & Calculus
Mr. P. Banda
Lecturer – Linear Algebra
Mr. B.P. Mwewa
Lecturer - Probability & Statistics
FOUNDATION MATHEMATICS- MAT 110
Contents
About this MODULE
6
How this module is structured ........................................................................................... 6
Module overview
8
Welcome to module 1 MAT 110 ....................................................................................... 8
Module 1 MAT 110 ........................................................................................................... 8
Is this module for you? ...................................................................................................... 8
Module outcomes .............................................................................................................. 8
Timeframe ......................................................................................................................... 9
Study skills ........................................................................................................................ 9
Need help? ....................................................................................................................... 10
Assignments .................................................................................................................... 11
Assessments ..................................................................................................................... 11
Getting around this Module
11
Margin icons .................................................................................................................... 11
Unit 1
13
SETS ................................................................................................................................ 13
1.0 Introduction ...................................................................................................... 13
1.1 Definition of a set. ............................................................................................ 13
1.1.2 Set notation and presentation ........................................................................ 14
1.1.3 Set operations ............................................................................................. 15
1.2 Types of sets..................................................................................................... 19
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FOUNDATION MATHEMATICS- MAT 110
1.2.1 The natural numbers.............................................................................................. 19
1.2.2 Integers ................................................................................................................. 19
1.2.3 Rational numbers .................................................................................................. 20
1.2.4
Irrational numbers… ....................................................................................... 170
1.3.0
Laws of sets ....................................................................................................... 22
1.4.0
Properties of real numbers ................................................................................. 23
1.4.1
Properties of addition and multiplication .......................................................... 24
1.4.2
Identity property ........................................................................................... 25
1.4.3
Closure property ........................................................................................... 25
1.4.4
Inverse properties ......................................................................................... 26
1.4.5
Zero-product property................................................................................... 26
summary .......................................................................................................................... 26
Unit 2
27
2. Introductory algebra .................................................................................................... 27
2.0 Introduction ..................................................................................................... 27
2.1 Functions ......................................................................................................... 27
2.1.1 Definition ……………………………………………………………27
2.1.2 Types of functions …………………………………………………….28
 One - to - one functions ………………………………………………28
 Composite functions …………………………………………………29
 Inverse functions ……………………………………………………..30
 Odd and Even functions ………………………………………………32
2.2 Quadratic Functions ……………………………………………………35
2.2.1 Methods of solving quadraticequations ........................................................ 35
2.2.2 Completing the square ................................................................................. 38
2.2.3 Quadratic formula . ....................................................................................... 41
2.2.4 Solutions to Quadratic equations .................................................................. 44
2.3.0 Polynomial functions .................................................................................... 49
2.3.1 Evaluation of Polynomials ........................................................................... 50
2.4.0 Binomial theorem ......................................................................................... 53
2.5.0 Mathematical induction ................................................................................ 61
Solutions to some Activities ............................................................................................ 66
Bibliography .................................................................................................................... 69
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FOUNDATION MATHEMATICS- MAT 110
About this module
MODULE 1 has been produced by COPPERBELT UNIVERSITY COLLEGE COPPERBELT
UNIVERSITY COLLEGE and structured in the same way as outlined below.
How this module is structured
The module overview
The module overview gives you a general introduction to the module. Information
contained in the module overview will help you determine:
 If the module is suitable for you.
 What you will already need to know.
 What you can expect from the module.
 How much time you will need to invest to complete the module.
The overview also provides guidance on:
 Study skills.
 Where to get help.
 Module assignments and assessments.
 Activity icons.
 Units.
We strongly recommend that you read the overview carefully before starting your study.
The module content
The module is broken down into two (2) units. Each unit comprises:
 An introduction to the unit content.
 Unit outcomes.
 New terminology.
 Core content of the unit with a variety of learning activities.
 A unit summary.
 Assignments and/or assessments, as applicable.
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FOUNDATION MATHEMATICS- MAT 110
Resources
For those interested in learning more on contents in this module, we provide you with a
list of additional resources at the end of this module these may be books, articles or web
sites.
Your comments
After completing this module we would appreciate if you would take a few moments to
give us your feedback on any aspect of this module. Your feedback might include
comments on:
 Module content and structure.
 Module reading materials and resources.
 Module assignments.
 Module assessments.
 Module duration.
 Module support (assigned tutors, technical help, etc.)
Your constructive feedback will help us to improve and enhance this module.
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FOUNDATION MATHEMATICS- MAT 110
Module overview
Welcome to module 1 MAT 110
This module is based on introductory concepts in Mathematics and is part of foundation
mathematics course.
Module 1 MAT 110
Is this module for you?
This module is intended for Diploma holder teachers of Mathematics that aim to further their
teaching skills.
.
Module outcomes
Upon completion of module 1 you will be able to:
i. Use and apply elements of sets properties,
ii. Represent Sets in form of a diagram.
iii. Manipulate
algebraic
properties
and
functions.
iv. Understand graphical representations of
functions.
Outcomes
.
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FOUNDATION MATHEMATICS- MAT 110
Time frame
The expected study time is 100 hours
How long?
Study skills
As an adult learner your approach to learning will be different to that from
your school days: you will choose what you want to study, you will have
professional and/or personal motivation for doing so and you will most
likely be fitting your study activities around other professional or domestic
responsibilities.
Essentially you will be taking control of your learning environment. As a
consequence, you will need to consider performance issues related to time
management, goal setting, stress management, etc. Perhaps you will also
need to reacquaint yourself in areas such as essay planning, coping with
exams and using the web as a learning resource.
Your most significant considerations will be time and space i.e. the time you
dedicate to your learning and the environment in which you engage in that
learning.
We recommend that you take time now—before starting your self-study—
to familiarize yourself with these issues. There are a number of excellent
resources on the web. A few suggested links are:
 http://www.how-to-study.com/
The “How to study” web site is dedicated to study skills resources. You
will find links to study preparation (a list of nine essentials for a good
study place), taking notes, strategies for reading text books, using
reference sources, test anxiety.
 http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student Affairs.
You will find links to time scheduling (including a “where does time
go?” link), a study skill checklist, basic concentration techniques,
control of the study environment, note taking, how to read essays for
analysis, memory skills (“remembering”).
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FOUNDATION MATHEMATICS- MAT 110
 http://www.howtostudy.org/resources.php
Another “How to study” web site with useful links to time management,
efficient reading, questioning/listening/observing skills, getting the most
out of doing (“hands-on” learning), memory building, tips for staying
motivated, developing a learning plan.
The above links are our suggestions to start you on your way. At the time of
writing these web links were active. If you want to look for more go to
www.google.com and type “self-study basics”, “self-study tips”, “self-study
skills” or similar.
Need help?
For any help contact the University College through:
Help
1. Copperbelt College of Education Secretary - Phone +260
212293003
2. email: [email protected]
3. The Mathematics HOD – 097 9190079 or 096 4024179 or
095 950084
4. District Resource Centre in your district –
All contacts during Office hours.
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FOUNDATION MATHEMATICS- MAT 110
Assignments
Two, Tutor marked assignments will be considered for assessment.
To be submitted to the university college through District Resource
Centres before or at stipulated times in the units.
Assignments
Assessments
There will be one Tutor marked test, at the end of this module, and
several self- marked tasks at the end of each unit.
Assessments
A three hour test will be administered at the University College
during a residential school as advised
Getting around this module
Margin icons
While working through this module you will notice the frequent use of margin icons. These icons
serve to “signpost” a particular piece of text, a new task or change in activity; they have been
included to help you to find your way around this module.
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FOUNDATION MATHEMATICS- MAT 110
A complete icon set is shown below. We suggest that you familiarize yourself with the icons and
their meaning before starting your study.
Activity
Assessment
Assignment
Case study
Discussion
Group activity
Help
Note it!
Outcomes
Reading
Reflection
Study skills
Summary
Terminology
Time
Tip
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FOUNDATION MATHEMATICS- MAT 110
Unit 1
SETS
1.0 Introduction
This Unit will apprise you of concepts on sets covered at the Mathematics Diploma level of your
study; however some of you (Science students) never covered this in your study. Since this level
of study is meant for both categories, we will take you through.
Upon completion of this unit you will be able to:
 Describe different sets.
 Use appropriate set operations.
Outcomes
 apply laws of sets
1.1. 1 Definition of a set.
Imagine your school undertakes a tour of Kafue Pational Park. During the tour you see an Owl, a
Zebra, Guinea Fowl, Whitestock, an Eagle, Impala, an Elephant, Mukwa Tree, Baobob tree and
Masuku tree.
Categorise the the things you saw during the tour.
How did you arrive at the categories?
Describe one of the cartegories you came up with.
In a few words, can you describe what a Set is?
Does your description fit with the following definition?
A set is a collection or list of well-defined objects called members or elements.
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FOUNDATION MATHEMATICS- MAT 110
1.1.2 Set notation and presentation
Sets are usually denoted by capital letters, whereas the elements represented by letters are
denoted by small letters. For instance, from the categories you made above, you would let A
represent animals you saw, B to represent Birds and C to represent trees as shown below:
A = {Elephant, Zebra, Impala}
B = {Guinea Fowl, Owl, Eagle, White stock}
C = {Mukwa, Masuku, Baobab}
The above way of presenting sets is called listing. The other two ways of set presentation are:

Set-builder notation form

Statement/ description
Can you now try to illustrate sets using set-builder notation and description.
For sets A, B and C, you would notice that they have limited number of elements and such sets
are said to be Finite. Sets which have unlimited number of elements are called Infinite.
Give two examples of infinite sets using any of the three methods of set presentations.
In summary, you should have appreciated the following:

use of capital letters to name sets

use of curly brackets in sets i.e. { }

Use of set-builder notation to represent sets. For example,
{x: x is a wild animal, x is found in Kafue National park}
 Describing sets. For instance,
A set of trees found in Kafue National park.
 Listing sets
A = {Elephant, Zebra, Impala}.
Attempt the activity below:
Activity 1.1
Identify finite and infinite sets from the following:
(i)
{x: x < 5, x  R}
(ii)
Set of persons on earth
(iii)
{1, 2, 3, 4, …}
(iv)
Set of lakes in Zambia
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FOUNDATION MATHEMATICS- MAT 110
Now, can you consider the following sets in reference to the Kafue National Park tour above:
(i)
The number of elements in set A and set B.
(ii)
Peter complained that he only saw the Zebra, can you describe Peter’s set?
(iii)
Jane and Mainza saw the same type of trees. What can you say about Jane and
Mainza’s sets?
(iv)
The Driver was left behind to look after the vehicle and other property, as the
team
used the Tour Guide for game viewing. Describe the driver’s set in terms
of animals
viewed?
(v)
Ms Joan saw only an Impala and a zebra during the tour. How would you
describe her set relative to the viewed animals?
Compare your solutions to the ones below
(a)
These are Equivalent sets, as they have equal number of elements.
(b)
It is a Singleton set as it has only one element.
(c)
These are Equal sets as they have identical elements
(d)
An Empty or Null set as it has no elements, denoted by { }, or 
(e)
A Subset as it is part of the other set, denoted by  .
The set containing All animals or Trees or Birds viewed is called a Universal set,
denoted by E, U or .
1.1.3 Set operations
You have already met division, multiplication and subtraction as operations on numbers in
arithmetic.
When these are performed on numbers they form new numbers. In sets also we have operations
that are performed on sets to form new ones. Some of these operations include:
(a)
Union
(b) Intersection
(c)
Complement
Union of a Set
In 2009 at Ndola Technical School for girls, there was an outbreak of two mysterious diseases,
Small pox and Swine flu. 17 were diagnosed with Small pox and 32 were diagnosed with Swine
Flu. It is given that there were 100 pupils in the school.
Would you establish how many pupils were diagnosed with the diseases?
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FOUNDATION MATHEMATICS- MAT 110
You should have realised that the total number of pupils that were attacked was 49 without
knowledge of additional information. To get this we should have added 17 and 32. This set is
called a Union set of pupils attacked by the diseases. This total number of students attacked by
the two diseases can be presented symbolically as:
n(Swine Flu  Small pox) = 49.
You could read this as ‘the union set of students with swine flu and Small pox is 49.’
Definition:
Let A and B be arbitrary non empty sets
then
Example One
Let
find
Solution:
Intersection set
After a careful examination was conducted, it was established that a total of 42 pupils
were attacked by either small pox or swine flu. Can you determine the number of pupils
attacked by more than one disease? I believe you noticed that at first we had 49 cases of
pupils who contracted the diseases while in the reassessment that followed, there were 42
cases. This meant that there was a reduction of 7 and it implied that the 7 had suffered
both diseases.
This set is called an Intersection set of pupils attacked by both diseases. The number of
students suffering from both diseases can symbolically be presented as:
n(Swine Flu  small pox) = 7.
You could read this as ‘the intersection set of students with swine flu and small pox is 7.’
You could read this as ‘the union set of students with swine flu and Small pox is 49.’
Definition:
Let A and B be arbitrary non empty sets
then
Diagrammatically
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FOUNDATION MATHEMATICS- MAT 110
Figure 1.1
Example two
Let
find
Solution:
then A and B are said to be Disjoint sets
Diagrammatically
Figure 1.2
Complement
(a) Determine number of those who were not attacked by swine flu but they belong to that
school?
To establish the number of those not affected by swine flue, we need to take a careful analysis of
the information. Taking into account the total number of pupils in the school less those attacked
by Swine Flu only, it can be shown that:
100 – 32 = 68
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FOUNDATION MATHEMATICS- MAT 110
The number 68 represents members belonging to the Complement set of those attacked by
Swine flu.
The information above can be summarised in the definition below.
We can express this symbolically as (Swine flu).
Definition:
Let
, then the complement of the set A is defined as
Example two
Let
and that
, list the elements of
.
Solution.
Lets begin by listing the elements in each of the sets:
Now from set E and set A we see that, the element 2 is in the universal set but not in set A. We
therefore write;
The complement of a complement is the set itself.
Consider the number of pupils for the set ((Swine flu)) = (100  32) = (68) = 32
This i the same as the number of pupils with swine flu only.
From the discussion above,
Can you now try to list the elements of set
Activity 1.2
Given the sets P = { x : 0 < x  3 , x  Z} and Q = { x : 2 < x  4 , x  Z },
where P and Q are subsets of M = { x : 0  x ,  10 x  Z }, Find
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FOUNDATION MATHEMATICS- MAT 110
(a) P  Q
(b) Q  P
(e) P'  P
(f) (Q )
(c) P
(d) P'  Q
You are advised to workout the question above before checking for solutions at the end
of this unit.
The above question can also be expressed in the form of Venn diagrams. Recall that a
Venn diagram is simply an illustration of sets in diagram form. For example, the sets P
and Q can be shown in a single diagram taking into account the universal set M.
Figure 1.3
The shaded part I believe could be linked to one of the parts to the questions above, can you
identify it. Using the same diagram can you equally shade the other parts of the solutions?
1.2 Types of sets of Numbers.
From your earlier study of sets of numbers, can you list the set of numbers and the symbol used?
Well, if you have forgotten don’t worry,
For a quick reminder, some sets of numbers from earlier levels of study are; set of natural
numbers, integers, rational number and irrational numbers.
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FOUNDATION MATHEMATICS- MAT 110
1.2.1
The natural numbers.
These are numbers we normally use for day today counting; they are represented by the symbol
N and N = {1, 2, 3, 4 …}.
1.2.2
Integers
When the set of natural numbers is extended to include 0, and negative whole numbers, we have
a set of Integers represented by the symbol Z and Z = {…,-2., -1, 0, 1, 2 …}.
1.2.3 Rational numbers
a
, where b  0 , then the set of these numbers is
b
1 2 
called rational numbers. Basically every integer is a rational e.g.  ,  the set is represented by
3 1 
Q.
If the number can be expressed in the form
1.2.4
Irrational numbers
Numbers that cannot be expressed in the form
numbers e. g
2,
a
, where b  0 , form a set called irrational
b
3 , , e.
Example three
Let A = {1, 2, 3, 4}, B = {2, 4, 5, 6}.
Then A  B = {2, 4}.
If A = {x: x is a positive integer, x is a multiple of 3} and
B = {x: x is a positive integer, x is a multiple of 4}
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FOUNDATION MATHEMATICS- MAT 110
then A  B = {x: x is a positive integer, x is a multiple of 12}.
Let A = {a, b, c,}, B = {d, g}.
Then A  B = .
Since A and B are disjoint.
However, A  B =  does not imply both A and B are empty.
Let A = {4, 5, 6},
B = {3, 4, 5, 6,}.
Then, A  B = A.
From the above example, A  B.
If A  B then A  B = A.
Given the universal set  and A, B as its subsets, the results above may be illustrated in
diagrammatic form as follows:
AB
AB
Figure 1.4
Figure 1.5
Figure 1.6
AB
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FOUNDATION MATHEMATICS- MAT 110
Figure 1.7
BA
You should know that if the universal set is not given, the rectangle may be omitted.
ve complement or difference
If P and Q are two sets, the set of elements do not belonging to Q but belong to P is
called the Relative complement or difference set denoted by
P – Q = {x: x  P and x  Q}
= {x: x  P and x  Q'}
= {x: x  (P  Q'}
P–Q
Similarly,
= (P  Q')
Q – P = (Q  P')  P – Q ≠ Q – P.
1.3.0 Laws of sets
After looking at specific examples we can now draw our attention to general laws that apply to
sets known as the Boolean laws of algebra.
1.
Idempotent laws
(i) P  P = P
2.
(ii) P  P = P
Commutative laws
(i) P  Q = Q  P
3.
(ii) Q  P = P  Q
Distributive laws
(i) P  ( Q  T) = (P  Q)  (P  T)
(ii) P  ( Q  T) = ( P  Q)  ( P  T)
4.
Associative laws
(i) ( P  Q)  T = P  (Q  T)
(ii) ( P  Q)  T = P  (Q  T)
5.
De Morgan’s laws
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FOUNDATION MATHEMATICS- MAT 110
6.
(i)
(P  Q)' = P'  Q'
(ii)
( P  Q)' = P'  Q'
Complement
(i) (P')' = P
7.
(ii) P  P' = ø
