Download Math Analysis-HP - Whittier Union High School District

Summer Assignment 2016-2017
Math Analysis Honors (PreCalculus Honors)
Mr. Mercado
Welcome to PreCalculus Honors. PreCalculus Honors is a very challenging course if you lack the basics skills
in Algebra 1, Geometry, and Algebra 2. Your summer assignment reviews the basics for each course. You can
do the assignment anytime during the summer. To receive credit for the assignment it must be done completely,
correctly with work shown on college/wide rule or graph paper. The assignment will be due on the first day of
class. The practice test will be due on the second day of class and there will be an exam based on the summer
assignmen. Please email me if you have any questions, [email protected].
Is PreCalculus Honors the right course for you? Answer truthfully the following questions for the answer.
Yes No
□
□
□
□
□
□
Did you have any missing assignments in Algebra 2?
Did you turn in any assignments late in Algebra 2?
Is your test average for the year a C or lower in Algebra 2?
If you answered yes to any of the above questions, PreCalculus Honors may not be the right course for you and
may result with you not passing PreCalculus Honors.
Mr. Mercado
Mr. Mercado
Summer Assignment (272 total questions)
Geometry
Chapter 1.7 Page 52 #1 – 33 all (33)
Chapter 3.2 Page 157 #1 – 35 odd (18)
Chapter 4.1 Page 221 #1 – 37 odd (19)
Chapter 7.1 Page 436 #1 – 23 odd (12)
Chapter 7.4 Page 461 #1 – 25 odd (13)
Algebra 2
Chapter A1 Page A12 #11 – 67 odd (29)
Chapter A2 Page A24 #11 – 115 e.o.o.* (27)
Chapter A3 Page A35 #9 – 201 e.o.o.* (49)
Chapter A4 Page A46 #7 – 83 e.o.o (20)
Chapter A5 Page A60 #5 – 97 e.o.o.*, #127 – 143 odd (33)
Chapter A6 Page A69 #7 – 79 e.o.o. (19)
*e.o.o. stands for every other odd, which means to do every second odd or keep adding 4;
Examples of e.o.o.: #1, 5, 9, 13, 17, 21, 25, 29, etc.
#5, 9, 13, 17, etc.
#9, 13, 17, 21, etc.
#11, 15, 19, 23, etc.
Before
Now
Why?
Key Vocabulary
• perimeter, p. 923
• circumference,
p. 923
• area, p. 923
• diameter, p. 923
• radius, p. 923
You classified polygons.
You will find dimensions of polygons.
So you can use measures in science, as in Ex. 46.
Recall that perimeter is the distance around a figure, circumference is the
distance around a circle, and area is the amount of surface covered by a
figure. Perimeter and circumference are measured in units of length, such
as meters (m) and feet (ft). Area is measured in square units, such as square
meters (m 2) and square feet (ft 2).
For Your Notebook
KEY CONCEPT
Formulas for Perimeter P, Area A, and Circumference C
Square
Rectangle
side length s
P 5 4s
A 5 s2
length l and width w
P 5 2l 1 2w
A 5 lw
s
Triangle
side lengths a, b,
and c, base b,
and height h
P5a1b1c
a
r
diameter d
and radius r
C 5 πd 5 2πr
A 5 πr 2
b
1
A5}
bh
d
Pi (p ) is the ratio of a circle’s
circumference to its diameter.
2
EXAMPLE 1
w
Circle
c
h
l
Find the perimeter and area of a rectangle
FT
BASKETBALL Find the perimeter and area of the
rectangular basketball court shown.
Perimeter
Area
P 5 2l 1 2w
A 5 lw
5 2(84) 1 2(50)
5 84(50)
5 268
5 4200
FT
1.7
Find Perimeter, Circumference,
and Area
c The perimeter is 268 feet and the area
is 4200 square feet.
1.7 Find Perimeter, Circumference, and Area
49
EXAMPLE 2
Find the circumference and area of a circle
TEAM PATCH You are ordering circular cloth patches for your soccer
team’s uniforms. Find the approximate circumference and area of
the patch shown.
Solution
APPROXIMATE p
First find the radius. The diameter is 9 centimeters,
The approximations
1
so the radius is }
(9) 5 4.5 centimeters.
2
22
3.14 and } are
7
commonly used as
approximations for the
irrational number π.
Unless told otherwise,
use 3.14 for π.
✓
CM
Then find the circumference and area.
Use 3.14 to approximate the value of π.
C 5 2πr ø 2(3.14)(4.5) 5 28.26
A 5 πr 2 ø 3.14(4.5)2 5 63.585
c The circumference is about 28.3 cm. The area is about 63.6 cm2.
GUIDED PRACTICE
for Examples 1 and 2
Find the area and perimeter (or circumference) of the figure. If necessary,
round to the nearest tenth.
1.
2.
3.
5.7 m
2 yd
1.6 cm
13 m
★
EXAMPLE 3
Standardized Test Practice
Triangle QRS has vertices Q(1, 2), R(4, 6), and S(5, 2). What
is the approximate perimeter of triangle QRS?
A 8 units
B 8.3 units
C 13.1 units
D 25.4 units
Solution
AVOID ERRORS
Write down your
calculations to make
sure you do not make
a mistake substituting
values in the Distance
Formula.
First draw triangle QRS in a coordinate plane.
Find the side lengths. Use the Distance Formula
to find QR and RS.
R(4, 6)
QS 5 ⏐5 2 1⏐ 5 4 units
}}
}
}}
}
QR 5 Ï(4 2 1)2 1 (6 2 2)2 5 Ï25 5 5 units
RS 5 Ï(5 2 4)2 1 (2 2 6)2 5 Ï17 ø 4.1 units
P 5 QS 1 QR 1 RS ø 4 1 5 1 4.1 5 13.1 units
c The correct answer is C.
Chapter 1 Essentials of Geometry
1
Œ(1, 2)
1
Then find the perimeter.
50
y
A B C D
S(5, 2)
x
EXAMPLE 4
Solve a multi-step problem
SKATING RINK An ice-resurfacing machine
is used to smooth the surface of the ice at a
skating rink. The machine can resurface about
270 square yards of ice in one minute.
About how many minutes does it take the
machine to resurface a rectangular skating
rink that is 200 feet long and 90 feet wide?
ANOTHER WAY
Solution
For an alternative
method for solving the
problem in Example 4,
turn to page 57 for
the Problem Solving
Workshop.
The machine can resurface the ice at a rate of 270 square yards per minute.
So, the amount of time it takes to resurface the skating rink depends on
its area.
STEP 1 Find the area of the rectangular skating rink.
Area 5 lw 5 200(90) 5 18,000 ft 2
The resurfacing rate is in square yards per minute. Rewrite
the area of the rink in square yards. There are 3 feet in 1 yard,
and 32 5 9 square feet in 1 square yard.
1 yd2
18,000 ft 2 p }
5 2000 yd 2
2
9 ft
Use unit analysis.
STEP 2 Write a verbal model to represent the situation. Then write and
solve an equation based on the verbal model.
Let t represent the total time (in minutes) needed to resurface
the skating rink.
Area of rink
(yd2)
5
Resurfacing rate
(yd2 per min)
2000 5 270 p t
3
Total time
(min)
Substitute.
7.4 ø t
Divide each side by 270.
c It takes the ice-resurfacing machine about 7 minutes to resurface
the skating rink.
✓
GUIDED PRACTICE
for Examples 3 and 4
4. Describe how to find the height from F
y
to }
EG in the triangle at the right.
E(1, 6)
5. Find the perimeter and the area of the
triangle shown at the right.
F(7, 3)
6. WHAT IF? In Example 4, suppose the
skating rink is twice as long and twice
as wide. Will it take an ice-resurfacing
machine twice as long to resurface the
skating rink? Explain your reasoning.
1
G(1, 2)
1
1.7 Find Perimeter, Circumference, and Area
x
51
EXAMPLE 5
Find unknown length
The base of a triangle is 28 meters. Its area is
308 square meters. Find the height of the triangle.
h
Solution
1
A5}
bh
Write formula for the area of a triangle.
2
1
308 5 }
(28)h
2
28 m
Substitute 308 for A and 28 for b.
22 5 h
Solve for h.
c The height is 22 meters.
✓
GUIDED PRACTICE
for Example 5
7. The area of a triangle is 64 square meters, and its height is 16 meters.
Find the length of its base.
1.7
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 7, 21, and 41
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 19, 26, 38, and 45
5 MULTIPLE REPRESENTATIONS
Ex. 44
SKILL PRACTICE
1. VOCABULARY How are the diameter and radius of a circle related?
2.
★ WRITING Describe a real-world situation in which you would need to
find a perimeter, and a situation in which you would need to find an area.
What measurement units would you use in each situation?
EXAMPLE 1
3. ERROR ANALYSIS Describe and correct the
on p. 49
for Exs. 3–10
error made in finding the area of a triangle
with a height of 9 feet and a base of 52 feet.
A 5 52(9) 5 468 ft2
PERIMETER AND AREA Find the perimeter and area of the shaded figure.
4.
5.
6.
8 ft
4.2 m
18 ft
7.
30 yd
78 yd
7m
8.
9 mm
72 yd
(FPNFUSZ
52
Chapter 1 Essentials of Geometry
15 in.
24 mm
at classzone.com
15 mm
9.
10 cm
17 cm
9 cm
8 cm
6 cm
10. DRAWING A DIAGRAM The base of a triangle is 32 feet. Its height is
1
16 }
feet. Sketch the triangle and find its area.
2
EXAMPLE 2
CIRCUMFERENCE AND AREA Use the given diameter d or radius r to find the
on p. 50
for Exs. 11–15
circumference and area of the circle. Round to the nearest tenth.
11. d 5 27 cm
12. d 5 5 in.
13. r 5 12.1 cm
14. r 5 3.9 cm
15. DRAWING A DIAGRAM The diameter of a circle is 18.9 centimeters.
Sketch the circle and find its circumference and area. Round your
answers to the nearest tenth.
EXAMPLE 3
DISTANCE FORMULA Find the perimeter of the figure. Round to the nearest
on p. 50
for Exs. 16–19
tenth of a unit.
16.
y
17.
R
18.
y
G
2
M
y
N
S
1
1
T
19.
E
x
1
x
1
F
Œ
x
1
★
MULTIPLE CHOICE What is the approximate
area (in square units) of the rectangle shown at
the right?
A
A 6.7
B 8.0
1
C 9.0
D 10.0
P
y
B
C
D
x
3
EXAMPLE 4
CONVERTING UNITS Copy and complete the statement.
on p. 51
for Exs. 20–26
20 . 187 cm 2 5 ? m 2
21. 13 ft 2 5 ? yd 2
22. 18 in.2 5 ? ft 2
23. 8 km 2 5 ? m 2
24. 12 yd 2 5 ? ft 2
25. 24 ft 2 5 ? in.2
26.
★
MULTIPLE CHOICE A triangle has an area of 2.25 square feet. What is
the area of the triangle in square inches?
A 27 in.2
B 54 in.2
C 144 in.2
D 324 in.2
EXAMPLE 5
UNKNOWN MEASURES Use the information about the figure to find the
on p. 52
for Exs. 27–30
indicated measure.
27. Area 5 261 m 2
Find the height h.
28. Area 5 66 in.2
29. Perimeter 5 25 in.
Find the base b.
8 14
h
Find the width w.
in.
w
36 m
b
8 in.
1.7 Find Perimeter, Circumference, and Area
53
30. UNKNOWN MEASURE The width of a rectangle is 17 inches. Its perimeter
is 102 inches. Find the length of the rectangle.
31.
ALGEBRA The area of a rectangle is 18 square inches. The length of the
rectangle is twice its width. Find the length and width of the rectangle.
32.
ALGEBRA The area of a triangle is 27 square feet. Its height is three
times the length of its base. Find the height and base of the triangle.
33.
ALGEBRA Let x represent the side length of a square. Find a regular
polygon with side length x whose perimeter is twice the perimeter of the
square. Find a regular polygon with side length x whose perimeter is three
times the length of the square. Explain your thinking.
FINDING SIDE LENGTHS Find the side length of the square with the given
area. Write your answer as a radical in simplest form.
34. A 5 184 cm 2
38.
35. A 5 346 in.2
36. A 5 1008 mi 2
37. A 5 1050 km 2
★
SHORT RESPONSE In the diagram, the diameter of the yellow
circle is half the diameter of the red circle. What fraction of the
area of the red circle is not covered by the yellow circle? Explain.
39. CHALLENGE The area of a rectangle is 30 cm 2 and its perimeter is
26 cm. Find the length and width of the rectangle.
PROBLEM SOLVING
EXAMPLES
1 and 2
on pp. 49–50
for Exs. 40–41
40. WATER LILIES The giant Amazon water lily
has a lily pad that is shaped like a circle. Find
the circumference and area of a lily pad with a
diameter of 60 inches. Round your answers to
the nearest tenth.
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
41. LAND You are planting grass on a rectangular plot of land.
You are also building a fence around the edge of the plot. The plot is
45 yards long and 30 yards wide. How much area do you need to cover
with grass seed? How many feet of fencing do you need?
GPSQSPCMFNTPMWJOHIFMQBUDMBTT[POFDPN
EXAMPLE 4
on p. 51
for Ex. 42
42. MULTI-STEP PROBLEM Chris is installing a solar panel. The maximum
amount of power the solar panel can generate in a day depends in part
on its area. On a sunny day in the city where Chris lives, each square
meter of the panel can generate up to 125 watts of power. The flat
rectangular panel is 84 centimeters long and 54 centimeters wide.
a. Find the area of the solar panel in square meters.
b. What is the maximum amount of power (in watts) that the panel could
generate if its area was 1 square meter? 2 square meters? Explain.
c. Estimate the maximum amount of power Chris’s solar panel can
generate. Explain your reasoning.
54
5 WORKED-OUT SOLUTIONS
on p. WS1
★ 5 STANDARDIZED
TEST PRACTICE
5 MULTIPLE
REPRESENTATIONS
3.2
Before
Now
Why?
Key Vocabulary
• corresponding
angles, p. 149
• alternate interior
angles, p. 149
• alternate exterior
angles, p. 149
• consecutive interior
angles, p. 149
Use Parallel Lines
and Transversals
You identified angle pairs formed by a transversal.
You will use angles formed by parallel lines and transversals.
So you can understand angles formed by light, as in Example 4.
ACTIVITY EXPLORE PARALLEL LINES
Materials: lined paper, tracing paper, straightedge
STEP 1 Draw a pair of parallel lines cut by a
nonperpendicular transversal on lined paper.
Label the angles as shown.
1 2
3 4
STEP 2 Trace your drawing onto tracing paper.
STEP 3 Move the tracing paper to position ∠ 1 of the
traced figure over ∠ 5 of the original figure.
Compare the angles. Are they congruent?
5 6
7 8
STEP 4 Compare the eight angles and list all the
congruent pairs. What do you notice about the
special angle pairs formed by the transversal?