Property of the universal ε
(i) P  ε = ε
8.
(iii) P  P' = ε
(ii) P  ε = P
(iii) ε' = ø
Property of the empty set.
(i) P  ø = P
(ii) P  ø = P
(iii) ø' = ε
Several statements can be proved using the above laws. Let us look at one of such statements.
(a)
Prove by Boolean laws given that P  ( P'  Q) = P  Q
Proof: P  ( P'  Q) = (P  (P')  (P  Q) - By distributive law
= ø  (P  Q)
By complement law
= P  Q.
(b)
Show that
P  (Q – P ) = P  Q
To work out this question you need to simplify one side and make it similar to the other. In our
case we will simplify the left hand side to show the right hand side using the laws.
Using the left hand side of the statement, we proceed as follows:
P  (Q – P) = P  (Q  P ' ) , by the difference property
= ( P  Q)  ( P  P ' ) by distributive law
= ( P  Q)  ε
= P Q
(c)
Show that
(P'  Q')’ = P  Q.
Similarly you would show the above question using the given laws.
Here you will notice that it is also easier to prove the right side using the left hand side. We will
proceed as follows:
(P'  Q')’ = [( P ' ) '  (Q ' ) ' ] by distributing the law
= P Q
By De Morgan’s laws.
Activity 3 [Self marked assessment]
Using the laws of sets above, can you show the following:
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FOUNDATION MATHEMATICS- MAT 110
(a) P  ( P  Q)  P  Q
(b) P  Q  Q '  P '
(c) P  (Q  P)  P  Q
(d) P  ( A  Q ' )  P  Q '
(e) [( P  Q) '  P]'  Q  P '
(f) P  (Q  R)  ( P  Q)  ( P  R)
(g) [( P  Q)  ( P  Q ' )] ' 
ø
1.4.0 Properties of real numbers
1.4.1 Introduction
Having looked at different sets of numbers, you now need to explore further some
properties that apply to numbers.
Can you list some of the properties of numbers which you know?
Compare your list with what is listed below:
1.4.1 Properties of addition and multiplication
Commutative law
Study the following cases:
(i)
3 +7 = 7 + 3
(ii)
0.8 + 0.2 = 0.2 + 0.8
(iii)
64=46
(iv)
-4  10 = 10  (-4)
What have you noticed in the cases above?
You might have observed that when you add or multiply two numbers in one order the answer is
the same when you add or multiply them in the reverse order.
In general, if a and b are real numbers, then
a+b=b+a
This property is called commutative law of addition.
Similarly, a  b = b  a or ab = ba.
This property is called commutative law of multiplication.
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FOUNDATION MATHEMATICS- MAT 110
Activity 4 [Self marked assessment]
Now can you try to apply the same procedure to subtraction and division? Comment on your
findings.
Associative law
Try to work out the following:
(i)
3 + (4 + 8)
(ii)
(3 + 4) + 8
(iii)
0.6 x (0.5 x 1.2)
(iv)
(0.6 x 0.5) x 1.2
Comment on your answers.
You should have noticed that the answers for parts (i) and (ii) are the same and that of parts (iii)
are (iv) are the same too.
In general, if a, b and c are real numbers, then
a + (b + c) = (a + b) + c
This property is called associative law of addition.
Similarly, a  (b  c) = (a  b)  c or a(bc) = (ab)c.
This property is called associative law of multiplication.
Associative law
Suppose a, b and c are real numbers, then
a(b + c) = ab + ac.
This law is called the distributive law of multiplication over addition.
Now, take any three numbers and verify this property.
1.4.2 Identity property
Addition
Let a be any real number then
a + 0 = 0 + a = a.
This is called the identity property of addition. 0 is referred to as an identity element of addition.
Multiplication
Let a be any real number then
a  1 = 1  a = a.
This is called the identity property of multiplication.
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FOUNDATION MATHEMATICS- MAT 110
1 is referred to as an identity element of multiplication.
1.4.3 Closure property
Addition
Let a and b be any two real numbers, then
a + b is a real number.
Example: If a = 20 and b = 15 then
a + b = 20 + 15 = 35 which is a real number too.
Multiplication
Let a and b be any two real numbers, then
a  b = ab is a real number.
Pick any two real numbers and verify this property.
1.4.4 Inverse properties
Try to write the inverse of each of these numbers 2, 5 and -3.
From your list which one do you think is the additive inverse 2, of 5 and of -3?
You could have probably generated the following list:
For 2 the additive inverse is -2;
For 5 the additive inverse is -5;
For -3 the additive inverse is 3.
In general, if a is any real number, then
a + (-a) = (-a) + a = 0.
The additive inverse for any real number a is –a.
From your list of inverses above write down the multiplicative inverses of each of 2, 5 and -3.
I am sure you could have come up with the following answers:
For 2 the multiplicative inverse is 1/2
For 5 multiplicative inverse is 1/5;
For -3 the multiplicative inverse is -1/3.
In general, if a is any real number, then its multiplicative inverse is 1/a, where a ≠ 0.
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FOUNDATION MATHEMATICS- MAT 110
1.4.5 Zero-product property
If ab = 0, then a = 0 or b = 0 or both a = 0 and b = 0.
summary
In this unit you learned ways of presenting sets, the notations
used, different types of sets of numbers, some operations on sets,
laws used in sets and application of properties of sets.
Unit 2
Introductory algebra
2.0 Introduction
In this unit we are going to explore different aspects of algebra. Some of you may have already
met some of these aspects in your earlier studies at Mathematics Diploma level. Since some of
you did not, we will take you through. This unit will cover relations and functions, quadratic
functions, polynomial functions, Mathematical induction and Binomial expansions.
Upon completion of this unit you will be able to:
24
FOUNDATION MATHEMATICS- MAT 110
 solve quadratic equations using different methods
 draw and interpret graphs of quadratic and polynomial
functions
Outcomes
 factorise polynomial functions using various methods
 Apply Pascal’s triangle law in binomial theorem.
 Apply binomial theorem to real life situations.
2.1 FUNCTIONS
2.1.1 Definition:
A function is a correspondence between two sets, the domain and the range, such that for each
value in the domain there corresponds exactly one value in the range.
Consider the diagram below
This can also be written as f : x  y
Domain and Range of a function
The domain is the set of inputs for example, {p, q, r} while the range is set of out puts for
example, {1, 2, 3}.
2.1.2 TYPES OF FUNCTIONS
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FOUNDATION MATHEMATICS- MAT 110
ONE TO ONE FUNCTIONS
A function f : x  y is one-to-one if each image y
or f : x  y is one-to-one if for a,b
f(a) = f(b)
Y has exactly one x
X assigned to it .
X.
a=b.
A function f which is one-to-one is called an injective.
Example
Show whether or not the following functions f : x  y are 1:1
(i)
(ii)
Solution:
Suppose that there is a pair of numbers in the domain f, say a and b such that f(a) = f(b). If we
can show that a = b then we have shown that the function is 1:1
Let a ,b
x then f(x) is 1:1 if f(a) = f(b)
a = b.
f(a) =5a +2
f(b) = 5b +2
Therefore
f(a) =f(b)
meaning that
5a +2 = 5b+2
5a = 5b
i.e. a = b
Hence
is a 1:1 function.
ii)
f(a) = a2+3
f(b) = b2+3
therefore
f(a) =f(b)
meaning that
26
FOUNDATION MATHEMATICS- MAT 110
a2+3= b2+3
implying that
a2 = b2
but a =
b
Since a = b or a = -b
Therefore
is not 1:1 function.
Definition: A function f : x  y is said to be onto if every y
is the image of some x
.
Or f : x  y is onto if and only if the range of f is the entire co-domain of f.
A function which is onto is sometimes called surjective.
COMPOSITE FUNCTION.
Let f and g be functions, the composite function denoted b fog is given by
Similarly
Example:
Let
find (i)
(ii)
(ii)
Solution.
(i)
=
=
=
=
+5.
(ii)
(iii)
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FOUNDATION MATHEMATICS- MAT 110
Now can you try to do this one?
Let
find (i)
(ii)
Compare your solutions of
.
and
? to the one above. What can you deduce?
You can see that
Meaning that, function composition is NOT commutative.
INVERSE FUNCTION
From your basic knowledge of numbers, you have come across the word inverse.
For example multiplicative inverse of 2, is ½ . The additive inverse of 2 is -2.
In function, the inverse of the function takes a similar form.
Consider the function,
And also, the inverse of
, then the inverse of
is ,
is 4x
To find the inverse of a function, written as
.we simply express in terms of
.
For instance
then replace x by f(x) then f(x) by x.
In summary you follow the steps bellow.
Steps to follow when finding the inverse of the function.
(i) Let y = f(x)
(ii) Solve for x
(iii) Replace y by x
(iv) Replace x by
You can now try to apply the above steps to example below.
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FOUNDATION MATHEMATICS- MAT 110
Find the inverse of the following functions
i.
ii.
Did you manage to get them correctly? If you did well done, if you didn’t go through it slowly.
Definition:
A function
has an inverse
Consider the function, f and its inverse
if it is one- to- one.
given bellow,
Activity
1.
the domain of the function
2. The domain of the function
is {0,1,2,3,4,5.}. find the range.
is a subset of . Write down the largest possible set
which is a suitable domain.
3. Given that
and that
in their simplest possible forms.
express the composite functions
i. A function can only have an inverse if and only it
one function.
and
it is a one to
29
FOUNDATION MATHEMATICS- MAT 110
ii. As not all functions are one to one, not all functions have an inverse.
Activity.
Given the function
(i)
(ii)
,
,
(iii)
(iv)
find
(v)
(vi)
Hint:

To find the answers to

To find
first work out what’s in the bracket, that is
, using the
knowledge of composite function and then find the inverse of the answer you will find. Do
the same to

and
.
You can now try to compare the solutions you have found.
Did you find

use or apply the idea of composite functions
If the inverse of
?, this does not distort the above caution that
is
then

The situation only arises if the inverse of f is g.
ODD AND EVEN FUNCTION
DEFINITION
A function
is said to be even, if
A function is even if it is symmetric about the y axis. A function of the form
integer is even.
.
, where n is even
30
FOUNDATION MATHEMATICS- MAT 110
The graph above is an example of the graph of an even function
A function is said to be odd if
or [
symmetry of b180 degrees. A function of the form
such a function has a rotational
, where n is odd integer is odd.
The graph above is an example of the graph of an odd function
Examples.
Discus the functions bellow.
(i)
(ii)
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FOUNDATION MATHEMATICS- MAT 110
Solution.
If f(x) is odd then
for any real number a then f(a) = -f(a)
When f(a) = -f(a)
a)
Hence f is odd.
If f is even then f\(a) = f(-a) for any a.
Hence f is even.
Activity:
1. Find the inverse of the following functions.
(i)
(ii)
(iii)
2. Using all the questions in question 1, verify that fg = gf= x.
3. Let
(i) Show that
32
FOUNDATION MATHEMATICS- MAT 110
(ii) Show that
4. Given that
and
(i)
(ii)
(iii)
5. Wirte down the equations for
sets.
,
,
and
in the following function
(i)
(ii)
(iii)
(iv)
2.2 Quadratic Functions
2.2.0 Introduction
In this section, we will look at what quadratic functions and equations are and
different methods of solving quadratic equations.
Can you identify the differences between the functions given below?

f ( x)  4 x  5

f ( x)  x 2  x  3
Amongst the things you should have noticed, is the difference in the degree (highest power) of
the function. The function, f ( x)  4 x  5 is called a linear function while the
function, f ( x)  x 2  x  3 is a non- linear function.
Any function of a degree of two (2) is referred to as a quadratic function. This is written in
general as
f ( x)  ax 2  bx  c , where x is the variable and a, b, and c are real numbers with
33
FOUNDATION MATHEMATICS- MAT 110
a ≠ 0.
When f ( x)  ax 2  bx  c is equated to zero, that is, f ( x)  ax 2  bx  c  0 then it changes
into a quadratic equation.
Activity 1
Which of the following are examples of quadratic equations?
x2 = 7
x3  x2  0
2x – 1 = 0
2x + 3x -1 =0.
We will now discuss some methods used for solving quadratic equations.
2.2.1
Methods of solving quadratic equations
Factorisation method – Zero product property
You may recall that from the zero product property of real numbers that when you are
given two real numbers, M and N such that MN = 0, then it is either M = 0 or N = 0 or
both are zeros. The first method is the zero product property.
For example, to solve x 2  x  0 , you need to reduce this to the form MN = 0.
To do this we factorize the given equation as follows:
x2  x  0
x( x  1)  0 and this is in the form MN = 0.
It follows that x  0 or x  1  0
Therefore,
x  0 or x  1 are the factors of x 2  x  0 .
This method is what is referred to as the factorisation method.
Now attempt this question,
Solve for x: x 2  x  12
After having worked out the problem, compare your solution with the one given below.
We write in
x 2  x  12
standard form and factor the trinomial.
x 2  x  12  0
x 2  4 x  3 x  12  0
( x  4)( x  3)  0
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FOUNDATION MATHEMATICS- MAT 110
x4 0
x3 0
or
Apply zero product property
Solving for x
x4
x  3
or
Thus, the roots or solutions of x 2  x  12 are 4 and  3.
We can know whether these are the correct solutions or not by substituting the roots into the
original equation.
Do not begin the factoring process until the quadratic equation has
been in standard form. For example, to write
x 2  x  12
i.e. x( x  1)  12
and x  12
or x  1  12 is WRONG!!!!
Remember we must have factors on one side of the equation and zero on the other side in
order to apply the zero product property.
Let us now look at the following examples.
Example 1
Write each of the following equations in standard quadratic form, ax 2  bx  c  0, and then
solve:
(a). ( x  3)( x  10)  x  10
(b).
1
6
 2
2
n3 n 9
Solution
(a). To solve this equation, we will proceed as follows:
( x  3)( x  10)  x  10
Expand the brackets
x  13 x  30  x  10
2
x 2  14 x  40  0
Writte n in standard form
( x  4)( x  10)  0
Factoring the trinomial
x  4  0 or x  10  0
Apply zero product property
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FOUNDATION MATHEMATICS- MAT 110
x4
x  10
or
Thus the roots of this equation are 4 and 10. You should check both roots.
(b) To find the solution of this quadratic equation we begin by factorising
n2 – 9, which factors to (n – 3)(n + 3).
1
6
 2
2
n3 n 9
1
6

2
n  3 (n  3)( n  3)
1(n + 3) – 6 = 2(n – 3)(n + 3) Multiply both sides by (n – 3)(n + 3)
n – 3 = 2n2 – 18
Simplify both sides
2n – n – 15 = 0
2
Write in standard form
2n2 – 6n + 5n – 15 = 0
2n (n – 3) + 5(n – 3) = 0
(n – 3)(2n + 5) = 0
n–3=0
Factor
or 2n + 5 = 0
n=3
Apply zero product property
or n = - 5/2
(n = 3 is invalid. Do you know the reason why 3 is an invalid solution?)
Thus the only root of this equation is –5/2.
When you have an expression such as (x – 3)(x – 10) = x – 10 it is wrong to divide both
sides by a variable expression ( in this case (x -10)) because you may lose one or more of
the roots.
Activity 2 (Self marked activity)
Solve the following quadratic equations by the factorisation method:
36
FOUNDATION MATHEMATICS- MAT 110
1.x 2  2 x  8
2.x 2  48  13 x
3.6 y 2  12 y
4.3t 2  t
5.n(2n  3)  5
6.( x  4)( x  6)  x  6
n 1 4

5
n
1 1 x 1
8.  
x 6 x 1
7.
2.2.2 Completing the square
Can you attempt solving the equation x 2  16 using the method above.
If correctly solved the solutions should have been x = 4 and x = -4
Can we look at this approach of solving the same equation.
You can express the equation as x( x)  16 , which shows that the same number has to multiply
itself.
To get the number which is multiplying itself you have to apply the square root on both sides.
x2 
16
And it shows that x = 4 or x = -4.
In this case x is a perfect square.
To solve questions of this form you need to reduce the equation to a form that is a perfect square.
For instance, observe the result of this expansion ( x  3) 2
( x  3) 2 = ( x  3)( x  3)  x 2  3x  3x  9
And this simplifies to x 2  6 x  9 .
To solve an equation such as x 2  6 x  16  0 you need to express it in perfect square form to
ease the calculation.
x 2  6 x  16  0 This can be expressed as
x 2  6 x  16 In order to obtain the perfect square of the expression x 2  6 x we compare this
expression with x 2  6 x  9 . You will realise that the difference between the two is 9. So we rewrite the
equation as x 2  6 x  9  16  9.
37
FOUNDATION MATHEMATICS- MAT 110
And this can now be solved easily by the similar approach as x 2  16 .
Before this stage one can realise that the equation can be expressed in the form
6
6
x 2  6x  
  16  
 .
 2 
 2 
2
2
So this can be solved as ( x  3) 2 = 25
( x  3) 2  25
( x  3)  5
x  35
x8
or
x  2
This approach requires that you reduce the left hand side into a perfect square. You can achieve this by
adding a constant to both sides of the equation x 2  bx  c . This now makes the left hand side a perfect
trinomial.
Note that if we add the square of half the coefficient of x, or (b / 2) 2 to the left hand side of the equation,
we obtain
b
x  bx   
2
2
2

b

x  
2

2
 Perfect square trinomial
After writing the left hand side of the equation as a perfect square, we then apply the square root property
to obtain the solutions. Let us illustrate this procedure using the next examples.
Example
Solve each quadratic equation by completing the square.
(a) x2 + 4 = 8x
(b) 3t2 + 4t = 2.
Solution:
You should remember that before completing the square we must always write the equation in the form
x2 + bx = c.
Thus we proceed as follows:
x2 + 4 = 8x
x2 – 8x = - 4
Write in the form x2 + bx = c.
x2 – 8x + 16 = - 4 + 16
Complete the square by adding 16 to both sides which is
38
FOUNDATION MATHEMATICS- MAT 110
 8
(x – 4) = 

 2 
2
2
( x  4) 2  12
x = 4 2 3
x = 4 +2 3 or 4–2 3
(b)To write 3t2 + 4t = 2 in the form x2 + bx = c, we divide both sides by 3, then
3t 2  4t  2
4
2
t
3
3
4
4
2 4
t2  t   
3
9 3 9
t2 
Divide both sides by 3
Complete the square by adding
4
to both sides
9
2
10
 2
t   
3
9


t
Factor the perfect square trinomial
2
10

3
9
Apply the square root property
2
10
-2

Add
on both sides and simplify t he radical
3
3
3
 2  10
t
Combine the fractions
3
- 2  10
 2  10
Thus the roots are
 0.387 and
 1.72.
3
3
t
Activity 3. Now try the following activity to determine your understanding of solving quadratic
equations using the method of completing the square.
1. x2 + 12x = 12
4. x2 – 10x + 16 = 0
2. n2 – 4n + 8 = 0
5. x2 + 3x = 10
3. x2 = 3x + 9
6 . 3n2 – 5n – 2 = 0
2.2.3 Quadratic formula
The third method that we can use to solve quadratic equations is by using the quadratic formula.
The method of completing the square enables us to derive the formula that can be used to solve any
quadratic equation that is written in standard form. Beginning with ax 2  bx  c  0 , we proceed as
follows;
39
FOUNDATION MATHEMATICS- MAT 110
ax 2  bx  c  0
ax 2  bx  c
subtract c from both sides
b
c
x
divide both sides by a
a
a
b
b2
b2
c
x2  x  2  2 
a
a
4a
4a
x2 