For Your Notebook
POSTULATE
POSTULATE 15 Corresponding Angles Postulate
If two parallel lines are cut by a transversal,
then the pairs of corresponding angles are
congruent.
t
2
p
6
q
∠2 > ∠6
EXAMPLE 1
Identify congruent angles
The measure of three of the numbered angles is
1208. Identify the angles. Explain your reasoning.
Solution
1208 2
3 4
5 6
7 8
By the Corresponding Angles Postulate, m∠ 5 5 1208.
Using the Vertical Angles Congruence Theorem, m∠ 4 5 1208.
Because ∠ 4 and ∠ 8 are corresponding angles, by the Corresponding Angles
Postulate, you know that m∠ 8 5 1208.
154
Chapter 3 Parallel and Perpendicular Lines
For Your Notebook
THEOREMS
THEOREM 3.1 Alternate Interior Angles Theorem
t
If two parallel lines are cut by a transversal, then
the pairs of alternate interior angles are congruent.
p
4
5
Proof: Example 3, p. 156
q
∠4 > ∠5
THEOREM 3.2 Alternate Exterior Angles Theorem
t
1
If two parallel lines are cut by a transversal, then
the pairs of alternate exterior angles are congruent.
p
q
8
Proof: Ex. 37, p. 159
∠1 > ∠8
THEOREM 3.3 Consecutive Interior Angles Theorem
t
If two parallel lines are cut by a transversal,
then the pairs of consecutive interior angles are
supplementary.
q
∠ 3 and ∠ 5 are
supplementary.
Proof: Ex. 41, p. 159
EXAMPLE 2
p
3
5
Use properties of parallel lines
ALGEBRA Find the value of x.
1158
4
(x 1 5)8
a
b
Solution
By the Vertical Angles Congruence Theorem, m∠ 4 5 1158. Lines a and b
are parallel, so you can use the theorems about parallel lines.
m∠ 4 1 (x 1 5)8 5 1808
1158 1 (x 1 5)8 5 1808
x 1 120 5 180
x 5 60
(FPNFUSZ
✓
GUIDED PRACTICE
Consecutive Interior Angles Theorem
Substitute 115 8 for m ∠ 4.
Combine like terms.
Subtract 120 from each side.
at classzone.com
for Examples 1 and 2
Use the diagram at the right.
1. If m∠ 1 5 1058, find m∠ 4, m∠ 5, and m∠ 8. Tell
which postulate or theorem you use in each case.
1 2
3 4
5 6
7 8
2. If m∠ 3 5 688 and m∠ 8 5 (2x 1 4)8, what is the
value of x? Show your steps.
3.2 Use Parallel Lines and Transversals
155
EXAMPLE 3
Prove the Alternate Interior Angles Theorem
Prove that if two parallel lines are cut by a transversal, then the pairs of
alternate interior angles are congruent.
Solution
WRITE PROOFS
You can use the
information from the
diagram in your proof.
Find any special angle
pairs. Then decide what
you know about those
pairs.
Draw a diagram. Label a pair of alternate interior
angles as ∠ 1 and ∠ 2. You are looking for an angle that
is related to both ∠ 1 and ∠ 2. Notice that one angle is a
vertical angle with ∠ 2 and a corresponding angle with
∠ 1. Label it ∠ 3.
GIVEN
PROVE
p
2
3
q
cpiq
c ∠1 > ∠2
STATEMENTS
1.
2.
3.
4.
t
1
REASONS
piq
∠1 > ∠3
∠3 > ∠2
∠1 > ∠2
EXAMPLE 4
1.
2.
3.
4.
Given
Corresponding Angles Postulate
Vertical Angles Congruence Theorem
Transitive Property of Congruence
Solve a real-world problem
SCIENCE When sunlight enters a drop of rain, different colors of light leave
the drop at different angles. This process is what makes a rainbow. For violet
light, m∠ 2 5 408. What is m∠ 1? How do you know?
2
1
Solution
Because the sun’s rays are parallel, ∠ 1 and ∠ 2 are alternate interior angles.
By the Alternate Interior Angles Theorem, ∠ 1 > ∠ 2. By the definition of
congruent angles, m∠ 1 5 m∠ 2 5 408.
✓
GUIDED PRACTICE
for Examples 3 and 4
3. In the proof in Example 3, if you use the third statement before the second
statement, could you still prove the theorem? Explain.
4. WHAT IF? Suppose the diagram in Example 4 shows yellow light leaving a
drop of rain. Yellow light leaves the drop at an angle of 418. What is m∠ 1
in this case? How do you know?
156
Chapter 3 Parallel and Perpendicular Lines
3.2
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 5, 9, and 39
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 3, 21, 33, 39, and 40
SKILL PRACTICE
1. VOCABULARY Draw a pair of parallel lines and a transversal. Label a pair
of corresponding angles.
EXAMPLES
1 and 2
2.
★ WRITING Two parallel lines are cut by a transversal. Which pairs
of angles are congruent? Which pairs of angles are supplementary?
3.
★
on pp. 154–155
for Exs. 3–16
MULTIPLE CHOICE In the figure at the right,
which angle has the same measure as ∠ 1?
A ∠2
B ∠3
C ∠4
D ∠5
1 2
3
4 5
USING PARALLEL LINES Find the angle measure.
Tell which postulate or theorem you use.
4. If m∠ 4 5 658, then m∠ 1 5 ? .
1 2
3 4
5 6
7 8
5. If m∠ 7 5 1108, then m∠ 2 5 ? .
6. If m∠ 5 5 718, then m∠ 4 5 ? .
7. If m∠ 3 5 1178, then m∠ 5 5 ? .
8. If m∠ 8 5 548, then m∠ 1 5 ? .
USING POSTULATES AND THEOREMS What postulate or theorem justifies the
statement about the diagram?
9. ∠ 1 > ∠ 5
10. ∠ 4 > ∠ 5
11. ∠ 2 > ∠ 7
12. ∠ 2 and ∠ 5 are supplementary.
13. ∠ 3 > ∠ 6
14. ∠ 3 > ∠ 7
15. ∠ 1 > ∠ 8
16. ∠ 4 and ∠ 7 are supplementary.
1
2
4
3
5
6
8
7
USING PARALLEL LINES Find m∠ 1 and m∠ 2. Explain your reasoning.
17.
18.
1508
19.
1408
1
2
2
1228
1
1 2
20. ERROR ANALYSIS A student concludes that
∠ 9 > ∠ 10 by the Corresponding Angles
Postulate. Describe and correct the error in
this reasoning.
9
∠ 9 > ∠ 10
10
3.2 Use Parallel Lines and Transversals
157
21.
SHORT RESPONSE Given p i q, describe two
methods you can use to show that ∠ 1 > ∠ 4.
t
★
1
p
2
3
q
4
USING PARALLEL LINES Find m∠ 1, m∠ 2, and m∠ 3. Explain your reasoning.
22.
23.
1
24.
1
2
808
1
2
1338
3
1158
3
ANGLES Use the diagram at the right.
2
‹]›
‹]›
A
25. Name two pairs of congruent angles if AB and DC are parallel.
‹]›
3
B
‹]›
26. Name two pairs of supplementary angles if AD and BC
D
are parallel.
C
ALGEBRA Find the values of x and y.
27.
28.
29.
3y 8
y8 x8
6y 8
458
30.
31.
608
3x 8
33.
★
32.
4x 8
(14x 2 10)8
(3y 1 2)8
(5y 2 5)8
2y 8
528
1358
B 75
C 110
D 115
t
1108
m
(y 2 5)8 1
34. DRAWING Draw a four-sided figure with sides }
MN and }
PQ, such that
}i }
MN
PQ, }
MP i }
NQ, and ∠ MNQ is an acute angle. Which angle pairs
formed are congruent? Explain your reasoning.
CHALLENGE Find the values of x and y.
35.
36.
608
1508
(5x 2 y)8
(5x 1 y)8
(2x 2 y)8
(2x 1 y)8
408
158
5x8
MULTIPLE CHOICE What is the value of y in the diagram?
A 70
5 WORKED-OUT SOLUTIONS
on p. WS1
658
558
2x 8
858
y8
x8
1308
★ 5 STANDARDIZED
TEST PRACTICE
n
4.1
Before
Now
Why?
Key Vocabulary
• triangle
scalene, isosceles,
equilateral, acute,
right, obtuse,
equiangular
• interior angles
• exterior angles
• corollary to a
Apply Triangle Sum
Properties
You classified angles and found their measures.
You will classify triangles and find measures of their angles.
So you can place actors on stage, as in Ex. 40.
A triangle is a polygon with three sides. A triangle with vertices A, B, and C is
called “triangle ABC” or “n ABC.”
For Your Notebook
KEY CONCEPT
Classifying Triangles by Sides
Scalene Triangle
Isosceles Triangle
Equilateral Triangle
No congruent sides
At least 2 congruent sides
3 congruent sides
theorem
READ VOCABULARY
Notice that an
equilateral triangle
is also isosceles. An
equiangular triangle is
also acute.
Classifying Triangles by Angles
Acute
Triangle
Right
Triangle
Obtuse
Triangle
Equiangular
Triangle
3 acute angles
1 right angle
1 obtuse angle
3 congruent angles
EXAMPLE 1
Classify triangles by sides and by angles
SUPPORT BEAMS Classify the triangular shape
of the support beams in the diagram by its sides
and by measuring its angles.
Solution
The triangle has a pair of congruent sides, so it is
isosceles. By measuring, the angles are 558, 558,
and 708. It is an acute isosceles triangle.
4.1 Apply Triangle Sum Properties
217
EXAMPLE 2
Classify a triangle in a coordinate plane
Classify nPQO by its sides. Then
determine if the triangle is a right
triangle.
y
Œ(6, 3)
P(21, 2)
O(0, 0)
x
Solution
STEP 1 Use the distance formula to find the side lengths.
}}
}}
}}
}}
}}
}}
OP 5 Ï (x2 2 x1)2 1 (y2 2 y1)2 5 Ï ((21) 2 0)2 1 (2 2 0)2 5 Ï5 ø 2.2
}
}
OQ 5 Ï (x2 2 x1)2 1 (y2 2 y1)2 5 Ï(6 2 0)2 1 (3 2 0)2 5 Ï45 ø 6.7
}
PQ 5 Ï(x2 2 x1)2 1 (y2 2 y1)2 5 Ï (6 2 (21))2 1 (3 2 2)2 5 Ï 50 ø 7.1
220
STEP 2 Check for right angles. The slope of }
OP is }
5 22. The slope
21 2 0
20
1
of }
OQ is 3}
5}
. The product of the slopes is 22
620
2
so }
OP ⊥ }
OQ and ∠ POQ is a right angle.
1 }12 2 5 21,
c Therefore, nPQO is a right scalene triangle.
✓
GUIDED PRACTICE
for Examples 1 and 2
1. Draw an obtuse isosceles triangle and an acute scalene triangle.
2. Triangle ABC has the vertices A(0, 0), B(3, 3), and C(23, 3). Classify it by
its sides. Then determine if it is a right triangle.
ANGLES When the sides of a polygon are extended, other angles are formed.
The original angles are the interior angles. The angles that form linear pairs
with the interior angles are the exterior angles.
READ DIAGRAMS
Each vertex has a pair
of congruent exterior
angles. However, it is
common to show only
one exterior angle at
each vertex.
interior angles
exterior angles
For Your Notebook
THEOREM
THEOREM 4.1 Triangle Sum Theorem
The sum of the measures of the interior
angles of a triangle is 1808.
Proof: p. 219; Ex. 53, p. 224
218
Chapter 4 Congruent Triangles
B
A
C
m∠ A 1 m∠ B 1 m∠ C 5 1808
AUXILIARY LINES To prove certain theorems, you may need to add a line, a
segment, or a ray to a given diagram. An auxiliary line is used in the proof of
the Triangle Sum Theorem.
PROOF
GIVEN
PROVE
Triangle Sum Theorem
c n ABC
c m∠ 1 1 m∠ 2 1 m∠ 3 5 1808
B
4
D
2 5
Plan a. Draw an auxiliary line through
1
3
A
C
}
for
B
and
parallel
to
AC
.
Proof
b. Show that m∠ 4 1 m∠ 2 1 m∠ 5 5 1808, ∠ 1 > ∠ 4, and ∠ 3 > ∠ 5.
c. By substitution, m∠ 1 1 m∠ 2 1 m∠ 3 5 1808.
STATEMENTS
‹]›
Plan a. 1. Draw BD
parallel to }
AC.
in
Action b. 2. m∠ 4 1 m∠ 2 1 m∠ 5 5 1808
3. ∠ 1 > ∠ 4, ∠ 3 > ∠ 5
4. m∠ 1 5 m∠ 4, m∠ 3 5 m∠ 5
c. 5. m∠ 1 1 m∠ 2 1 m∠ 3 5 1808
REASONS
1. Parallel Postulate
2. Angle Addition Postulate and
definition of straight angle
3. Alternate Interior Angles
Theorem
4. Definition of congruent
angles
5. Substitution Property of
Equality
For Your Notebook
THEOREM
THEOREM 4.2 Exterior Angle Theorem
The measure of an exterior angle of a triangle
is equal to the sum of the measures of the
two nonadjacent interior angles.
B
Proof: Ex. 50, p. 223
EXAMPLE 3
1
C
A
m∠ 1 5 m∠ A 1 m∠ B
Find an angle measure
J
ALGEBRA Find m∠ JKM.
x8
Solution
STEP 1 Write and solve an equation
to find the value of x.
(2x 2 5)8 5 708 1 x8
x 5 75
708
L
(2x 2 5)8
K
M
Apply the Exterior Angle Theorem.
Solve for x.
STEP 2 Substitute 75 for x in 2x 2 5 to find m∠ JKM.
2x 2 5 5 2 p 75 2 5 5 145
c The measure of ∠ JKM is 1458.
4.1 Apply Triangle Sum Properties
219
A corollary to a theorem is a statement that can be proved easily using the
theorem. The corollary below follows from the Triangle Sum Theorem.
For Your Notebook
COROLLARY
Corollary to the Triangle Sum Theorem
The acute angles of a right triangle are
complementary.
Proof: Ex. 48, p. 223
EXAMPLE 4
C
A
B
m∠ A 1 m∠ B 5 908
Find angle measures from a verbal description
ARCHITECTURE The tiled staircase shown
forms a right triangle. The measure of one
acute angle in the triangle is twice the
measure of the other. Find the measure of
each acute angle.
Solution
First, sketch a diagram of the situation. Let the
measure of the smaller acute angle be x8. Then
the measure of the larger acute angle is 2x8. The
Corollary to the Triangle Sum Theorem states
that the acute angles of a right triangle are
complementary.
2x 8
x8
Use the corollary to set up and solve an equation.
x8 1 2x8 5 908
x 5 30
Corollary to the Triangle Sum Theorem
Solve for x.
c So, the measures of the acute angles are 308 and 2(308) 5 608.