(Half of b ) 2
a
{x 
b 2 b 2  4ac
Factor the perfect square trinomial and add fractions on the right-hand side.
) 
2a
4a 2
x
b
b 2  4ac

2a
4a 2
x
b
b 2  4ac

2a
2a
Apply square root property
Simply the radical expression
b
b 2  4ac
x

2a
2a
Add –b/(2a) to both sides.
This last equation presents the quadratic formula that can be used to find the roots of any quadratic
equation that is written in standard form.
 If ax 2  bx  c  0(a  0), then
 b  b 2  4ac
x
2a
Example: Solve the quadratic equation x 2  6 x  16  0 .
Solution: You may notice that this equation is already in standard form with
a = 1, b = -6, c = -16.
 b  b 2  4ac
So, x 
2a
40
FOUNDATION MATHEMATICS- MAT 110

 (6) 
 62  4(1)(16)
2(1)

6  36  64
2

6  10
2
= 8 or -2
 The solutions are 8 or -2.
Activity 4. Self marked assessment
(i) x 2  2 3x  2  0
(iii)
(ii)  x 2  6 x  14  0
1 2
x  3x  2  0
2
Note that in the quadratic formula the quantity b 2  4ac is called the discriminant. It indicates the
nature of the roots of the quadratic equation ax 2  bx  c  0 , when a, b and c the real numbers. We
consider three cases.
Case 1
If b 2  4ac  0 then
roots:
b 2  4ac is a real number and the quadratic formula gives two distinct real
 b  b 2  4ac
2a
and
 b  b 2  4ac
2a
If a, b and c are rational numbers and b 2  4ac is the square of the rational numbers, then these two
distinct real roots are rational numbers . Otherwise, the roots are irrational numbers.
Case 2
If b 2  4ac  0, then b 2  4ac are pure imaginary numbers and the quadratic formula gives two
complex conjugate roots. That is, it has no real roots.

b
b 2  4ac
b
b 2  4ac
and 


2a
2a
2a
2a
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FOUNDATION MATHEMATICS- MAT 110
Case 3
If b 2  4ac  0, then

b 2  4ac is zero and the quadratic formula gives only one real root:The quantity
b
is called a double root or repeated root of multiplicity two.
2a
Now let us summarise the nature of roots of the quadratic equation
ax2 + bx + c = 0 where a, b and c are real numbers.
EQUATION
DISCRIMINANT
b2- 4ac = (-3)2 – 4(1)(7)
x – 3x + 7 = 0
= 9 – 28
2
= - 19
b2- 4ac = (-12)2 –
4(9)(4)
9x2 –12x + 4 = 0
NATURE OF ROOTS
No real solution
( complex conjugate
roots)
One real repeated
solution
= 144 – 144
=0
b2- 4ac = (5)2 – 4(2)(-3)
2x2 + 5x – 3 = 0
= 25 + 24
Two distinct real
solutions
= 49
Activity 5.
Can you determine the nature of roots for the equations
1. x 2  4 x  5  0
2. x 2  4 x  5  0
3. 2 x 2  4 x  35  0
5.  2 x 2  4 x  5  0
2.2.4 Solutions to Quadratic equations
If
 and  are the roots of the quadratic equation ax2 + bx + c = 0, then we can write
(x -
 )(x -  ) = 0
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FOUNDATION MATHEMATICS- MAT 110
i.e. x2 - x  -
x+  =0
or x2 – x(  +  ) +
  =0
Let us now compare this last equation with ax2 + bx + c = 0,
We see that – (  +  ) =
and that
  =
(i.e. x2 +
bx c
  0 ).
a a
b
b
or (  +  ) = a
a
c
.
a
Let us now look at another way of establishing the relationship between the roots and the coefficients of a
quadratic equation. This time we will use the two roots generated by the quadratic equation. Let them (
the two roots) be x1 and x2. Thus we have
 b  b 2  4ac
x1 
2a
and
 b  b 2  4ac
.
x2 
2a
Now let us consider the sum and product of the two roots.
Sum x1 + x2 =
 b  b 2  4ac
 b  b 2  4ac
 2b
b

+
=
2a
a
2a
2a
  b  b 2  4ac    b  b 2  4ac 
 

Product (x1)(x2) = 

 

2
a
2
a

 

2
2
2
2
b  (b  4ac) b  b  4ac
=
=
4a 2
4a 2
=
4ac
4a 2
=
c
.
a
We ca now state that if  and  are the roots of the quadratic equation
ax2 + bx + c = 0, then,
   ( the sum of roots)  
and  ( the product of roots ) 
b
a
c
.
a
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FOUNDATION MATHEMATICS- MAT 110
These relationships provide us with ways of checking potential solutions when solving quadratic
equations.
Now let us look at some examples involving the sum and product relationships.
Example1. If  and  are the roots of the equation x 2 - 3x - 3 = 0 state the values of (   ) and 
and find the values of
(b) (   ) 2
(a)  2   2
(c)  3   3
Solutions: If we compare x 2 - 3x - 3 = 0 with ax2 + bx + c = 0, we see that a = 1, b = -3 and c = -3.
Therefore,     
b
c
 3 and    3 .
a
a
(b) (   ) 2
(a)  2   2
= (   ) 2  2
= (3)2 – 2(-3)
=  2  2   2
= (  2   2 )  2
=9+6
= (   ) 2  2  2
= (3)2 – 2(-3) –2(-3)
= 15
=9+6+6
= 21.
 3   3  (   ) 3  3 2   3 2
 (   ) 3  3 (   )
(c)
 (3) 3  3(3)(3)
 27  27
 54.
Example 2: If  and  are the roots of the equation x 2 + 3x - 5 = 0, find the quadratic equation that has
roots of
44
FOUNDATION MATHEMATICS- MAT 110
(a)   2,   2
(b)
1
1
,
 1  1
(c)
1

2
,
1
2
Solution: From the given equation, we see that a = 1, b = 3 and c = -5. The sum of the roots which is
 
b
3
c 5
   3 and the product of roots which is   
 5 .
a
1
a
1
Now we need to find the sum and product of the required roots and then write down the required
equation.
(a) Sum of new roots(given) = (  2)  (   2)
= (   )  4
= ( 3)  4
=  7.
Product of new roots = (  2)(   2)
=   2  2   4
=   2(   )  4
= (5)  2(3)  4
= 5.
Now since in the required equation the sum of roots 
write down the required equation as
(b) sum of new roots =
b
c
 7 , and the product of roots  5 , we can
a
a
x2+ 7x +5 = 0.
1
1
(   1)  (  1)


 1  1
(  1)(   1)
=
=
=
1
7
=
1
.
7
(   )  2
      1
(3)  2
(5)  (3)  1
1
1
1
 1  1 
 

 .

7
   1    1        1 (5)  (3)  1
Product of roots 
1
7
Therefore the required equation is x2 – ( ) x 
1
 0 or 7x2 –x +1 = 0.
7
45
FOUNDATION MATHEMATICS- MAT 110
(c)
Sum of new roots =
1
1 2   2


 2 2
 2 2
=
(   ) 2  2
=
 2 2
(   ) 2  2
(  ) 2
(3) 2  2(5)
=
(5) 2
=
 1  1

2  2
   
Product of roots = 
19
.
25

1
 = 2 2
  
=
1
() 2
=
1
( 5) 2
=
1
.
25
Now the required equation can be written as
x2  (
19
1
)x 
0
25
25
or 25x2 –19x + 1 = 0.
Generally, if any two roots of an equation,  and  are the roots of a quadratic equation, then we can
write
(x -  )(x -  ) = 0
x2 -  x -  x +  = 0
or x2 – (   )    0
From this last equation we can conclude that any quadratic equation can be expressed as
x 2  (sum of the roots)x  (product of the roots)  0.
Now it is time for you to try out what we have discussed in this section by way of the exercise that
follows.
46
FOUNDATION MATHEMATICS- MAT 110
Activity 6.
1.
2.
State (i) the sum and (ii) the product of the roots of each of the following quadratics:
(a) x 2 + 9x + 4 = 0
(b) x 2 + 2x – 5 = 0
(c) x 2 + 7x + 2 = 0
(d) x 2 + 9x – 3 = 0
(e) 2x 2 - 7x + 1 =0
(f) 7 x 2 + x – 1 =0
(g) 3x 2 + 10x – 2 =0
(h) 5x 2 + 10x + 1 = 0
In each part of this question, you are given the sum and the product of the roots of the quadratic. In
each case, find the quadratic equation in the form ax 2 + bx + c = 0 with a, b and c taking integer
values.
(a)
(b)
(c)
(d)
Sum
-3
6
7
-
2
3
Product
-1
-4
-5
-
7
3
-
(e)
(f)
5
2
-
-2
(g)
(h)
-1
3
2
-
1
4
-5
-
1
3
2
3
1
2
3. If  and  are the roots of the equation x 2 + 11x + 2 = 0 find the values of
(b)   
(a) 
(a)   
2
(e) 3  (    )
(d)  2 +  2
(f)  2  +  2
4. If  and  are the roots of the equation. 2x 2 - 5x + 1 = 0 find the values of
(a) 
(b)   
(c)  2 + 3  +  2
5. If  and  are the roots of the equation 6x 2 + 2x – 3 = 0, find the values of
(a)