✓
GUIDED PRACTICE
for Examples 3 and 4
3. Find the measure of ∠ 1 in the diagram
3x8
shown.
408
1 (5x 2 10)8
4. Find the measure of each interior angle of n ABC, where m∠ A 5 x8,
m∠ B 5 2x8, and m∠ C 5 3x8.
5. Find the measures of the acute angles of
the right triangle in the diagram shown.
2x8
(x 2 6)8
6. In Example 4, what is the measure of the obtuse angle formed between
the staircase and a segment extending from the horizontal leg?
220
Chapter 4 Congruent Triangles
4.1
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 9, 15, and 41
★ 5 STANDARDIZED TEST PRACTICE
Exs. 7, 20, 31, 43, and 51
SKILL PRACTICE
VOCABULARY Match the triangle description with the most specific name.
1. Angle measures: 308, 608, 908
A. Isosceles
2. Side lengths: 2 cm, 2 cm, 2 cm
B. Scalene
3. Angle measures: 608, 608, 608
C. Right
4. Side lengths: 6 m, 3 m, 6 m
D. Obtuse
5. Side lengths: 5 ft, 7 ft, 9 ft
E. Equilateral
6. Angle measures: 208, 1258, 358
F. Equiangular
7.
★
WRITING Can a right triangle also be obtuse? Explain why or why not.
EXAMPLE 1
CLASSIFYING TRIANGLES Copy the triangle and measure its angles. Classify
on p. 217
for Exs. 8–10
the triangle by its sides and by its angles.
8. X
9.
Y
EXAMPLE 2
on p. 218
for Exs. 11–13
Z
10. J
M
L
K
H
N
COORDINATE PLANE A triangle has the given vertices. Graph the triangle
and classify it by its sides. Then determine if it is a right triangle.
11. A(2, 3), B(6, 3), C(2, 7)
12. A(3, 3), B(6, 9), C(6, 23)
13. A(1, 9), B(4, 8), C(2, 5)
EXAMPLE 3
FINDING ANGLE MEASURES Find the value of x. Then classify the triangle by
on p. 219
for Exs. 14–19
its angles.
14.
15.
608
x8
x8
608
16.
3x8
648
608
708
x8
ALGEBRA Find the measure of the exterior angle shown.
17.
18.
19.
(2x 2 2)8
x8
EXAMPLE 4
on p. 220
for Ex. 20
20.
458
x8
248
(3x 1 6)8
(2x 1 18)8
1
(3x 1 2)8
★
SHORT RESPONSE Explain how to use the
Corollary to the Triangle Sum Theorem to find
the measure of each angle.
3x8
2x8
4.1 Apply Triangle Sum Properties
221
ANGLE RELATIONSHIPS Find the measure of the numbered angle.
21. ∠ 1
22. ∠ 2
23. ∠ 3
24. ∠ 4
25. ∠ 5
26. ∠ 6
1
408
2
3
4
208
5
6
27.
ALGEBRA In nPQR, ∠ P > ∠ R and the measure of ∠ Q is twice the
measure of ∠ R. Find the measure of each angle.
28.
ALGEBRA In nEFG, m∠ F 5 3(m∠ G), and m∠ E 5 m∠ F 2 308. Find
the measure of each angle.
ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error.
29.
31.
30.
All equilateral triangles
are also isosceles. So, if
n ABC is isosceles, then
it is equilateral as well.
★
m∠ 1 1 808 1 508 5 1808
80°
1
50°
MULTIPLE CHOICE Which of the following is not possible?
A An acute scalene triangle
B A triangle with two acute exterior angles
C An obtuse isosceles triangle
D An equiangular acute triangle
ALGEBRA In Exercises 32–37, find the values of x and y.
32.
438
33.
y8
34.
258
1188
758
x8
x8
y8
35.
x8
228
y8
36.
y8
37.
y8
x8
x8
208
158
648
358
478
188
38. VISUALIZATION Is there an angle measure that is so small that any
triangle with that angle measure will be an obtuse triangle? Explain.
39. CHALLENGE Suppose you have the equations y 5 ax 1 b, y 5 cx 1 d,
and y 5 ex 1 f.
a. When will these three lines form a triangle?
b. Let c 5 1, d 5 2, e 5 4, and f 5 27. Find values of a and b so that no
triangle is formed by the three equations.
4
3
4
1
41
c. Draw the triangle formed when a 5 }
,b5}
, c 5 2}, d 5 }
, e 5 0,
3
3
and f 5 21. Then classify the triangle by its sides.
222
5 WORKED-OUT SOLUTIONS
on p. WS1
★ 5 STANDARDIZED
TEST PRACTICE
3
x8
y8
7.1
Before
Apply the Pythagorean
Theorem
You learned about the relationships within triangles.
Now
You will find side lengths in right triangles.
Why?
So you can find the shortest distance to a campfire, as in Ex. 35.
Key Vocabulary
• Pythagorean triple
• right triangle,
p. 217
• leg of a right
triangle, p. 241
• hypotenuse, p. 241
One of the most famous theorems in mathematics
is the Pythagorean Theorem, named for the ancient
Greek mathematician Pythagoras (around 500 B.C.).
This theorem can be used to find information about
the lengths of the sides of a right triangle.
hypotenuse
leg
leg
For Your Notebook
THEOREM
THEOREM 7.1 Pythagorean Theorem
In a right triangle, the square of the length
of the hypotenuse is equal to the sum of the
squares of the lengths of the legs.
b
Proof: p. 434; Ex. 32, p. 455
EXAMPLE 1
c
a
c 2 5 a 2 1 b2
Find the length of a hypotenuse
Find the length of the hypotenuse of the right triangle.
8
Solution
ABBREVIATE
In the equation for the
Pythagorean Theorem,
“length of hypotenuse”
and “length of leg”
was shortened to
“hypotenuse” and “leg”.
✓
(hypotenuse)2 5 (leg)2 1 (leg)2
x
Pythagorean Theorem
x 2 5 62 1 8 2
Substitute.
x 2 5 36 1 64
Multiply.
2
Add.
x 5 10
Find the positive square root.
x 5 100
GUIDED PRACTICE
6
for Example 1
Identify the unknown side as a leg or hypotenuse. Then, find the unknown
side length of the right triangle. Write your answer in simplest radical form.
1.
x
3
5
2.
x
4
6
7.1 Apply the Pythagorean Theorem
433
★
EXAMPLE 2
Standardized Test Practice
A 16 foot ladder rests against the side of the
house, and the base of the ladder is 4 feet
away. Approximately how high above the
ground is the top of the ladder?
A 240 feet
B 20 feet
C 16.5 feet
D 15.5 feet
FT
XFT
FT
Solution
1ofLength
ladder2
2
APPROXIMATE
In real-world
applications, it is
usually appropriate
to use a calculator to
approximate the square
root of a number.
Round your answer to
the nearest tenth.
✓
Distance
1from
house2
2
5
1ofHeight
ladder2
2
1
162 5 42 1 x2
Substitute.
256 5 16 1 x2
Multiply.
240 5 x
2
Subtract 16 from each side.
}
Ï240 5 x
Find positive square root.
15.491 ø x
Approximate with a calculator.
The ladder is resting against the house at about 15.5 feet above the ground.
c The correct answer is D.
GUIDED PRACTICE
A B C D
for Example 2
3. The top of a ladder rests against a wall, 23 feet above the ground. The base
of the ladder is 6 feet away from the wall. What is the length of the ladder?
4. The Pythagorean Theorem is only true for what type of triangle?
PROVING THE PYTHAGOREAN THEOREM There are many proofs
of the Pythagorean Theorem. An informal proof is shown below.
You will write another proof in Exercise 32 on page 455.
In the figure at the right, the four right triangles are congruent,
and they form a small square in the middle. The area of the large
square is equal to the area of the four triangles plus the area of
the smaller square.
REVIEW AREA
Recall that the area of a
square with side
length s is A 5 s 2.
The area of a triangle
with base b and
1
height h is A 5 }bh.
2
Area of
large square
1
(a 1 b)2 5 41 }
ab 2 1 c 2
2
a 2 1 2ab 1 b 2 5 2ab 1 c 2
2
2
a 1b 5c
(FPNFUSZ
434
Area of
four triangles
5
2
at classzone.com
Chapter 7 Right Triangles and Trigonometry
1
Area of
smaller square
Use area formulas.
Multiply.
Subtract 2ab from each side.
a
b
b
a
c
c
c
a
b
c
b
a
EXAMPLE 3
Find the area of an isosceles triangle
Find the area of the isosceles triangle with side lengths 10 meters,
13 meters, and 13 meters.
Solution
STEP 1 Draw a sketch. By definition, the length of an altitude
is the height of a triangle. In an isosceles triangle, the
altitude to the base is also a perpendicular bisector.
So, the altitude divides the triangle into two right
triangles with the dimensions shown.
13 m
STEP 2 Use the Pythagorean Theorem to find the height
h
5m
13 m
5m
of the triangle.
c 2 5 a2 1 b 2
2
2
13 5 5 1 h
2
169 5 25 1 h2
144 5 h
READ TABLES
You may find it helpful
to use the Table of
Squares and Square
Roots on p. 924.
2
Pythagorean Theorem
Substitute.
Multiply.
Subtract 25 from each side.
12 5 h
Find the positive square root.
STEP 3 Find the area.
1
1
Area 5 }
(base)(height) 5 }
(10)(12) 5 60 m 2
2
2
c The area of the triangle is 60 square meters.
✓
GUIDED PRACTICE
for Example 3
Find the area of the triangle.
5.
18 ft
26 m
6.
30 ft
18 ft
20 m
26 m
PYTHAGOREAN TRIPLES A Pythagorean triple is a set of three positive
integers a, b, and c that satisfy the equation c 2 5 a2 1 b2.
For Your Notebook
KEY CONCEPT
STANDARDIZED
TESTS
You may find it helpful
to memorize the basic
Pythagorean triples,
shown in bold, for
standardized tests.
Common Pythagorean Triples and Some of Their Multiples
3, 4, 5
5, 12, 13
8, 15, 17
7, 24, 25
6, 8, 10
9, 12, 15
30, 40, 50
3x, 4x, 5x
10, 24, 26
15, 36, 39
50, 120, 130
5x, 12x, 13x
16, 30, 34
24, 45, 51
80, 150, 170
8x, 15x, 17x
14, 48, 50
21, 72, 75
70, 240, 250
7x, 24x, 25x
The most common Pythagorean triples are in bold. The other triples are the
result of multiplying each integer in a bold face triple by the same factor.
7.1 Apply the Pythagorean Theorem
435
EXAMPLE 4
Find the length of a hypotenuse using two methods
Find the length of the hypotenuse of the right triangle.
24
Solution
10
x
Method 1: Use a Pythagorean triple.
A common Pythagorean triple is 5, 12, 13. Notice that if you multiply
the lengths of the legs of the Pythagorean triple by 2, you get the lengths
of the legs of this triangle: 5 p 2 5 10 and 12 p 2 5 24. So, the length of the
hypotenuse is 13 p 2 5 26.
Method 2: Use the Pythagorean Theorem.
x2 5 102 1 242
2
x 5 100 1 576
Multiply.
x2 5 676
Add.
x 5 26
✓
Pythagorean Theorem
Find the positive square root.
GUIDED PRACTICE
for Example 4
Find the unknown side length of the right triangle using the Pythagorean
Theorem. Then use a Pythagorean triple.
7.
9 in.
7.1
8.
12 in.
x
x
14 cm
48 cm
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 9, 11, and 33
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 17, 27, 33, and 36
5 MULTIPLE REPRESENTATIONS
Ex. 35
SKILL PRACTICE
1. VOCABULARY Copy and complete: A set of three positive integers a, b,
and c that satisfy the equation c 2 5 a2 1 b2 is called a ? .
2.
ALGEBRA Find the length of the hypotenuse of the right triangle.
EXAMPLE 1
on p. 433
for Exs. 3–7
★ WRITING Describe the information you need to have in order to use
the Pythagorean Theorem to find the length of a side of a triangle.
3.
4.
50
x
120
436
Chapter 7 Right Triangles and Trigonometry
5.
x
56
33
42
40
x
ERROR ANALYSIS Describe and correct the error in using the Pythagorean
Theorem.
6.
7.
26
x
7
10
24
24
x2 5 72 1 242
a 2 1 b2 5 c 2
x2 5 (7 1 24) 2
102 1 262 5 242
x2 5 312
x 5 31
EXAMPLE 2
on p. 434
for Exs. 8–10
FINDING A LENGTH Find the unknown leg length x.
8.
9.
10.
13.4 in.
x
16.7 ft
9.8 in.
x
5.7 ft
4.9 ft
x
8.9 ft
EXAMPLE 3
FINDING THE AREA Find the area of the isosceles triangle.
on p. 435
for Exs. 11–13
11.
12.
17 m
h
17 m
20 ft
20 ft
h
13.
10 cm
on p. 436
for Exs. 14–17
10 cm
32 ft
12 cm
16 m
EXAMPLE 4
h
FINDING SIDE LENGTHS Find the unknown side length of the right triangle
using the Pythagorean Theorem or a Pythagorean triple.
14.
x
17.
15.
72
21
16.
50
30
60
x
x
68
★
MULTIPLE CHOICE What is the length of the hypotenuse of a right
triangle with leg lengths of 8 inches and 15 inches?
A 13 inches
B 17 inches
C 21 inches
D 25 inches
PYTHAGOREAN TRIPLES The given lengths are two sides of a right triangle.
All three side lengths of the triangle are integers and together form a
Pythagorean triple. Find the length of the third side and tell whether it is a
leg or the hypotenuse.
18. 24 and 51
19. 20 and 25
20. 28 and 96
21. 20 and 48
22. 75 and 85
23. 72 and 75
7.1 Apply the Pythagorean Theorem
437
7.4
Before
Now
Why?
Key Vocabulary
• isosceles triangle,
Special Right Triangles
You found side lengths using the Pythagorean Theorem.
You will use the relationships among the sides in special right triangles.
So you can find the height of a drawbridge, as in Ex. 28.
A 458-458-908 triangle is an isosceles right triangle that
can be formed by cutting a square in half as shown.
p. 217
For Your Notebook
THEOREM
THEOREM 7.8 458-458-908 Triangle Theorem
USE RATIOS
The extended ratio of
the side lengths of a
458-458-908 triangle is
}
1: 1: Ï2.
}
In a 458-458-908 triangle, the hypotenuse is Ï 2 times
as long as each leg.
}
hypotenuse 5 leg p Ï 2
x 2
458
x
Proof: Ex. 30, p. 463
EXAMPLE 1
x 458
Find hypotenuse length in a 458-458-908 triangle
Find the length of the hypotenuse.
a.
b.
8
3 2
3 2
458
Solution
a. By the Triangle Sum Theorem, the measure of the third angle must be
458. Then the triangle is a 458-458-908 triangle, so by Theorem 7.8, the
}
hypotenuse is Ï2 times as long as each leg.