+


(b)
1
2
+
1
2
(c) 2
2
1


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FOUNDATION MATHEMATICS- MAT 110
(d)
1

-
1

-
1
3
(e)  3 +

(f)
1

3
+
1
3
6. If the square roots of ax 2 + bx – c = 0 differ by 3 show that b 2 =9a 2 + 4ac
7. If  and  are the roots of the quadratic equation x 2 - 4x – 2 = 0 find the quadratic equation
that has roots of:
1 1
(a)   2,   2
(b)
(c)  2 ,  2
(d)  2  ,  2 
(e)  3 ,  3
(f)
(g)   2 ,   2

(h)
,
 
1

2
,
1
2
 
,
 
2.3.0 Polynomial functions
Definition.
A polynomial function in x is an expression involving powers of x, normally arranged in descending or
ascending powers. The degree of the polynomial is given by the highest power of x occurring in the
expression.
For instance 3 x 4  6 x 3  2 x  1 is a polynomial of the 4th degree
5 x 3  4 x 2  x  3 is a polynomial of the 3rd degree.
Thus from the foregoing discussion, you can see that quadratic equations are also polynomial functions of
2nd degree.
In general polynomial functions in x are functions of the form f ( x)  ax n  bx n1  .....  k
where a, b, k are real numbers and n is a positive integer.
2.3.1 Evaluation of Polynomials
For the quadratic function f (x ) = x 2  5 x  6 , ( x  3) and
2, then f (x ) = 0.
( x  2) are factors. Now if x = 3 and x =
f ( x)  7 x 5  2 x 2  x  6 would require substituting 2 in all the terms of f (x) individually and getting
the sum.
f (2)  7(2) 5  2(2) 2  2  6 and this gives us
48
FOUNDATION MATHEMATICS- MAT 110
228. Hence the solution for f (2)  228
Let f (x ) be a quadratic f (x ) = x 2  5 x  6 then f (x ) has a factors ( x  3) and
then f (x ) = 0 and also when x = 2.
( x  2) . If x = 3
It can be observed that when f (x ) = 0 then the substituted value are roots for the function . for
polynomials with degree greater than two would not be easy to find factors. For example
f ( x)  3 x 3  2 x 2  x  6 .
Now we can extend the method of factorising quadratic functions to polynomial functions as follows.
To find factors for f ( x)  3x 3  2 x 2  x  6 , we assume f (x ) = 0 and find roots for the polynomial
When we test for x = -1, or 1 or any other value, until f (x ) = 0, then the
We would have
f (1)  3(1) 3  2(1) 2  1  6  3  2  7  8
Since f (x )  0 , then x +1 is not a factor.
f (1)  3(1) 3  2(1) 2  1  6  3  2  1  6  0
Since f (x )  0 , then x 1 is not a factor.
This would now mean dividing f ( x)  3x 3  2 x 2  x  6 by x 1 and further subject what remains to
the same procedure as we need to find the other factors.
This is done by either continuing by checking for possible or by a method
called Synthetic division , first let us show the long division method
Long division method
Can you follow through this division:
f ( x)
 ( x  1) 3x 3  2 x 2  x  6
x 1
Using the long division procedure as in ordinary numbers, we have
( x  1) 3x 3  2 x 2  x  6
3x 2  5 x  6
( x  1) 3x 3  2 x 2  x  6
- (3x 3  3x 2 )
(5x 2  x)
- (5x 2  5 x)
6x - 6
- (6x - 6)
0
49
FOUNDATION MATHEMATICS- MAT 110
We again subject the f ( x)  3x 2  5 x  6 to further factorization if it has factors otherwise this could be
expressed as f ( x)  ( x  1)(3x 2  5 x  6)
Can you attempt dividing
f ( x)  x 3  x 2  11x  12 by x – 2 and compare
your solution to what follows below:
x 2  3x  5
x  2 x 3  x 2  11x  12
x3  2x 2
 3 x 2  11x
 3x 2  6 x
 5 x  12
 5 x  10
2
This implies that x 3  x 2  11x  12  ( x  2)( x 2  3x  5) +2 where x-2 is the divisor, x 2  3 x  5 is
the quotient, x 3  x 2  11x  12 is the dividend and 2 is the remainder. This method is called the long
division.
In the example above, the divisor is a first degree polynomial of the form g(x) = x-c. In all such cases, the
remainder will either be the zero polynomial or it will have degree zero. In other words, the remainder
will be a constant. This particular type of division can be carried out by a short cut method called
synthetic division
Synthetic division
In order for you to understand what is involved in using this method, let’s work the example above using
synthetic division as follows:
Procedure
Write the coefficients of each term of the dividend in a row.
For x 3 the coefficient is 1
For x 2 the coefficient is 1
For x the coefficient is -11 and of course the constant itself 12.
2│1 1 -11 12
_____2__6_-10____
1 3 -5 R 2
50
FOUNDATION MATHEMATICS- MAT 110
Thus the quotient polynomial is x 2  3 x  5 and the remainder is R = 2
that can also be arrived at by substituting 2 in f ( x)  x 3  x 2  11x  12 :
f (2)  2 3  2 2  11(2)  12  8  4  22  12
f (2)  2
We can generally summarise the discussion as follows;
i. If f (x ) is a polynomial function and c is a constant, then f (c )  R , where R is a
remainder upon division of f (x ) by x – c . This is what is referred to as the
Remainder Theorem
ii. If f (x ) is a polynomial function and c is a constant, then f (c)  0 , if and only if x
– c is a factor . This is what is referred to as the Factor Theorem.
Can you try to workout the following:
Activity 7
1. Factorise the given polynomials
i. f ( x)  2 x 4  5x 3  15x 2  10 x  8
ii. f ( x)  6 x 3  5x 2  34 x  40
2. Determine the reminder when f (x ) is divided by g (x )
a.
f ( x)  5 x 3  4 x 2  3 x  6
: g ( x)  x  2
b.
f ( x)  3x 3  11x 2  10 x  12
: g ( x)  x  3
c.
f ( x)  9 x 4  6 x 3  4 x  5
: g ( x)  3x  1
3. Find the value of k given that when 2 x 3  2kx 2  3x  2 is divided by x  2 the remainder is 40
4. Find the value of k such that x +5 is a factor of x 3  kx  125
5. Find the value of k such that x 
1
is a factor of 3 x 3  x 2  kx  5
3
2.4.0 Binomial expansions
Introduction
You have already met expressions composed of two terms connected by + or  . Such expressions are
called binomials. In this section you are going to learn how to find expansions of binomials raised to
positive integers using Pascal’s triangle. Additionally, you will be exposed to the binomial theorem,
51
FOUNDATION MATHEMATICS- MAT 110
binomial series and their respective applications. Among the many examples of binomials are the
following: 2x + y, 4a – 2b, 1+y, 3x-4y, etc.
Binomial expansion for a positive integral power
Pascal’s triangle
Let us consider the expansion of a + b raised to a few positive integers, including zero:
(a + b)0 =
1
(a + b)1 =
a+b
(a + b)2 =
a2 + 2ab + b2
(a + b)3 =
a3 + 3a2b + 3a b2 + b3
(a + b)4 =
a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 =
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
…
…
…
…
…
…
…
Now, study the coefficients of each of the terms in the expansions. Do you recognise any pattern? Your
pattern should look like the one below:
1
1
1
1
1
1
1
1
2
3
4
5
3
6
10
1
1
4
10
6
15
20
… …
…
…
1
5
15
…
1
6
1
…
…
This pattern forms a triangular array of coefficients called Pascal’s triangle and is named after Blaise
Pascal (1623 – 1662), a French mathematician.
Activity 1
Consider the following expansion of (a + b)4
52
FOUNDATION MATHEMATICS- MAT 110
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Upon close examination of this expansion, what do you notice about:
i. the sum of exponents in each term?
ii. the number of terms?
iii. the exponents of a and b as they progress in the expansion?
Did you observe the following results?
i.
that the sum of exponents is always 4
ii.
that there are 4+1=5 terms in the expansion.
iii.
that the exponents of a are decreasing by one unit while those of b are increasing by one unit.
Arising from the observations above, we can generalise that in the expansion of (a + b)n , where n  Z  :
the sum of exponents in each term is equal to n.
the expansion has n+1 terms.
the exponents of a reduce by one as one moves from left to right, while those of b increase by
one as one moves from left to right.
Example: Use Pascal’s triangle to find the expansion of 2 x  3 y 
3
Solution:
From the Pascal’s triangle, you take note that the coefficients for the expansion are:
1 3 3 1.
Therefore, 2 x  3 y   (1)( 2 x) 3 (3 y ) 0  (3)( 2 x) 2 (3 y )1  (3)( 2 x)1 (3 y ) 2  (1)( 2 x) 0 (3 y ) 3
3
= 8x 3  24 x 2 y  54 xy 2  27 y 3
Activity 2: (Self check). This activity is intended to give you practice in using Pascal’s triangle.
Expand the following using Pascal’s triangle:
(i) (x + 2y)
4
(ii) (1 – z)
7
 x 2
(iii)   
2 x
4
You may have realised that it is not easy to use Pascal’s triangle to expand binomial expressions when the
exponent is large. For this reason, you need an alternative method which is suitable for cases where the
exponent is ‘large’. However, to start with, let us go through the following fundamentals:
53
FOUNDATION MATHEMATICS- MAT 110
Factorial notation
Given that n  Z  , we define n! read as n factorial, by n! = n (n – 1) (n – 2) (n – 3)… (3)(2)(1).
Additionally, 0! =1! = 1 and Proof of this is beyond the scope of this discussion.
In words, the factorial of a positive integer n is the product of n and all smaller positive integers. For
instance, 5! = 5.4.3.2.1 = 120.
Binomial coefficients
Do you still remember the Pascal’s triangle? Interestingly, the entries in a Pascal’s triangle can be found
by using the expression:
n
n!
  