}
hypotenuse 5 leg p Ï2
}
5 8Ï2
REVIEW ALGEBRA
Remember the
following properties of
radicals:
Ïa p
}
}
}
Ïb 5 Ïa p b
Ïa p a 5 a
}
For a review of radical
expressions, see p. 874.
458-458-908 Triangle Theorem
Substitute.
b. By the Base Angles Theorem and the Corollary to the Triangle Sum
Theorem, the triangle is a 458-458-908 triangle.
}
hypotenuse 5 leg p Ï2
}
}
458-458-908 Triangle Theorem
5 3Ï2 p Ï 2
Substitute.
53p2
Product of square roots
56
Simplify.
7.4 Special Right Triangles
457
EXAMPLE 2
Find leg lengths in a 458-458-908 triangle
Find the lengths of the legs in the triangle.
5 2
x
x
Solution
By the Base Angles Theorem and the Corollary to the Triangle Sum Theorem,
the triangle is a 458-458-908 triangle.
}
hypotenuse 5 leg p Ï 2
}
}
5Ï 2 5 x p Ï 2
}
Substitute.
}
5Ï2
Ï2
xÏ 2
Ï2
}
Divide each side by Ï 2 .
}
} 5 }
}
55x
★
45 8-45 8-90 8 Triangle Theorem
EXAMPLE 3
Simplify.
Standardized Test Practice
Triangle WXY is a right triangle.
Find the length of }
WX .
Y
25 cm
458
X
ELIMINATE CHOICES
You can eliminate
choices C and D because
the hypotenuse has to
be longer than the leg.
W
}
A 50 cm
B 25Ï2 cm
C 25 cm
25Ï2
D }
cm
}
2
Solution
By the Corollary to the Triangle Sum Theorem, the triangle is a
458-458-908 triangle.
}
hypotenuse 5 leg p Ï2
}
WX 5 25Ï 2
c The correct answer is B.
✓
GUIDED PRACTICE
458-458-908 Triangle Theorem
Substitute.
A B C D
for Examples 1, 2, and 3
Find the value of the variable.
1.
2.
2 2
2
2
x
y
x
3.
8
8
d
8
8
4. Find the leg length of a 458-458-908 triangle with a hypotenuse length of 6.
458
Chapter 7 Right Triangles and Trigonometry
A 308-608-908 triangle can be formed by dividing an equilateral triangle in half.
For Your Notebook
THEOREM
THEOREM 7.9 308-608-908 Triangle Theorem
USE RATIOS
The extended ratio of
the side lengths of a
308-608-908 triangle is
}
1 : Ï3 : 2.
In a 308-608-908 triangle, the hypotenuse is twice as long as the shorter
}
leg, and the longer leg is Ï 3 times as long as the shorter leg.
hypotenuse 5 2 p shorter leg
}
longer leg 5 shorter leg p Ï3
x
608
2x
308
Proof: Ex. 32, p. 463
EXAMPLE 4
x 3
Find the height of an equilateral triangle
LOGO The logo on the recycling bin at the right resembles
an equilateral triangle with side lengths of 6 centimeters.
What is the approximate height of the logo?
Solution
REVIEW MEDIAN
Remember that in an
equilateral triangle, the
altitude to a side is also
the median to that side.
So, altitude }
BD bisects
}
AC .
Draw the equilateral triangle described. Its altitude forms
the longer leg of two 308-608-908 triangles. The length h of
the altitude is approximately the height of the logo.
B
}
longer leg 5 shorter leg p Ï 3
6 cm
h
}
h 5 3 p Ï3 ø 5.2 cm
EXAMPLE 5
A
6 cm
608
608
C
3 cm D 3 cm
Find lengths in a 308-608-908 triangle
Find the values of x and y. Write your answer
in simplest radical form.
608
y
x
308
9
STEP 1 Find the value of x.
}
longer leg 5 shorter leg p Ï 3
}
9 5 xÏ 3
Substitute.
9
Ï3
}
}
} 5 x
9
Ï3
308-608-908 Triangle Theorem
Divide each side by Ï 3 .
}
Multiply numerator and
}
denominator by Ï 3 .
Ï3
Ï3
}
} p }
} 5 x
}
9Ï 3
3
}5x
Multiply fractions.
}
3Ï 3 5 x
Simplify.
STEP 2 Find the value of y.
hypotenuse 5 2 p shorter leg
}
}
y 5 2 p 3Ï 3 5 6Ï 3
30 8-60 8-90 8 Triangle Theorem
Substitute and simplify.
7.4 Special Right Triangles
459
EXAMPLE 6
Find a height
DUMP TRUCK The body of a dump truck is raised to
empty a load of sand. How high is the 14 foot body
from the frame when it is tipped upward at the
given angle?
a. 458 angle
b. 608 angle
Solution
a. When the body is raised 458 above the frame,
the height h is the length of a leg of a 458-458-908
triangle. The length of the hypotenuse is 14 feet.
}
REWRITE
MEASURES
To write 9.9 ft in feet
and inches, multiply the
decimal part by 12.
12 p 0.9 5 10.8
So, 9.9 ft is about
9 feet 11 inches.
14 5 h p Ï 2
458-458-908 Triangle Theorem
}
} 5 h
14
Ï2
Divide each side by Ï 2 .
9.9 ø h
Use a calculator to approximate.
458
}
14 ft
c When the angle of elevation is 458, the body is about 9 feet 11 inches
above the frame.
b. When the body is raised 608, the height h is the length of the longer
leg of a 308-608-908 triangle. The length of the hypotenuse is 14 feet.
hypotenuse 5 2 p shorter leg
14 5 2 p s
308-608-908 Triangle Theorem
608
Substitute.
75s
Divide each side by 2.
}
longer leg 5 shorter leg p Ï 3
}
308-608-908 Triangle Theorem
h 5 7Ï 3
Substitute.
h ø 12.1
Use a calculator to approximate.
c When the angle of elevation is 608, the body is about 12 feet 1 inch
above the frame.
(FPNFUSZ
✓
GUIDED PRACTICE
at classzone.com
for Examples 4, 5, and 6
Find the value of the variable.
5.
6.
608
4
3
x
4
h
308
2
2
7. WHAT IF? In Example 6, what is the height of the body of the dump truck
if it is raised 308 above the frame?
8. In a 308-608-908 triangle, describe the location of the shorter side. Describe
the location of the longer side?
460
Chapter 7 Right Triangles and Trigonometry
14 ft
7.4
EXERCISES
HOMEWORK
KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 5, 9, and 27
★ 5 STANDARDIZED TEST PRACTICE
Exs. 2, 6, 19, 22, 29, and 34
SKILL PRACTICE
1. VOCABULARY Copy and complete: A triangle with two congruent sides
and a right angle is called ? .
2.
EXAMPLES
1 and 2
on pp. 457–458
for Exs. 3–5
★ WRITING Explain why the acute angles in an isosceles right triangle
always measure 458.
45 8-45 8-90 8 TRIANGLES Find the value of x. Write your answer in simplest
radical form.
3.
4.
7
458
EXAMPLE 3
6.
on p. 458
for Exs. 6–7
★
5.
3 2
x
5 2
x
x
5 2
x
}
MULTIPLE CHOICE Find the length of AC.
}
C
}
A 7Ï2 in.
B 2Ï 7 in.
}
7Ï 2
C }
in.
458
}
D Ï 14 in.
2
A
B
7 in.
7. ISOSCELES RIGHT TRIANGLE The square tile shown
has painted corners in the shape of congruent
458-458-908 triangles. What is the value of x? What
is the side length of the tile?
EXAMPLES
4 and 5
on p. 459
for Exs. 8–10
30 8-60 8-90 8 TRIANGLES Find the value of each variable. Write your answers
in simplest radical form.
8.
9.
10.
3 3
12 3
308
y
9
x
x
308
608
y
y
x
SPECIAL RIGHT TRIANGLES Copy and complete the table.
11.
12.
b
458
c
d
608
458
a
a
b
C
7
?
?
f
e
?
11
?
?
?
10
}
Ï5
?
?
?
}
6Ï2
?
d
e
f
5
?
?
308
?
?
14
?
?
}
8Ï 3
?
?
?
12
}
18Ï3
7.4 Special Right Triangles
?
461
ALGEBRA Find the value of each variable. Write your answers in
simplest radical form.
13.
14.
15
608
24
y
x
16.
m
n
17.
r
18.
t
458
608
18
(FPNFUSZ
★
308
3
s
608
19.
15.
6
q
9 3
e
608
458
4
p
f
g
u
at classzone.com
MULTIPLE CHOICE Which side lengths do not represent a
308-608-908 triangle?
}
1 Ï3
A }
, }, 1
2
}
}
}
5 5Ï 3
C }
, }, 10
2
}
B Ï 2 , Ï 6 , 2Ï 2
2
}
D 3, 3Ï 3 , 6
2
ERROR ANALYSIS Describe and correct the error in finding the length of
the hypotenuse.
20.
21.
7 3
7
5
45º
5
30º
22.
★
5 2
WRITING Abigail solved Example 5 on page 459 in a different way.
}
}
Instead of dividing each side by Ï 3 , she multiplied each side by Ï3 . Does
her method work? Explain why or why not.
ALGEBRA Find the value of each variable. Write your answers in
simplest radical form.
23.
24.
308
g
f
10
25.
y
x
8
1508
308
4 2
26. CHALLENGE n ABC is a 308-608-908 triangle.
y
Find the coordinates of A.
A
x
1
462
5 WORKED-OUT SOLUTIONS
on p. WS1
y
1
C (23, 21)
★ 5 STANDARDIZED
TEST PRACTICE
B(3, 21)
608
x
APPENDIX A
REVIEW OF FUNDAMENTAL CONCEPTS OF ALGEBRA
A.1 REAL NUMBERS AND THEIR PROPERTIES
What you should learn
• Represent and classify real numbers.
• Order real numbers and use
inequalities.
• Find the absolute values of real
numbers and find the distance
between two real numbers.
• Evaluate algebraic expressions.
• Use the basic rules and
properties of algebra.
Real Numbers
Real numbers are used in everyday life to describe quantities such as age, miles per
gallon, and population. Real numbers are represented by symbols such as
4
3
!5, 9, 0, , 0.666 . . . , 28.21, "2, ", and "
!32.
3
Here are some important subsets (each member of subset B is also a member of set A)
of the real numbers. The three dots, called ellipsis points, indicate that the pattern
continues indefinitely.
!1, 2, 3, 4, . . .#
Why you should learn it
Real numbers are used to represent
many real-life quantities. For example,
in Exercises 83–88 on page A12, you
will use real numbers to represent the
federal deficit.
Set of natural numbers
!0, 1, 2, 3, 4, . . .#
Set of whole numbers
!. . . , !3, !2, !1, 0, 1, 2, 3, . . .#
Set of integers
A real number is rational if it can be written as the ratio p)q of two integers, where
q $ 0. For instance, the numbers
1
1
125
# 0.3333 . . . # 0.3, # 0.125, and
# 1.126126 . . . # 1.126
3
8
111
are rational. The decimal representation of a rational number either repeats (as in
173
1
55 # 3.145 ' or terminates (as in 2 # 0.5'. A real number that cannot be written as the
ratio of two integers is called irrational. Irrational numbers have infinite nonrepeating
decimal representations. For instance, the numbers
"2 # 1.4142135 . . . & 1.41
" # 3.1415926 . . . & 3.14
and
are irrational. (The symbol & means “is approximately equal to.”) Figure A.1 shows
subsets of real numbers and their relationships to each other.
Real
numbers
Classifying Real Numbers
Determine which numbers in the set
Irrational
numbers
Rational
numbers
Integers
Negative
integers
Natural
numbers
FIGURE
Example 1
$!13, !
Noninteger
fractions
(positive and
negative)
1
3
5
8
%
"5, !1, ! , 0, , "2, ", 7
are (a) natural numbers, (b) whole numbers, (c) integers, (d) rational numbers, and
(e) irrational numbers.
Solution
a. Natural numbers: !7#
b. Whole numbers: !0, 7#
c. Integers: !!13, !1, 0, 7#
Whole
numbers
Zero
A.1 Subsets of real numbers
$
%
1
5
d. Rational numbers: !13, !1, ! , 0, , 7
3
8
!
#
e. Irrational numbers: ! "5, "2, "
Now try Exercise 11.
A1
A2
Appendix A
Review of Fundamental Concepts of Algebra
Real numbers are represented graphically on the real number line. When you
draw a point on the real number line that corresponds to a real number, you are plotting
the real number. The point 0 on the real number line is the origin. Numbers to the right
of 0 are positive, and numbers to the left of 0 are negative, as shown in Figure A.2. The
term nonnegative describes a number that is either positive or zero.
Origin
Negative
direction
FIGURE
−4
−3
−2
−1
0
1
2
3
Positive
direction
4
A.2 The real number line
As illustrated in Figure A.3, there is a one-to-one correspondence between real
numbers and points on the real number line.
− 53
−3
−2
−1
0
−2.4
π
0.75
1
2
−3
3
Every real number corresponds to exactly
one point on the real number line.
FIGURE
−2
2
−1
0
1
2
3
Every point on the real number line
corresponds to exactly one real number.
A.3 One-to-one correspondence
Example 2
Plotting Points on the Real Number Line
Plot the real numbers on the real number line.
a. !
7
4
b. 2.3
c.
2
3
d. !1.8
Solution
All four points are shown in Figure A.4.
− 1.8 − 74
−2
FIGURE
2
3
−1
0
2.3
1
2
3
A.4
a. The point representing the real number ! 74 # !1.75 lies between !2 and !1, but
closer to !2, on the real number line.
b. The point representing the real number 2.3 lies between 2 and 3, but closer to 2, on
the real number line.
c. The point representing the real number 23 # 0.666 . . . lies between 0 and 1, but
closer to 1, on the real number line.
d. The point representing the real number !1.8 lies between !2 and !1, but closer to
!2, on the real number line. Note that the point representing !1.8 lies slightly to
the left of the point representing ! 74.
Now try Exercise 17.
Appendix A.1
Real Numbers and Their Properties
A3
Ordering Real Numbers
One important property of real numbers is that they are ordered.
Definition of Order on the Real Number Line
a
−1
If a and b are real numbers, a is less than b if b ! a is positive. The order of a
and b is denoted by the inequality a < b. This relationship can also be described
by saying that b is greater than a and writing b > a. The inequality a ≤ b means
that a is less than or equal to b, and the inequality b ≥ a means that b is greater
than or equal to a. The symbols <, >, %, and & are inequality symbols.
b
0
1
2
Geometrically, this definition implies that a < b if and only if a lies to the left of
b on the real number line, as shown in Figure A.5.
FIGURE A.5 a < b if and only if a lies to
the left of b.
Example 3
−4
−3
FIGURE
−4
−2
a. !3, 0
−2
−1
0
1
1 1
,
4 3
1 1
d. ! , !
5 2
1
c. Because 41 lies to the left of 3 on the real number line, as shown in Figure A.8, you
A.8
1
4
can say that is less than
− 12 − 15
−1
FIGURE
c.
a. Because !3 lies to the left of 0 on the real number line, as shown in Figure A.6, you
can say that !3 is less than 0, and write !3 < 0.
b. Because !2 lies to the right of !4 on the real number line, as shown in Figure A.7,
you can say that !2 is greater than !4, and write !2 > !4.
1
3
0
b. !2, !4
Solution
A.7
1
4
FIGURE
Place the appropriate inequality symbol (< or >' between the pair of real numbers.
0
A.6
−3
FIGURE
−1
Ordering Real Numbers
1
3,
and write 41 < 13.
d. Because ! 15 lies to the right of ! 12 on the real number line, as shown in Figure A.9,
you can say that ! 15 is greater than ! 12, and write ! 15 > ! 12.
0
A.9
Now try Exercise 25.
Example 4
Interpreting Inequalities
Describe the subset of real numbers represented by each inequality.
a. x % 2
x≤2
x
0
FIGURE
1
2
3
4
A.10
−2 ≤ x < 3
x
−2
−1
FIGURE
A.11
0
1
2
3
b. !2 % x < 3
Solution
a. The inequality x ≤ 2 denotes all real numbers less than or equal to 2, as shown in
Figure A.10.
b. The inequality !2 ≤ x < 3 means that x ≥ !2 and x < 3. This “double inequality”
denotes all real numbers between !2 and 3, including !2 but not including 3, as
shown in Figure A.11.
Now try Exercise 31.
A4
Appendix A
Review of Fundamental Concepts of Algebra
Inequalities can be used to describe subsets of real numbers called intervals. In the
bounded intervals below, the real numbers a and b are the endpoints of each interval.
The endpoints of a closed interval are included in the interval, whereas the endpoints of
an open interval are not included in the interval.
Bounded Intervals on the Real Number Line
Notation
+a, b*
The reason that the four types
of intervals at the right are called
bounded is that each has a finite
length. An interval that does not
have a finite length is unbounded
(see below).
WARNING / CAUTION
Whenever you write an interval
containing ' or ! ', always
use a parenthesis and never a
bracket. This is because ' and
! ' are never an endpoint of an
interval and therefore are not
included in the interval.
Interval Type
Closed
(a, b'
Open
+a, b'
Inequality
a % x % b
Graph
x
a
b
a
b
a
b
a
b
a < x < b
x
a % x < b
(a, b*
x
a < x % b
x
The symbols ', positive infinity, and ! ', negative infinity, do not represent
real numbers. They are simply convenient symbols used to describe the unboundedness
of an interval such as (1, '' or (! ', 3*.
Unbounded Intervals on the Real Number Line
Notation
+a, ''
Interval Type
Inequality
x & a
Graph
x
a
(a, ''
Open
x > a
x
a
(! ', b*
x % b
x
b
(! ', b'
Open
(! ', ''
Entire real line
x < b
x
b
Example 5
!' < x <
'
Using Inequalities to Represent Intervals
Use inequality notation to describe each of the following.
a. c is at most 2.
b. m is at least !3.
c. All x in the interval (!3, 5*
Solution
a. The statement “c is at most 2” can be represented by c ≤ 2.
b. The statement “m is at least !3” can be represented by m ≥ !3.
c. “All x in the interval (!3, 5*” can be represented by !3 < x ≤ 5.
Now try Exercise 45.
x
Appendix A.1
Example 6
Real Numbers and Their Properties
A5
Interpreting Intervals
Give a verbal description of each interval.
a. (!1, 0'
b. + 2, ''
c. (! ', 0'
Solution
a. This interval consists of all real numbers that are greater than !1 and less than 0.
b. This interval consists of all real numbers that are greater than or equal to 2.
c. This interval consists of all negative real numbers.
Now try Exercise 41.
Absolute Value and Distance
The absolute value of a real number is its magnitude, or the distance between the
origin and the point representing the real number on the real number line.
Definition of Absolute Value
If a is a real number, then the absolute value of a is
,a, # $!a,
a,
if a ≥ 0
.
if a < 0
Notice in this definition that the absolute value of a real number is never negative.
For instance, if a # !5, then !5 # ! (!5' # 5. The absolute value of a real
number is either positive or zero. Moreover, 0 is the only real number whose absolute
value is 0. So, 0 # 0.
, ,
,,
Example 7
Finding Absolute Values
, ,
,!4.3, # 4.3
,,
2
2
#
3
3
a. !15 # 15
b.
c.
d. ! !6 # ! (6' # !6
, ,
Now try Exercise 51.
Example 8
Evaluate
Evaluating the Absolute Value of a Number
,x, for (a) x > 0 and (b) x < 0.
x
Solution
,,
a. If x > 0, then x # x and
,,
,x, # x # 1.
x
b. If x < 0, then x # !x and
x
, , # !x # !1.
x
x
Now try Exercise 59.
x
A6
Appendix A
Review of Fundamental Concepts of Algebra
The Law of Trichotomy states that for any two real numbers a and b, precisely one
of three relationships is possible:
a # b,
a < b,
Example 9
or
a > b.
Law of Trichotomy
Comparing Real Numbers
Place the appropriate symbol (<, >, or =) between the pair of real numbers.
, ,!,3,
,
a. !4
,!,10,
, ,!,!7,
b. !10
Solution
, , ,,
, , , ,
, , , ,
c. ! !7
, ,
,,
, ,
, ,
, ,
, ,
a. !4 > 3 because !4 # 4 and 3 # 3, and 4 is greater than 3.
b. !10 # 10 because !10 # 10 and 10 # 10.
c. ! !7 < !7 because ! !7 # !7 and !7 # 7, and !7 is less than 7.
Now try Exercise 61.
Properties of Absolute Values
,,
2. !a # a
, , , ,, ,
4.
1. a & 0
3. ab # a b
−2
−1
0
,,
Absolute value can be used to define the distance between two points on the real
number line. For instance, the distance between !3 and 4 is
7
−3
, , ,,
a
,a,, b $ 0
#
b
,b,
1
2
3
4
A.12 The distance between !3
and 4 is 7.
,!3 ! 4, # ,!7,
#7
FIGURE
as shown in Figure A.12.
Distance Between Two Points on the Real Number Line
Let a and b be real numbers. The distance between a and b is
,
, ,
,
d(a, b' # b ! a # a ! b .
Example 10
Finding a Distance
Find the distance between !25 and 13.
Solution
The distance between !25 and 13 is given by
,!25 ! 13, # ,!38, # 38.
Distance between !25 and 13
The distance can also be found as follows.
,13 ! (!25', # ,38, # 38
Now try Exercise 67.
Distance between !25 and 13
Appendix A.1
Real Numbers and Their Properties
A7
Algebraic Expressions
One characteristic of algebra is the use of letters to represent numbers. The letters are
variables, and combinations of letters and numbers are algebraic expressions. Here
are a few examples of algebraic expressions.
5x,
2x ! 3,
x2
4
,
(2
7x ( y
Definition of an Algebraic Expression
An algebraic expression is a collection of letters (variables) and real numbers
(constants) combined using the operations of addition, subtraction, multiplication,
division, and exponentiation.
The terms of an algebraic expression are those parts that are separated by addition.
For example,
x 2 ! 5x ( 8 # x 2 ( (!5x' ( 8
has three terms: x 2 and !5x are the variable terms and 8 is the constant term. The
numerical factor of a term is called the coefficient. For instance, the coefficient of !5x
is !5, and the coefficient of x 2 is 1.
Example 11
Identifying Terms and Coefficients
Algebraic Expression
1
7
b. 2x2 ! 6x ( 9
3 1
c. ( x4 ! y
x
2
a. 5x !
Terms
Coefficients
1
7
2x2, !6x, 9
3 1 4
, x , !y
x 2
1
7
2, !6, 9
1
3, , !1
2
5x, !
5, !
Now try Exercise 89.
To evaluate an algebraic expression, substitute numerical values for each of the
variables in the expression, as shown in the next example.
Example 12
Evaluating Algebraic Expressions
Expression
a. !3x ( 5
b. 3x 2 ( 2x ! 1
2x
c.
x(1
Value of
Variable
x#3
x # !1
x # !3
Substitute
Value of
Expression
!3(3' ( 5
3(!1'2 ( 2(!1' ! 1
2(!3'
!3 ( 1
!9 ( 5 # !4
3!2!1#0
!6
#3
!2
Note that you must substitute the value for each occurrence of the variable.
Now try Exercise 95.
When an algebraic expression is evaluated, the Substitution Principle is used. It
states that “If a # b, then a can be replaced by b in any expression involving a.” In
Example 12(a), for instance, 3 is substituted for x in the expression !3x ( 5.
A8
Appendix A
Review of Fundamental Concepts of Algebra
Basic Rules of Algebra
There are four arithmetic operations with real numbers: addition, multiplication,
subtraction, and division, denoted by the symbols (, + or * , !, and ) or /. Of
these, addition and multiplication are the two primary operations. Subtraction and
division are the inverse operations of addition and multiplication, respectively.
Definitions of Subtraction and Division
Subtraction: Add the opposite.
a ! b # a ( (!b'
Division: Multiply by the reciprocal.
If b $ 0, then a)b # a
-b. # b .
1
a
In these definitions, !b is the additive inverse (or opposite) of b, and 1)b is
the multiplicative inverse (or reciprocal) of b. In the fractional form a)b,
a is the numerator of the fraction and b is the denominator.
Because the properties of real numbers below are true for variables and
algebraic expressions as well as for real numbers, they are often called the Basic Rules
of Algebra. Try to formulate a verbal description of each property. For instance, the
first property states that the order in which two real numbers are added does not affect
their sum.
Basic Rules of Algebra
Let a, b, and c be real numbers, variables, or algebraic expressions.
Property
Commutative Property of Addition:
Commutative Property of Multiplication:
Associative Property of Addition:
Associative Property of Multiplication:
Distributive Properties:
Additive Identity Property:
Multiplicative Identity Property:
Additive Inverse Property:
Multiplicative Inverse Property:
Example
a(b#b(a
ab # ba
(a ( b' ( c # a ( (b ( c'
(ab' c # a(bc'
a(b ( c' # ab ( ac
(a ( b'c # ac ( bc
a(0#a
a*1#a
a ( (!a' # 0
1
a * # 1, a $ 0
a
x2
4x ( # x 2 ( 4x
(4 ! x' x 2 # x 2(4 ! x'
(x ( 5' ( x 2 # x ( (5 ( x 2'
(2x * 3y'(8' # (2x'(3y * 8'
3x(5 ( 2x' # 3x * 5 ( 3x * 2x
( y ( 8' y # y * y ( 8 * y
5y 2 ( 0 # 5y 2
(4x 2'(1' # 4x 2
5x 3 ( (!5x 3' # 0
1
(x 2 ( 4' 2
#1
x (4
-
.
Because subtraction is defined as “adding the opposite,” the Distributive Properties
are also true for subtraction. For instance, the “subtraction form” of
a(b ( c' # ab ( ac is a(b ! c' # ab ! ac. Note that the operations of subtraction
and division are neither commutative nor associative. The examples
7 ! 3 $ 3 ! 7 and
20 ) 4 $ 4 ) 20
show that subtraction and division are not commutative. Similarly
5 ! (3 ! 2' $ (5 ! 3' ! 2 and
16 ) (4 ) 2) $ (16 ) 4) ) 2
demonstrate that subtraction and division are not associative.
Appendix A.1
Example 13
Real Numbers and Their Properties
A9
Identifying Rules of Algebra
Identify the rule of algebra illustrated by the statement.
a. (5x3'2 # 2(5x3'
b.
-4x ( 31. ! -4x ( 31. # 0
1
# 1, x $ 0
7x
d. (2 ( 5x2' ( x2 # 2 ( (5x2 ( x2'
c. 7x *
Solution
a. This statement illustrates the Commutative Property of Multiplication. In other
words, you obtain the same result whether you multiply 5x3 by 2, or 2 by 5x3.
b. This statement illustrates the Additive Inverse Property. In terms of subtraction, this
property simply states that when any expression is subtracted from itself the result
is 0.
c. This statement illustrates the Multiplicative Inverse Property. Note that it is
important that x be a nonzero number. If x were 0, the reciprocal of x would be
undefined.
d. This statement illustrates the Associative Property of Addition. In other words, to
form the sum
2 ( 5x2 ( x2
it does not matter whether 2 and 5x2, or 5x2 and x2 are added first.
Now try Exercise 101.
Properties of Negation and Equality
Let a, b, and c be real numbers, variables, or algebraic expressions.
Notice the difference between
the opposite of a number and a
negative number. If a is already
negative, then its opposite, !a,
is positive. For instance, if
a # !5, then
!a # !(!5) # 5.
Property
1. (!1' a # !a
Example
(!1'7 # !7
2. ! (!a' # a
! (!6' # 6
3. (!a'b # ! (ab' # a(!b'
(!5'3 # ! (5 * 3' # 5(!3'
4. (!a'(!b' # ab
(!2'(!x' # 2x
5. ! (a ( b' # (!a' ( (!b'
! (x ( 8' # (!x' ( (!8'
6. If a # b, then a ± c # b ± c.
1
2
7. If a # b, then ac # bc.
42
8. If a ± c # b ± c, then a # b.
1.4 ! 1 # 75 ! 1 ⇒ 1.4 # 75
9. If ac # bc and c $ 0, then a # b.
3x # 3
# !x ! 8
( 3 # 0.5 ( 3
* 2 # 16 * 2
*4
⇒ x#4
A10
Appendix A
Review of Fundamental Concepts of Algebra
Properties of Zero
The “or” in the Zero-Factor
Property includes the possibility
that either or both factors may be
zero. This is an inclusive or, and
it is the way the word “or” is
generally used in mathematics.
Let a and b be real numbers, variables, or algebraic expressions.
1. a ( 0 # a and a ! 0 # a
3.
0
# 0,
a
2. a
4.
a$0
*0#0
a
is undefined.
0
5. Zero-Factor Property: If ab # 0, then a # 0 or b # 0.
Properties and Operations of Fractions
Let a, b, c, and d be real numbers, variables, or algebraic expressions such that
b $ 0 and d $ 0.
1. Equivalent Fractions:
2. Rules of Signs: !
a
c
# if and only if ad # bc.
b d
a !a
a
!a a
#
#
and
#
b
b
!b
!b b
3. Generate Equivalent Fractions:
a ac
# , c$0
b bc
4. Add or Subtract with Like Denominators:
a c
a±c
± #
b b
b
5. Add or Subtract with Unlike Denominators:
In Property 1 of fractions, the
phrase “if and only if” implies
two statements. One statement
is: If a)b # c)d, then ad # bc.