, where n is the power of a binomial, r  0,1,2,........, n and
 r  r!(n  r )!
n
  is a binomial
r
coefficient.
At times, you will be required to find a specific term e.g. k th term, in the expansion of a given binomial
raised to power n . The value of r for the k
th
term is given by r = k – 1. e. g. a third term in an expansion
will have a value of r = 3 – 1 = 2.
 5
5!
5.4.3.2.1
5 .4
Using part of the knowledge acquired here, observe that   
=
=
=10.
 3  3!(5  3)! 3.2.1.2.1 2.1
Activity 2: Evaluate the following without use of a calculator:
7
 4
5
1.  
5
2.  
0
3.  
5
 
 
n
4.  
n
 
BINOMIAL THEOREM
You are now ready to learn of the binomial theorem which is a suitable way for finding expansions of
binomials with ‘large’ powers. Given that n  Z  ,(a + b)n can be expanded as follows:
n
n
n
n
n
 
 
 
 
 
(a + b)n =   an +   an-1b +   an-2b2 +   an-3b3 + … ……..+   bn
0
1
2
3
n
n
n
n
 
 
 
= an +   an-1b +   an-2b2 +   an-3b3 + … + bn
1
2
3
54
FOUNDATION MATHEMATICS- MAT 110
= an + nan-1b +
n(n  1) n-2 2 n(n  1)( n  2) n-3 3
a b +
a b + … + bn
2!
3!
n
This is referred to as the binomial theorem and it can also be briefly stated as (a + b)n =
r o
The symbol

n
  r a
nr
 
br .
r
n
implies summing up and   a n  r b is the general term for the expansion in the
r
 
binomial theorem.
Example: Use the binomial theorem to find the expansion of (x + 2)5
5
5
5
5
 
 
 
 
Solution: (x + 2)5 =   x5-0 .20 +   x5-1 21+   x5-2.22 +   x5-3.23
0
1
2
3
5
 4
5
5
+   x5-4.24 +   x0.25
Using your knowledge of factorials learned earlier, you will notice that the right hand side reduces to x5
+ 10x4 +40x3+ 80x2+80x+32, and this is the required expansion.
Activity 3:
Use the binomial theorem to expand and simplify the following:
(i) (x + 3y)3
(ii) (2x – y)4
(iv) (2x – y)5
(v) (3x – 4y)4
(iii) (2a + b2)3
Example:
Find the 7th term in the expansion of (4x – y3)9.
n
Solution: Recall that in the expansion of a  b  , n  Z  , each term is of the form:   an – rbr . We
r
n
 
th
shall now use this expression to calculate the 7 term of the expansion:
You should be able to remember that the value of r for the 7 th term is given by r =7-1= 6. Now,




9
comparing a  b n and 4 x  y 3 , shows that n = 9, a = 4x and b = -y3 .
Consequently, you will notice that the required 7 th term is given by:
55
FOUNDATION MATHEMATICS- MAT 110
9!
 9
  (4x)3(-y3)6 =
(4x)3(-y3)6
6!3!
6
= 84.64x3y18
= 5376 x3y18.
Activity 4: Now try to work out these problems:
Find the terms indicated, in the expansions of the following:
1. (1 + 2x)10 ; 6th term 2. (1 – 5x)7 ; 4th term 3. 4  14 x  ; middle term
6

4 . 2x 
2
1
3x
;5
9
th
term
5. (a +
1
2a
)9 ; 5th term.
6.
2  2x 12 ;
middle term
Finding an independent (constant) term
Just like we calculated for specific terms of binomial expansions, we can also compute a term
independent of a variable (generally called the constant term). To help you understand how this can be
done, consider the examples below:
Examples:
Find the terms independent of x for the following expressions:
1

1.  x  
x

4
2

2.  x 2  
x

6
Solutions:
1

1.  x  
x

4
56
FOUNDATION MATHEMATICS- MAT 110
4
1 
1
Are you able to recognise that n = 4, a = x, b =
in  x   ? What do you think is the value of r? You
x
x 
may have noticed that the value of r is not obvious at this stage. Nevertheless, you still need to find
r
1
 4
 4
this value so as to use it in the general term   x4 – r  x  =   x4 – r . x – r =
 
r 
r 
 4 4 – 2 r
  x
r 
If the last expression above is to be independent of x, the following should hold:
x4 – 2r = x0 i.e. 4 – 2r = 0  r = 2. With this value of r, you can now compute the required independent
term as follows:
 4
4!
Term independent of x is   x0 =
= 6.
2!2!
2
2

2.  x 2  
x

6
It may be clear to you that n = 6, r = ?, a = x2, b = 
2
.
x
r
 6  2(6 – r)   2 
6
  x
=   x12 – 2r . x – r. (-2)r =
x


r 
r 
Consequently, the general term is:
 6  12 – 3 r
  (x
)(-2)r .
r
 
For this last expression to be independent of x, the following should hold:
x12 – 3r = x0 i.e. 12 – 3r = 0  r = 4.
6
6!
Hence, the term independent of x is   (-2)4 =
 16 = 240
2!4!
 4
Activity 5: This activity will enhance your skill in calculating for a constant term
without
computing all the terms of an expansion.
Find the terms independent of x in the expansions of each of the following expressions:
1.
1

x  
x

8
2

2.  x 2  
x

9
2

3.  x 2  
x

6
1 

4.  2 x  2 
x 

18
57
FOUNDATION MATHEMATICS- MAT 110
 2 1
x  
x

5.
9

1 
6.  x 

x

12
Approximations
Among several other applications, you can use the knowledge acquired so far to solve problems relating
to approximations. The examples which follow will enable you to understand how that is done.
Example: 1
a)
Write down the first three terms in the binomial expansion of (1 – 3x)8.
b)
Hence evaluate (0.997)8 correct to five decimal places.
Solution:
(a)
1  3x 8 = 1 + 8(-3x) + 8  7 (-3x) + …
2
2 1
= 1 – 24x + 252x2 + …
(b)
Now let
1  3x8 = (0.997) . Then
8
1 – 3x = 0.997  3x = 0.003
 x = 0.001.
We can now use the expansion in part ‘a’ with x = 0.001 as follows:
(0.997)8 = 1 – 24(0.001) + 252(0.001)2 + …
= 1 – 0.024 + 0.000252 + …
= 0.97625(5 d.p.).
Example 2:
Obtain the first four terms in the expansion of
 1 
1  x 
2 

10
in ascending powers
of x.
Hence find the approximation of (1.005)10 to 4 decimal places.
Solution:
10
2
3
10  9  8  x 
1 

 x  10  9  x 
 1  x  = 1 + 10   +
  +
  +…
2 
21 2
32  1  2 

2
58
FOUNDATION MATHEMATICS- MAT 110
= 1 + 5x +
Now you should let
x
2
45 2
x + 15x3 + ……….
4
= 0.005  x = 0.010.
With the value of x computed, you should be able to now use the earlier expansion above, with x = 0.010,
to obtain:
(1.005)10 = 1 + 5(0.010) +
45
(0.010)2+ 15(0.010)3 + …
4
= 1 + 0.05 + 0.001125 + 0.000015 + …
= 1.0511 ( correct to 4 d.p.)
Example: 3
Find the first three terms of the expansion of the expression (4 – 2x)12 and use the resulting
binomial expansion to evaluate (3.998)12 correct to the nearest whole number.
Solution:


(4 – 2x)12 = 4 1 2  1 
12
2
x
 x  12  11  x 
12
 = 4 [ 1 + 12    +
  + …]
2
21 2
 2
= 16777216( 1 – 6x +
33 2
x + …)
2
Let 0.001 = x in the expansion of (4 – 2x)12 as follows:
[4 – 2(0.001)]12 = (3.998)12 =16777216[ 1 – 6(0.001) +
33
(0.001)2 +……….]
2
= 16777216(1.0000165 – 0.006 + …)
= 16676829.
Activity 6:
1.
(a) Expand (1 – x)10 in ascending powers of x up to and including the term in x4.
(b) Hence evaluate (0.99)10 correct to six decimal places.
59
FOUNDATION MATHEMATICS- MAT 110
4
2.
1 