The other statement is: If
ad # bc, where b $ 0 and
d $ 0, then a)b # c)d.
6. Multiply Fractions:
7. Divide Fractions:
Example 14
a
b
c
a c
ad ± bc
± #
b d
bd
ac
* d # bd
a
c
a
) #
b d b
d
ad
* c # bc ,
c$0
Properties and Operations of Fractions
a. Equivalent fractions:
x 3 * x 3x
#
#
5 3 * 5 15
c. Add fractions with unlike denominators:
b. Divide fractions:
7 3 7 2 14
) # * #
x 2 x 3 3x
x
2x 5 * x ( 3 * 2x 11x
(
#
#
3
5
3*5
15
Now try Exercise 119.
If a, b, and c are integers such that ab # c, then a and b are factors or divisors of c.
A prime number is an integer that has exactly two positive factors—itself and 1—such
as 2, 3, 5, 7, and 11. The numbers 4, 6, 8, 9, and 10 are composite because each can be
written as the product of two or more prime numbers. The number 1 is neither prime
nor composite. The Fundamental Theorem of Arithmetic states that every
positive integer greater than 1 can be written as the product of prime numbers in
precisely one way (disregarding order). For instance, the prime factorization of 24 is
24 # 2 * 2 * 2 * 3.
Appendix A.1
A.1
EXERCISES
Real Numbers and Their Properties
A11
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
p
of two integers, where q $ 0.
q
________ numbers have infinite nonrepeating decimal representations.
The point 0 on the real number line is called the ________.
The distance between the origin and a point representing a real number on the real number line is
the ________ ________ of the real number.
A number that can be written as the product of two or more prime numbers is called a ________ number.
An integer that has exactly two positive factors, the integer itself and 1, is called a ________ number.
An algebraic expression is a collection of letters called ________ and real numbers called ________.
The ________ of an algebraic expression are those parts separated by addition.
The numerical factor of a variable term is the ________ of the variable term.
The ________ ________ states that if ab # 0, then a # 0 or b # 0.
1. A real number is ________ if it can be written as the ratio
2.
3.
4.
5.
6.
7.
8.
9.
10.
SKILLS AND APPLICATIONS
In Exercises 11–16, determine which numbers in the set are
(a) natural numbers, (b) whole numbers, (c) integers,
(d) rational numbers, and (e) irrational numbers.
In Exercises 25–30, plot the two real numbers on the real
number line. Then place the appropriate inequality symbol
(< or >) between them.
11.
12.
13.
14.
15.
16.
25. !4, !8
27. 32, 7
26. !3.5, 1
16
28. 1, 3
2
29. 56, 3
8
3
30. ! 7, ! 7
!!9, ! 72, 5, 23, "2, 0, 1, !4, 2, !11#
!"5, !7, ! 73, 0, 3.12, 54 , !3, 12, 5#
!2.01, 0.666 . . . , !13, 0.010110111 . . . , 1, !6#
!2.3030030003 . . . , 0.7575, !4.63, "10, !75, 4#
!! ", ! 13, 63, 12"2, !7.5, !1, 8, !22#
!25, !17, ! 125, "9, 3.12, 12", 7, !11.1, 13#
In Exercises 17 and 18, plot the real numbers on the real
number line.
17. (a) 3 (b) 72
18. (a) 8.5 (b)
5
(c) ! 2 (d) !5.2
4
8
3 (c) !4.75 (d) ! 3
In Exercises 19–22, use a calculator to find the decimal form
of the rational number. If it is a nonterminating decimal,
write the repeating pattern.
19.
21.
5
8
41
333
20.
22.
1
3
6
11
24.
−3
−7
−2
−6
−1
−5
0
−4
1
−3
−2
31.
33.
35.
37.
39.
41.
x % 5
x < 0
+4, ''
!2 < x < 2
!1 ≤ x < 0
+!2, 5'
32.
34.
36.
38.
40.
42.
x & !2
x > 3
(! ', 2'
0 ≤ x ≤ 5
0 < x ≤ 6
(!1, 2*
In Exercises 43–50, use inequality notation and interval
notation to describe the set.
In Exercises 23 and 24, approximate the numbers and place
the correct symbol (< or >) between them.
23.
In Exercises 31– 42, (a) give a verbal description of the
subset of real numbers represented by the inequality or the
interval, (b) sketch the subset on the real number line, and (c)
state whether the interval is bounded or unbounded.
2
3
−1
0
43.
44.
45.
46.
47.
48.
49.
50.
y is nonnegative.
y is no more than 25.
x is greater than !2 and at most 4.
y is at least !6 and less than 0.
t is at least 10 and at most 22.
k is less than 5 but no less than !3.
The dog’s weight W is more than 65 pounds.
The annual rate of inflation r is expected to be at least
2.5% but no more than 5%.
Appendix A
Review of Fundamental Concepts of Algebra
In Exercises 51–60, evaluate the expression.
51.
52.
53.
54.
55.
56.
,!10,
,0,
,3 ! 8,
,4 ! 1,
,!1, ! ,!2,
!3 ! ,!3,
!5
!5
58. !3 !3
57.
59.
, ,
, ,
x
(
, 2,,
x < !2
x(2
x!1
60.
, x > 1
x!1
,
,
In Exercises 61–66, place the correct symbol (<, >, or #'
between the two real numbers.
61.
62.
63.
64.
65.
66.
,!3,!! ,!3,
,!4,!,4,
!5!! ,5,
! ,!6,!,!6,
! ,!2,!! ,2,
BUDGET VARIANCE In Exercises 79–82, the accounting
department of a sports drink bottling company is checking to
see whether the actual expenses of a department differ from
the budgeted expenses by more than $500 or by more than
5%. Fill in the missing parts of the table, and determine
whether each actual expense passes the “budget variance
test.”
79.
80.
81.
82.
In Exercises 73–78, use absolute value notation to describe
the situation.
73.
74.
75.
76.
77.
The distance between x and 5 is no more than 3.
The distance between x and !10 is at least 6.
y is at least six units from 0.
y is at most two units from a.
While traveling on the Pennsylvania Turnpike, you pass
milepost 57 near Pittsburgh, then milepost 236 near
Gettysburg. How many miles do you travel during that
time period?
78. The temperature in Bismarck, North Dakota was 60,F
at noon, then 23,F at midnight. What was the change in
temperature over the 12-hour period?
,a ! b,
!
!
!
!
2600
!(!2)!!2
a # 126, b # 75
a # !126, b # !75
a # ! 52, b # 0
a # 14, b # 11
4
112
a # 16
5 , b # 75
a # 9.34, b # !5.65
Budgeted
Actual
Expense, b Expense, a
$112,700
$113,356
$9,400
$9,772
$37,640
$37,335
$2,575
$2,613
0.05b
!
!
!
!
FEDERAL DEFICIT In Exercises 83–88, use the bar graph,
which shows the receipts of the federal government (in
billions of dollars) for selected years from 1996 through
2006. In each exercise you are given the expenditures of the
federal government. Find the magnitude of the surplus or
deficit for the year. (Source: U.S. Office of Management
and Budget)
In Exercises 67–72, find the distance between a and b.
67.
68.
69.
70.
71.
72.
Wages
Utilities
Taxes
Insurance
Receipts
(in billions of dollars)
A12
2407.3
2400
2200
2025.5
2000
1800
1600
1400
1722.0
1853.4 1880.3
1453.2
1200
1996 1998 2000 2002 2004 2006
Year
Year Receipts
83.
84.
85.
86.
87.
88.
1996
1998
2000
2002
2004
2006
!
!
!
!
!
!
Expenditures
|Receipts !
Expenditures|
$1560.6 billion
$1652.7 billion
$1789.2 billion
$2011.2 billion
$2293.0 billion
$2655.4 billion
!
!
!
!
!
!
In Exercises 89–94, identify the terms. Then identify the
coefficients of the variable terms of the expression.
89. 7x ( 4
91. "3x 2 ! 8x ! 11
x
93. 4x 3 ( ! 5
2
90. 6x 3 ! 5x
92. 3"3x 2 ( 1
x2
94. 3x 4 !
4
Appendix A.1
In Exercises 95–100, evaluate the expression for each value
of x. (If not possible, state the reason.)
95.
96.
97.
98.
99.
100.
Expression
4x ! 6
9 ! 7x
x 2 ! 3x ( 4
!x 2 ( 5x ! 4
x(1
x!1
x
x(2
(a)
(a)
(a)
(a)
Values
x # !1
(b)
x # !3
(b)
x # !2
(b)
x # !1
(b)
x#0
x#3
x#2
x#1
(a) x # 1
(b) x # !1
(a) x # 2
(b) x # !2
In Exercises 101–112, identify the rule(s) of algebra illustrated
by the statement.
101. x ( 9 # 9 ( x
102. 2( 12 ' # 1
1
103.
(h ( 6' # 1, h $ !6
h(6
104. (x ( 3' ! (x ( 3' # 0
105. 2(x ( 3' # 2 * x ( 2 * 3
106. (z ! 2' ( 0 # z ! 2
107. 1 * (1 ( x' # 1 ( x
108. (z ( 5'x # z * x ( 5 * x
109. x ( ( y ( 10' # (x ( y' ( 10
110. x(3y' # (x * 3'y # (3x' y
111. 3(t ! 4' # 3 * t ! 3 * 4
1
112. 7(7 * 12' # ( 17 * 7'12 # 1 * 12 # 12
In Exercises 113–120, perform the operation(s). (Write
fractional answers in simplest form.)
5
3
113. 16
( 16
5
115. 58 ! 12
( 16
1
117. 12 ) 4
119.
114. 76 ! 47
6
13
116. 10
11 ( 33 ! 66
4
118. ! (6 * 8 '
2x
x
!
3
4
120.
5x
6
2
*9
EXPLORATION
In Exercises 121 and 122, use the real numbers A, B, and C
shown on the number line. Determine the sign of each
expression.
C B
A
0
121. (a) !A
(b) B ! A
122. (a) !C
(b) A ! C
A13
Real Numbers and Their Properties
123. CONJECTURE
(a) Use a calculator to complete the table.
1
n
0.5
0.01
0.0001
0.000001
5)n
(b) Use the result from part (a) to make a conjecture
about the value of 5)n as n approaches 0.
124. CONJECTURE
(a) Use a calculator to complete the table.
1
n
10
100
10,000
100,000
5)n
(b) Use the result from part (a) to make a conjecture
about the value of 5)n as n increases without
bound.
TRUE OR FALSE? In Exercises 125–128, determine
whether the statement is true or false. Justify your answer.
125. If a > 0 and b < 0, then a ! b > 0.
126. If a > 0 and b < 0, then ab > 0.
127. If a < b, then
128. Because
1 1
< , where a $ 0 and b $ 0.
a b
a(b a b
c
c
c
# ( , then
# ( .
c
c
c
a(b a b
,
,
,, ,,
129. THINK ABOUT IT Consider u ( v and u ( v ,
where u $ 0 and v $ 0.
(a) Are the values of the expressions always equal? If
not, under what conditions are they unequal?
(b) If the two expressions are not equal for certain
values of u and v, is one of the expressions always
greater than the other? Explain.
130. THINK ABOUT IT Is there a difference between
saying that a real number is positive and saying that a
real number is nonnegative? Explain.
131. THINK ABOUT IT Because every even number is
divisible by 2, is it possible that there exist any even
prime numbers? Explain.
132. THINK ABOUT IT Is it possible for a real number to
be both rational and irrational? Explain.
133. WRITING Can it ever be true that a # !a for a
real number a? Explain.
,,
134. CAPSTONE Describe the differences among the
sets of natural numbers, whole numbers, integers,
rational numbers, and irrational numbers.
A14
Appendix A
Review of Fundamental Concepts of Algebra
A.2
What you should learn
• Use properties of exponents.
• Use scientific notation to represent
real numbers.
• Use properties of radicals.
• Simplify and combine radicals.
• Rationalize denominators and
numerators.
• Use properties of rational exponents.
EXPONENTS AND RADICALS
Integer Exponents
Repeated multiplication can be written in exponential form.
Repeated Multiplication
a*a*a*a*a
Exponential Form
a5
(!4'(!4'(!4'
(!4'3
(2x'(2x'(2x'(2x'
(2x'4
Why you should learn it
Exponential Notation
Real numbers and algebraic expressions
are often written with exponents and
radicals. For instance, in Exercise 121
on page A26, you will use an expression
involving rational exponents to find the
times required for a funnel to empty for
different water heights.
If a is a real number and n is a positive integer, then
an # a * a
*a.
. .a
n factors
where n is the exponent and a is the base. The expression an is read “a to the nth
power.”
An exponent can also be negative. In Property 3 below, be sure you see how to use a
negative exponent.
Properties of Exponents
T E C H N O LO G Y
You can use a calculator to
evaluate exponential expressions.
When doing so, it is important to
know when to use parentheses
because the calculator follows the
order of operations. For instance,
evaluate (!2'4 as follows.
Scientific:
/
2
yx
0
#$!
4
"
/
/!0
2
0
>
Graphing:
4
Let a and b be real numbers, variables, or algebraic expressions, and let m and n
be integers. (All denominators and bases are nonzero.)
Example
# 32(4 # 36 # 729
1.
32
2.
am
# am!n
an
x7
# x7! 4 # x 3
x4
3. a!n #
-.
1
1
#
an
a
4. a0 # 1,
*
34
n
a$0
y!4 #
-.
1
1
#
y4
y
4
(x 2 ( 1'0 # 1
5. (ab'm # am bm
(5x'3 # 53x3 # 125x3
6. (am'n # amn
( y3'!4 # y3(!4) # y!12 #
ENTER
The display will be 16. If you
omit the parentheses, the display
will be !16.
Property
# a m(n
a ma n
7.
-b.
a
m
#
am
bm
, , ,,
8. a2 # a 2 # a2
-x.
2
3
#
1
y12
23
8
#
x3 x3
,(!2'2, # ,!2,2 # (2'2 # 4
Appendix A.2
Exponents and Radicals
A15
It is important to recognize the difference between expressions such as (!2'4 and
In (!2'4, the parentheses indicate that the exponent applies to the negative sign
as well as to the 2, but in !24 # ! (24', the exponent applies only to the 2. So,
(!2'4 # 16 and !24 # !16.
The properties of exponents listed on the preceding page apply to all integers m and
n, not just to positive integers, as shown in the examples in this section.
!24.
Example 1
Evaluating Exponential Expressions
a. (!5'2 # (!5'(!5' # 25
Negative sign is part of the base.
b. !52 # ! (5'(5' # !25
Negative sign is not part of the base.
c. 2
Property 1
d.
* 2 4 # 21(4 # 25 # 32
4
4
1
1
# 44!6 # 4!2 # 2 #
46
4
16
Properties 2 and 3
Now try Exercise 11.
Example 2
Evaluating Algebraic Expressions
Evaluate each algebraic expression when x # 3.
a. 5x!2
b.
1
(!x'3
3
Solution
a. When x # 3, the expression 5x!2 has a value of
5x!2 # 5(3'!2 #
5
5
# .
2
3
9
1
b. When x # 3, the expression (!x'3 has a value of
3
1
1
1
(!x'3 # (!3'3 # (!27' # !9.
3
3
3
Now try Exercise 23.
Example 3
Using Properties of Exponents
Use the properties of exponents to simplify each expression.
a. (!3ab4'(4ab!3'
b. (2xy2'3
c. 3a(!4a2'0
Solution
a. (!3ab4'(4ab!3' # (!3'(4'(a'(a'(b4'(b!3' # !12a 2b
b. (2xy 2'3 # 23(x'3( y 2'3 # 8x3y6
c. 3a(!4a 2'0 # 3a(1' # 3a, a $ 0
5x 3 2 52(x 3'2 25x 6
d.
-y.
#
y2
#
y2
Now try Exercise 31.
d.
-5xy .
3 2
A16
Appendix A
Review of Fundamental Concepts of Algebra
Example 4
Rarely in algebra is there only
one way to solve a problem.
Don’t be concerned if the steps
you use to solve a problem are
not exactly the same as the steps
presented in this text. The
important thing is to use steps
that you understand and, of
course, steps that are justified
by the rules of algebra. For
instance, you might prefer the
following steps for Example 4(d).
- .
3x 2
y
!2
- .
y
#
3x 2
2
y2
# 4
9x
Note how Property 3 is used
in the first step of this solution.
The fractional form of this
property is
-.
a
b
!m
Rewrite each expression with positive exponents.
a. x!1
b.
1
3x!2
c.
12a3b!4
4a!2b
d.
-3xy .
2 !2
Solution
1
x
a. x!1 #
Property 3
b.
1
1(x 2' x 2
#
#
3x!2
3
3
The exponent !2 does not apply to 3.
c.
12a3b!4 12a3 * a2
#
4a!2b
4b * b4
Property 3
d.
-3xy .
#
3a5
b5
Property 1
#
3!2(x 2'!2
y!2
Properties 5 and 7
#
3!2x!4
y!2
Property 6
#
y2
32x 4
Property 3
#
y2
9x 4
Simplify.
2 !2
-..
b
#
a
Rewriting with Positive Exponents
m
Now try Exercise 41.
Scientific Notation
Exponents provide an efficient way of writing and computing with very large (or very
small) numbers. For instance, there are about 359 billion billion gallons of water on
Earth—that is, 359 followed by 18 zeros.
359,000,000,000,000,000,000
It is convenient to write such numbers in scientific notation. This notation has the form
± c + 10n, where 1 ≤ c < 10 and n is an integer. So, the number of gallons of water on
Earth can be written in scientific notation as
3.59
+
100,000,000,000,000,000,000 # 3.59 + 1020.
The positive exponent 20 indicates that the number is large (10 or more) and that
the decimal point has been moved 20 places. A negative exponent indicates that the
number is small (less than 1). For instance, the mass (in grams) of one electron is
approximately
9.0
+
10!28 # 0.0000000000000000000000000009.
28 decimal places
Appendix A.2
Example 5
Exponents and Radicals
A17
Scientific Notation
Write each number in scientific notation.
a. 0.0000782
b. 836,100,000
Solution
a. 0.0000782 # 7.82 + 10!5
b. 836,100,000 # 8.361 + 108
Now try Exercise 45.
Example 6
Decimal Notation
Write each number in decimal notation.
a. !9.36 + 10!6
b. 1.345 + 102
Solution
a. !9.36 + 10!6 # !0.00000936
b. 1.345
+
102 # 134.5
Now try Exercise 55.
T E C H N O LO G Y
Most calculators automatically switch to scientific notation when they are showing
large (or small) numbers that exceed the display range.
To enter numbers in scientific notation, your calculator should have an exponential
entry key labeled
EE
or EXP .
Consult the user’s guide for your calculator for instructions on keystrokes and how
numbers in scientific notation are displayed.
Example 7
Evaluate
Using Scientific Notation
(2,400,000,000'(0.0000045'
.
(0.00003'(1500'
Solution
Begin by rewriting each number in scientific notation and simplifying.
(2,400,000,000'(0.0000045' (2.4 + 109'(4.5 + 10!6'
#
(0.00003'(1500'
(3.0 + 10!5'(1.5 + 103'
#
(2.4'(4.5'(103'
(4.5'(10!2'
# (2.4'(105'
# 240,000
Now try Exercise 63(b).
A18
Appendix A
Review of Fundamental Concepts of Algebra
Radicals and Their Properties
A square root of a number is one of its two equal factors. For example, 5 is a square
root of 25 because 5 is one of the two equal factors of 25. In a similar way, a cube root
of a number is one of its three equal factors, as in 125 # 53.
Definition of nth Root of a Number
Let a and b be real numbers and let n & 2 be a positive integer. If
a # bn
then b is an nth root of a. If n # 2, the root is a square root. If n # 3, the root
is a cube root.
Some numbers have more than one nth root. For example, both 5 and !5 are square
roots of 25. The principal square root of 25, written as "25, is the positive root, 5. The
principal nth root of a number is defined as follows.
Principal nth Root of a Number
Let a be a real number that has at least one nth root. The principal nth root of a
is the nth root that has the same sign as a. It is denoted by a radical symbol
n a.
"
Principal nth root
The positive integer n is the index of the radical, and the number a is the radicand.
2 a. (The plural of index is
If n # 2, omit the index and write "a rather than "
indices.)
A common misunderstanding is that the square root sign implies both negative and
positive roots. This is not correct. The square root sign implies only a positive root.
When a negative root is needed, you must use the negative sign with the square root
sign.
Incorrect: "4 # ± 2
Example 8
Correct: ! "4 # !2
and "4 # 2
Evaluating Expressions Involving Radicals
a. "36 # 6 because 62 # 36.
b. ! "36 # !6 because ! ("36' # ! ("62' # ! (6' # !6.
5 3 53 125
125 5
c. 3
# because
# 3#
.
64
4
4
4
64
5 !32 # !2 because (!2'5 # !32.
d. "
4 !81 is not a real number because there is no real number that can be raised to the
e. "
fourth power to produce !81.
"
-.
Now try Exercise 65.
Appendix A.2
Exponents and Radicals
A19
Here are some generalizations about the nth roots of real numbers.
Generalizations About nth Roots of Real Numbers
Real Number a
Integer n
Root(s) of a
Example
a > 0
n > 0, n is even.
n a,
n a
"
!"
4 81 # 3,
4 81 # !3
"
!"
a > 0 or a < 0
n is odd.
n
"
a
3
"
!8 # !2
a < 0
n is even.
No real roots
"!4 is not a real number.
a#0
n is even or odd.
n 0 # 0
"
5 0 # 0
"
Integers such as 1, 4, 9, 16, 25, and 36 are called perfect squares because they
have integer square roots. Similarly, integers such as 1, 8, 27, 64, and 125 are called
perfect cubes because they have integer cube roots.
Properties of Radicals
Let a and b be real numbers, variables, or algebraic expressions such that the
indicated roots are real numbers, and let m and n be positive integers.
1.
Property
m
n
a # ("
a'
n
2. "
a
3.
4.
n
"
a
n
"
b
n
n
b#"
ab
*"
#
"
n
a
,
b
Example
2
8 ' # (2'2 # 4
8 #(
3 2
"
n m
"
"5
* "7 # "5 * 7 # "35
4
"
27
b$0
4
"
9
m n
"a # mn
"a
"
3
"
#
"279 #
4
4
"
3
3
6
"10 # "
"
10
n
5. ("
a' # a
("3 '2 # 3
n
,,
,
,
"(!12'2 # !12 # 12
n n
6. For n even, "
a # a.
3
"
(!12'3 # !12
n n
For n odd, "
a # a.
,,
A common special case of Property 6 is "a2 # a .
Example 9
Using Properties of Radicals
Use the properties of radicals to simplify each expression.
a. "8
* "2
3 5
b. ("
'
3
Solution
a.
b.
c.
d.
"8
(
* "2 # "8 * 2 # "16 # 4
3 5
"
'3 # 5
3 x3 # x
"
,,
6 y6 # y
"
Now try Exercise 77.
3 x3
c. "
6 y6
d. "
A20
Appendix A
Review of Fundamental Concepts of Algebra
Simplifying Radicals
An expression involving radicals is in simplest form when the following conditions are
satisfied.
1. All possible factors have been removed from the radical.
2. All fractions have radical-free denominators (accomplished by a process called
rationalizing the denominator).
3. The index of the radical is reduced.
To simplify a radical, factor the radicand into factors whose exponents are
multiples of the index. The roots of these factors are written outside the radical, and the
“leftover” factors make up the new radicand.
WARNING / CAUTION
Example 10
When you simplify a radical, it
is important that both expressions
are defined for the same values
of the variable. For instance,
in Example 10(b), "75x3 and
5x"3x are both defined only
for nonnegative values of x.
Similarly, in Example 10(c),
4 (5x'4 and 5 x are both
"
defined for all real values of x.
Perfect
4th power
,,
Simplifying Even Roots
Leftover
factor
4
4
a. "
48 # "
16
4 4
4
2 * 3 # 2"
3
*3#"
Perfect
square
Leftover
factor
b. "75x3 # "25x 2 * 3x
# "(5x'2 * 3x
# 5x"3x
4
4
c. "(5x' # 5x # 5 x
, ,
Find largest square factor.
Find root of perfect square.
,,
Now try Exercise 79(a).
Example 11
Simplifying Odd Roots
Perfect
cube
Leftover
factor
3
3
a. "
24 # "
8
3 3
3
2 * 3 # 2"
3
*3#"
Perfect
cube
Leftover
factor
3
3
b. "
24a4 # "
8a3 * 3a
3 (2a'3
#"
* 3a
3 3a
# 2a "
3 !40x6 # "
3 (!8x6'
c. "
*5
3 (!2x 2'3
#"
3 5
# !2x 2 "
Find largest cube factor.
Find root of perfect cube.
Find largest cube factor.
*5
Find root of perfect cube.
Now try Exercise 79(b).
Appendix A.2
Exponents and Radicals
A21
Radical expressions can be combined (added or subtracted) if they are like
radicals—that is, if they have the same index and radicand. For instance, "2, 3"2,
and 12"2 are like radicals, but "3 and "2 are unlike radicals. To determine whether
two radicals can be combined, you should first simplify each radical.
Example 12
Combining Radicals
a. 2"48 ! 3"27 # 2"16
* 3 ! 3"9 * 3
Find square factors.
Find square roots and
multiply by coefficients.
# 8"3 ! 9"3
# (8 ! 9'"3
# ! "3
3
3
4
3 8
3 27
b. "16x ! "54x # "
* 2x ! "
3 2x ! 3x"
3 2x
# 2"
3
# (2 ! 3x' "2x
Combine like terms.
Simplify.
*
x3
* 2x
Find cube factors.
Find cube roots.
Combine like terms.
Now try Exercise 87.
Rationalizing Denominators and Numerators
To rationalize a denominator or numerator of the form a ! b"m or a ( b"m,
multiply both numerator and denominator by a conjugate: a ( b"m and a ! b"m
are conjugates of each other. If a # 0, then the rationalizing factor for "m is itself,
"m. For cube roots, choose a rationalizing factor that generates a perfect cube.
Example 13
Rationalizing Single-Term Denominators
Rationalize the denominator of each expression.
5
2
a.
b. 3
"
"
2 3
5
Solution
a.
5
2"3
#
5
"3
*"
2"3
3
"
5 3
#
2(3'
5"3
#
6
3 52
2
2 "
b. 3 # 3 * 3 2
"5
"5 "5
3 52
2"
# 3 3
"5
3 25
2"
#
5
"3 is rationalizing factor.
Multiply.
Simplify.
3 52 is rationalizing factor.
"
Multiply.
Simplify.
Now try Exercise 95.
A22
Appendix A
Review of Fundamental Concepts of Algebra
Example 14
Rationalizing a Denominator with Two Terms
2
2
#
3 ( "7 3 ( "7
Multiply numerator and
denominator by conjugate
of denominator.
3 ! "7
* 3 ! "7
#
2(3 ! "7 '
3(3' ( 3(!"7 ' ( "7(3' ! ("7 '("7 '
Use Distributive Property.
#
2(3 ! "7 '
(3'2 ! ("7 '2
Simplify.
#
2(3 ! "7 '
9!7
Square terms of
denominator.
#
2(3 ! "7 '
# 3 ! "7
2
Simplify.
Now try Exercise 97.
Sometimes it is necessary to rationalize the numerator of an expression. For
instance, in Appendix A.4 you will use the technique shown in the next example to
rationalize the numerator of an expression from calculus.
WARNING / CAUTION
Do not confuse the expression
"5 ( "7 with the expression
"5 ( 7. In general, "x ( y
does not equal "x ( "y.
Similarly, "x 2 ( y 2 does not
equal x ( y.
Example 15
"5 ! "7
2
Rationalizing a Numerator
#
"5 ! "7
"5 ( "7
* "5 ( "7
2
Multiply numerator and
denominator by conjugate
of numerator.
#
("5 '2 ! ("7 '2
2("5 ( "7'
Simplify.
#
5!7
2("5 ( "7 '
Square terms of numerator.
#
!2
!1
#
2("5 ( "7 ' "5 ( "7
Simplify.
Now try Exercise 101.
Rational Exponents
Definition of Rational Exponents
If a is a real number and n is a positive integer such that the principal nth root of
a exists, then a1)n is defined as
n a, where 1)n is the rational exponent of a.
a1)n # "
Moreover, if m is a positive integer that has no common factor with n, then
n a
a m)n # (a1)n'm # ("
'
m
The symbol
and
n a m.
a m)n # (a m'1)n # "
indicates an example or exercise that highlights algebraic techniques specifically
used in calculus.
Appendix A.2
WARNING / CAUTION
Rational exponents can be
tricky, and you must remember
that the expression bm)n is not
n b is a real
defined unless "
number. This restriction
produces some unusual-looking
results. For instance, the number
(!8'1)3 is defined because
3
"
!8 # !2, but the number
(!8'2)6 is undefined because
6 !8 is not a real number.
"
Exponents and Radicals
A23
The numerator of a rational exponent denotes the power to which the base is raised,
and the denominator denotes the index or the root to be taken.
Power
Index
n b
n bm
b m)n # ("
' #"
m
When you are working with rational exponents, the properties of integer exponents
still apply. For instance, 21)221)3 # 2(1)2)((1)3) # 25)6.
Example 16
Changing From Radical to Exponential Form
a. "3 # 31)2
2 (3xy'5 # (3xy'5)2
b. "(3xy'5 # "
4 x3 # (2x'(x3)4' # 2x1((3)4) # 2x7)4
c. 2x "
Now try Exercise 103.