Expand  1  x  and use it to evaluate (1.025)4 correct to 3 decimal places.
4 

3.
(a) Write down the first three terms in the binomial expansion of (2 + 5x)9.
(b) Hence evaluate (2.005)9 correct to two decimal places.
4.
Use binomial expansions to evaluate the following to the stated degree of accuracy:
(a) (1.998)8, correct to two places of decimal.
(b) (6.03)5, correct to three places of decimal.
2.5.0 Mathematical induction
In this section you are going to learn how to prove mathematical statements involving natural numbers.
You can try to test the truth of any given generalization by substituting all possible cases but you could
never try all the possible cases in an infinite sequence. The principle of mathematical induction is a
process that can be used to prove conjectures about infinite sequences.
Here is the formal statement of the principle of mathematical induction:
Let P(n) be a mathematical statement involving positive integers n. Such a statement is true for all n  a. If
1.
It is true for n = a
2.
The assumption that P(n) is true for n=k, implies that it is true for n=k+1.
Let us now consider some examples based on the use of the principle of mathematical induction:
Examples:
Prove by mathematical induction that the following mathematical statements are true for all positive
integers n.
1.
1 + 3 + 5 + … + (2n – 1) = n2.
60
FOUNDATION MATHEMATICS- MAT 110
12 + 22 + 32 + … + n2 =
2.
n(n  1)( 2n  1)
.
6
Proofs:
1.
Let P(n) = 1 + 3 + 5 + … + (2n – 1) = n2. Then
(i)
We test the conjecture for n =1.
When n = 1, (2  1)-1 = 2-1=1=12
Thus P(n) is true for n =1.
(ii)
We now assume that P(n) is true for the positive integer n = k.
Thus we have
P(k) = 1 + 3 + 5 + … + (2k– 1) = k2.
(iii)
With the assumption above, we need to prove that P(n) is also true for
n = k + 1.
By adding the (k + 1)th term i.e. [2(k + 1) – 1] = 2k + 1 to both sides of the assumed equality we
obtain
P(k) = 1 + 3 + 5 + … + (2k – 1) + (2k + 1) = k2 + 2k + 1
= k2 + k + k + 1
= (k + 1)2.
The last expression is identical to the expression n2 when n is substituted for k + 1. Thus P(k+1) is
true whenever P(k) is true.
Hence, P(n) is true for all positive integers n.
2. Let P(n) be the statement 12 + 22 + 32 + … + n2 =
n(n  1)( 2n  1)
. Then
6
(i) We test the conjecture for n =1:
When n = 1, 12 = 1=
1(1  1)[( 2  1)  1] (2)(3)

1
6
6
Thus P(n) is true for n = 1.
(ii) We now assume that P(n) is true for a positive integer n = k.
Thus we assume that:
P(k) = 12 + 22 + 32 + … + k2 =
k (k  1)( 2k  1)
.
6
(iii) With the assumption above, we need to prove that P(n) is also true for
n = k + 1.
Adding the (k + 1) th term i.e. (k + 1)2 to both sides of the assumed equality we obtain:
61
FOUNDATION MATHEMATICS- MAT 110
P(k) = 12 + 22 + 32 + … + k2 + (k + 1)2 =
k (k  1)( 2k  1)
 (k + 1)2
6
=
k (k  1)( 2k  1)  6(k  1) 2
6
=
(k  1)[ k (2k  1)  6(k  1) 2 ]
6
=
(k  1)( 2k 2  7k  6) 2
6
=
(k  1)( k  2)( 2k  3)
6
The last expression is identical to the given expression with n = k + 1.
Thus P(k +1) is true whenever P(k) is true.
Hence, P(n) is true for all positive integers n.
Activity 1: Use mathematical induction to prove that following conjectures are true for all
natural numbers n.
n(n  1)
.
2
1.
1+2+3+…+n=
2.
1
1
1
1
1


 ......... 

.
1 2 2  3 3  4
n(n  1) n  1
3.
n 2 (n  1) 2
1 +2 +3 +…+n =
.
4
3
3
3
3
You can also apply mathematical induction to prove that a certain expression is divisible by a given
integer. The following example illustrates this point:
Example:
Prove that 23n – 1 is divisible by 7 for all positive integers n.
Proof: Let P(n) be the conjecture 23n – 1 is divisible by 7 for all positive integers n.
(i)
P(1) = 23 – 1 =7 which is clearly divisible by 7.
So the statement is true for n =1.
(ii) We now assume that P(n) is true for any positive integer n = k, thus,
62
FOUNDATION MATHEMATICS- MAT 110
P(k) = 23k – 1 is divisible by 7.
(iii) We now prove the statement is true for n = k + 1.
From the assumption above, if 23k – 1 is divisible by 7, then there exists an integer p such that
2 3k  1
 p  23k – 1 = 7p
7
 23k = 1 + 7p
Multiplying both sides of the preceding equation by 23, we obtain:
23k .23 = 23(1 + 7p)
 23k+3 = 8(1 + 7p)
= 8 + 56p
= 1 + 7 + 56p
 2
3k+3
– 1 = 7 + 56p
i.e.23(k+1) – 1 = 7(1 + 8p).
Take note that 7(1 + 8p) is divisible by 7. Consequently, 23(k+1) – 1 is also divisible by 7.
Hence, 23n – 1 is divisible by 7 for all positive integers n.
.
Activity 2: Prove the following conjectures by mathematical induction for all positive
integers n.
1.
4n – 1 is divisible by 3.
2.
9 n  7 is divisible by 8.
Unit summary
63
FOUNDATION MATHEMATICS- MAT 110
In this unit you have learned how to prove mathematical statements
involving positive integers, say, n where n  a , using the principle of
mathematical induction. It should be clear to you by now that the first
step involved in proving that a given mathematical statement is true, is
to show that it is true for the first step i.e. for n=a. The second step as
was shown is to assume that a given statement is true for an arbitrary
positive integer, say n=k. Lastly, you need to prove that the assumption
implies that the statement is equally true for the next integer, n=k+1.
SOLUTIONS
64
FOUNDATION MATHEMATICS- MAT 110
UNIT 1
ACTIVITY 2
a) {3}
b) {1, 2, 3, 4}
c) {0, 4, 5, 6, 7, 8, 9, 10}
d) {4}
e) {}
f) {3, 4}
UNIT 2
ACTIVITY 1
x2 = 7
ACTIVITY 2
1.) x = 2, – 4
2.) x = – 3, 16
3.) y = 0, 2
4.) t = 0,
5.) n = – 1,
6.) x = – 6, – 3
7.) n = – 4, 5
8.) x = 2,
ACTIVITY 3
1.) x = – 6
2
2.) x = 2
2
or 2(1
)
3.) x =
)
or
4.) x = 2, 8
5.) x = 2, – 5
6.) n = 2, –
ACTIVITY 4
65
FOUNDATION MATHEMATICS- MAT 110
i.) x =
ii) x =
iii) x =
ACTIVITY 5
1.) No real roots
2.) Two distinct real roots
3.) No real roots
4.) No real roots
ACTIVITY 6
1.) a) i) – 9
b) i) – 2
c) i) – 7
d) i) – 9
ii) 4
ii) – 5
ii) 2
ii) – 3
e) i)
ii)
f) i)
ii)
g) i)
ii)
h) i) – 2
ii)
2.) a.) x2 + 3x – 1 = 0
b) x2 – 6x – 4 = 0
c) x2 – 7x – 5 = 0
d) 3x2 + 2x – 7 = 0
e) 2x2 + 5x – 4 = 0
f) 2x2 + 3x – 10 = 0
3.) a) 2
b) – 11
c) 121
d) 117
e) 14
f) – 22
4.) a)
b)
c)
66
FOUNDATION MATHEMATICS- MAT 110
5.) a.)
b.)
c.) 5
d.)
e.)
f.)
7.) a) x2 – 8x + 10 = 0
b) 2x2 + 4x – 1 = 0
c) x2 – 20x + 4 = 0
d) x2 + 8x – 8 = 0
e) x2 – 80x – 8 = 0
f) 4x2 – 20x + 1 = 0
g) x2 – 12x + 30 = 0
h) x2 + 10x – 1 = 0
ACTIVITY 7
1.) i) (x – 1)(x + 2)(x – 4)(2x + 1)
ii) (x – 2)(2x – 5)(3x + 4)
2.) a) 24
b) 0
c) 4
3.) k = 3
4.) k = 0
5.) k = 15
BIBLIOGRAPHY
67
FOUNDATION MATHEMATICS- MAT 110
1. Backhouse J. K. Houldsworth, Pure Mathematics Book 1; (2002), Longman, London
2. Bostock L. and Chadler S. Mathematics – The Core Course for A – level; (1994), Stanley
Thornes Publisher Ltd
3. Emmanuel R., Wood J. and Crawshaw J. Advanced Pure Mathematics Book 1&2; (2002),
Longman, London.
4. Kaufmann J. E. and Schmitters K. L. Mathematics – The Core Course for A – level; (1994),
Stanley Thornes Publisher Ltd, London.
5. Kearney A. P. An Introduction to Sets; (1996), Blackie and Sons Ltd. London.
6. Sadler A. J. Understanding Pure Mathematics; (2002), Oxfoed University Press, Oxford.
7. Solomon R.C. Pure Mathematics; (2002), John Murray Publisher Ltd, London.
8. Smedley R. and Garly W. Introducing Pure Mathematics; (2001), Oxford Unversity Press,
Oxford.
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