Example 17
T E C H N O LO G Y
Changing From Exponential to Radical Form
a. (x 2 ( y 2'3)2 # ("x 2 ( y 2 ' # "(x 2 ( y 2'3
3
>
There are four methods of evaluating radicals on most graphing
calculators. For square roots, you
can use the square root key " .
For cube roots, you can use the
cube root key "3 . For other
roots, you can first convert the
radical to exponential form and
then use the exponential key ,
or you can use the xth root key
x
"
(or menu choice). Consult
the user’s guide for your calculator
for specific keystrokes.
4 y3z
b. 2y3)4z1)4 # 2( y3z'1)4 # 2 "
c. a!3)2 #
1
1
#
a3)2 "a3
5 x
d. x 0.2 # x1)5 # "
Now try Exercise 105.
Rational exponents are useful for evaluating roots of numbers on a calculator, for
reducing the index of a radical, and for simplifying expressions in calculus.
Example 18
Simplifying with Rational Exponents
5 !32
a. (!32'!4)5 # ("
'
!4
# (!2'!4 #
1
1
#
(!2'4 16
b. (!5x5)3'(3x!3)4' # !15x(5)3)!(3)4) # !15x11)12,
c.
9 a3
"
d.
3 "
6 125 # "
6 (5'3 # 53)6 # 51)2 # "5
"
125 # "
#
a3)9
#
a1)3
#
3 a
"
x$0
Reduce index.
e. (2x ! 1'4)3(2x ! 1'!1)3 # (2x ! 1'(4)3)!(1)3)
# 2x ! 1,
x$
1
2
Now try Exercise 115.
The expression in Example 18(e) is not defined when x #
-2 * 12 ! 1.
!1)3
is not a real number.
# (0'!1)3
1
because
2
A24
Appendix A
A.2
Review of Fundamental Concepts of Algebra
EXERCISES
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
In the exponential form an, n is the ________ and a is the ________.
A convenient way of writing very large or very small numbers is called ________ ________.
One of the two equal factors of a number is called a __________ __________ of the number.
n
The ________ ________ ________ of a number a is the nth root that has the same sign as a, and is denoted by "
a.
n
In the radical form "
a, the positive integer n is called the ________ of the radical and the number a is called the
________.
When an expression involving radicals has all possible factors removed, radical-free denominators, and a reduced index,
it is in ________ ________.
Radical expressions can be combined (added or subtracted) if they are ________ ________.
The expressions a ( b"m and a ! b"m are ________ of each other.
The process used to create a radical-free denominator is known as ________ the denominator.
In the expression bm)n, m denotes the ________ to which the base is raised and n denotes the ________ or root to be taken.
SKILLS AND APPLICATIONS
In Exercises 11–18, evaluate each expression.
In Exercises 31–38, simplify each expression.
11. (a) 32 * 3
55
12. (a) 2
5
13. (a) (33'0
14. (a) (23 * 32'2
(b) 3 * 33
32
(b) 4
3
(b) !32
3
2
(b) (! 35 ' (53 '
31. (a) (!5z'3
32. (a) (3x'2
3
15. (a) !4
3
(b) 48(!4)!3
4 * 3!2
2!2 * 3!1
17. (a) 2!1 ( 3!1
18. (a) 3!1 ( 2!2
16. (a)
33. (a) 6y 2(2y0'2
34. (a) (!z'3(3z4'
(b) (!2'0
(b) (2!1'!2
(b) (3!2'2
37. (a) +(x2y!2'!1*!1
In Exercises 19 –22, use a calculator to evaluate the
expression. (If necessary, round your answer to three decimal
places.)
19. (!4' ( '
36
21. 3
7
3
52
7x 2
x3
r4
36. (a) 6
r
35. (a)
20. ( '(
43
22. !4
3
8!4
'
103
38. (a) (6x7'0,
x$0
39. (a) (x ( 5'0, x $ !5
40. (a) (2x5'0, x $ 0
41. (a) (!2x 2'3(4x3'!1
23.
25.
27.
29.
42. (a) (4y!2'(8y4'
24.
26.
28.
30.
7x!2, x # 4
5(!x'3, x # 3
!3x 4, x # !2
12(!x'3, x # ! 13
- .- .
a
b
(b) .b
a.
!2
3
!2
(b) (5x2z6'3(5x2z6'!3
In Exercises 39–44, rewrite each expression with positive
exponents and simplify.
In Exercises 23–30, evaluate the expression for the given
value of x.
!3x 3, x # 2
6x 0, x # 10
2x 3, x # !3
!20x2, x # ! 12
(b) 5x4(x2'
(b) (4x 3'0, x $ 0
3x 5
(b) 3
x
25y8
(b)
10y4
12(x ( y'3
(b)
9(x ( y'
4 3 3 4
(b)
y y
43. (a) 3n
44. (a)
* 32n
x 2 * xn
x 3 * xn
(b) (2x 2'!2
(b) (z ( 2'!3(z ( 2'!1
x !1
(b)
10
x!3y 4 !3
(b)
5
!2
a
b 3
(b)
b!2 a
a!3 a 3
(b)
b!3 b
- .
- .
- .- .
- .- .
Appendix A.2
In Exercises 45–52, write the number in scientific notation.
45.
47.
49.
50.
51.
10,250.4
46. !7,280,000
!0.000125
48. 0.00052
Land area of Earth: 57,300,000 square miles
Light year: 9,460,000,000,000 kilometers
Relative density of hydrogen: 0.0000899 gram per
cubic centimeter
52. One micron (millionth of a meter): 0.00003937 inch
In Exercises 53– 60, write the number in decimal notation.
53. 1.25 + 105
54. !1.801 + 105
55. !2.718 + 10!3
56. 3.14 + 10!4
57. Interior temperature of the sun: 1.5 + 107 degrees
Celsius
58. Charge of an electron: 1.6022 + 10!19 coulomb
59. Width of a human hair: 9.0 + 10!5 meter
60. Gross domestic product of the United States in 2007:
1.3743021 + 1013 dollars (Source: U.S. Department
of Commerce)
In Exercises 61 and 62, evaluate each expression without
using a calculator.
61. (a) (2.0 + 109'(3.4 + 10!4'
(b) (1.2 + 107'(5.0 + 10!3'
6.0 + 108
62. (a)
3.0 + 10!3
2.5 + 10!3
(b)
5.0 + 102
In Exercises 63 and 64, use a calculator to evaluate each
expression. (Round your answer to three decimal places.)
-
.
0.11 800
365
67,000,000 ( 93,000,000
(b)
0.0052
63. (a) 750 1 (
64. (a) (9.3
+
106'3(6.1
+
10!4'
(b)
(2.414 + 104'6
(1.68 + 105'5
In Exercises 65–70, evaluate each expression without using a
calculator.
65.
66.
67.
68.
(a)
(a)
(a)
(a)
"9
271)3
32!3)5
100!3)2
1 !1)3
69. (a) !
64
125 !1)3
70. (a) !
27
-
.
.
3 27
(b) "
8
(b) 363)2
!3)4
(b) (16
81 '
!1)2
(b) (94 '
1 !2)5
(b)
"32
1 !4)3
(b) !
125
-
.
- .
A25
Exponents and Radicals
In Exercises 71–76, use a calculator to approximate the
number. (Round your answer to three decimal places.)
71. (a) "57
3 452
72. (a) "
73. (a) (!12.4'!1.8
74. (a)
7 ! (4.1'!3.2
2
75. (a) "4.5 + 109
76. (a) (2.65 + 10!4'1)3
5 !273
(b) "
6
(b) "
125
!2.5
(b) (5"3'
(b)
- .
13
3
!3)2
- .
! !
3
2
13)3
3
(b) "
6.3 + 104
(b) "9 + 10!4
In Exercises 77 and 78, use the properties of radicals to
simplify each expression.
5 2 5
77. (a) ("
'
78. (a) "12 * "3
5 96x5
(b) "
4 (3x2'4
(b) "
In Exercises 79–90, simplify each radical expression.
79. (a) "20
3 16
80. (a) "
27
81. (a) "72x3
82. (a) "54xy4
83.
84.
85.
86.
87.
88.
(a)
(a)
(a)
(a)
(a)
(a)
(b)
89. (a)
(b)
90. (a)
(b)
3 16x5
"
4 3x 4 y 2
"
2"50 ( 12"8
4"27 ! "75
5"x ! 3"x
8"49x ! 14"100x
!3"48x 2 ( 7" 75x 2
3"x ( 1 ( 10"x ( 1
7"80x ! 2"125x
! "x 3 ! 7 ( 5"x 3 ! 7
11"245x 3 ! 9"45x 3
3 128
(b) "
(b) "75
4
182
(b)
z3
32a4
(b)
b2
(b) "75x2y!4
5 160x 8z 4
(b) "
(b) 10"32 ! 6"18
3 16 ( 3"
3 54
(b) "
(b) !2"9y ( 10"y
"
"
In Exercises 91–94, complete the statement with <, #, or >.
"113 !
"3
91. "5 ( "3 !"5 ( 3
92.
93. 5!"32 ( 22
94. 5!"32 ( 42
"11
In Exercises 95–98, rationalize the denominator of the
expression. Then simplify your answer.
95.
97.
1
96.
"3
5
"14 ! 2
98.
8
3
"
2
3
"5 ( "6
A26
Appendix A
Review of Fundamental Concepts of Algebra
In Exercises 99 –102, rationalize the numerator of the
expression. Then simplify your answer.
99.
"8
2
"5 ( "3
101.
3
100.
t # 0.03+125)2 ! (12 ! h'5)2*, 0 % h % 12
"2
3
"7 ! 3
102.
4
In Exercises 103 –110, fill in the missing form of the
expression.
Radical Form
103. "2.5
3 64
104. "
105.!
106.!
3 !216
107. "
108.!
4 81 3
109. ("
'
110.!
Rational Exponent Form
!
!
811)4
! (1441)2'
!
(!243'1)5
!
165)4
In Exercises 111–114, perform the operations and simplify.
( '
21)2x4
x!3 * x1)2
113. 3)2 !1
x *x
111.
2x2 3)2
112.
(xy'1)3
5!1)2 * 5x5)2
114.
(5x'3)2
115. (a)
6 x3
116. (a) "
(b) (x ( 1'
4 (3x2'4
(b) "
6
"
4
In Exercises 117 and 118, write each expression as a single
radical. Then simplify your answer.
117. (a)
118. (a)
""32
""243(x ( 1'
(b)
(b)
4 2x
""
3 10a7b
""
119. PERIOD OF A PENDULUM The period T (in
seconds) of a pendulum is T # 2""L)32, where L is
the length of the pendulum (in feet). Find the period of
a pendulum whose length is 2 feet.
120. EROSION A stream of water moving at the rate of v feet
per second can carry particles of size 0.03"v inches.
Find the size of the largest particle that can be carried
by a stream flowing at the rate of 34 foot per second.
The symbol
represents the amount of time t (in seconds) that it will
take for the funnel to empty.
(a) Use the table feature of a graphing utility to find
the times required for the funnel to empty for
water heights of h # 0, h # 1, h # 2, . . . h # 12
centimeters.
(b) What value does t appear to be approaching as the
height of the water becomes closer and closer to
12 centimeters?
122. SPEED OF LIGHT The speed of light is approximately
11,180,000 miles per minute. The distance from the
sun to Earth is approximately 93,000,000 miles. Find
the time for light to travel from the sun to Earth.
EXPLORATION
TRUE OR FALSE? In Exercises 123 and 124, determine
whether the statement is true or false. Justify your answer.
123.
x 4)3y 2)3
In Exercises 115 and 116, reduce the index of each radical.
4 32
"
121. MATHEMATICAL MODELING A funnel is filled
with water to a height of h centimeters. The formula
indicates an example or exercise that highlights
algebraic techniques specifically used in calculus.
The symbol
indicates an exercise or a part of an exercise in which
you are instructed to use a graphing utility.
x k(1
# xk
x
124. (a n' k # a n
k
125. Verify that a0 # 1, a $ 0. (Hint: Use the property of
exponents am)a n # am!n.)
126. Explain why each of the following pairs is not equal.
(a) (3x'!1 $
3
x
(b) y 3
* y2 $ y6
(c) (a 2b 3'4 $ a6b7
(d) (a ( b'2 $ a 2 ( b2
(e) "4x 2 $ 2x
(f) "2 ( "3 $ "5
127. THINK ABOUT IT Is 52.7 + 105 written in scientific
notation? Why or why not?
128. List all possible digits that occur in the units place
of the square of a positive integer. Use that list to determine whether "5233 is an integer.
129. THINK ABOUT IT Square the real number 5)"3
and note that the radical is eliminated from the
denominator. Is this equivalent to rationalizing the
denominator? Why or why not?
130. CAPSTONE
(a) Explain how to simplify the expression (3x3 y!2'!2.
(b) Is the expression
or why not?
"x4 in simplest form? Why
3
Appendix A.3
A27
Polynomials and Factoring
A.3 POLYNOMIALS AND FACTORING
What you should learn
• Write polynomials in standard form.
• Add, subtract, and multiply
polynomials.
• Use special products to multiply
polynomials.
• Remove common factors from
polynomials.
• Factor special polynomial forms.
• Factor trinomials as the product of
two binomials.
• Factor polynomials by grouping.
Why you should learn it
Polynomials can be used to model
and solve real-life problems. For
instance, in Exercise 224 on page A37,
a polynomial is used to model the
volume of a shipping box.
Polynomials
The most common type of algebraic expression is the polynomial. Some examples are
2x ( 5, 3x 4 ! 7x 2 ( 2x ( 4, and 5x 2y 2 ! xy ( 3. The first two are polynomials in x
and the third is a polynomial in x and y. The terms of a polynomial in x have the form
ax k, where a is the coefficient and k is the degree of the term. For instance, the
polynomial
2x 3 ! 5x 2 ( 1 # 2x 3 ( (!5' x 2 ( (0' x ( 1
has coefficients 2, !5, 0, and 1.
Definition of a Polynomial in x
Let a0, a1, a2, . . . , an be real numbers and let n be a nonnegative integer.
A polynomial in x is an expression of the form
an x n ( an!1x n!1 ( . . . ( a1x ( a 0
where an $ 0. The polynomial is of degree n, an is the leading coefficient, and
a0 is the constant term.
Polynomials with one, two, and three terms are called monomials, binomials, and
trinomials, respectively. In standard form, a polynomial is written with descending
powers of x.
Example 1
Writing Polynomials in Standard Form
Standard Form
Degree
Leading
Coefficient
!5x 7 ( 4x 2 ( 3x ! 2
!9x 2 ( 4
8 (8 # 8x 0'
7
2
0
!5
!9
8
Polynomial
a. 4x 2 ! 5x 7 ! 2 ( 3x
b. 4 ! 9x 2
c. 8
Now try Exercise 19.
A polynomial that has all zero coefficients is called the zero polynomial, denoted
by 0. No degree is assigned to this particular polynomial. For polynomials in more than
one variable, the degree of a term is the sum of the exponents of the variables in the
term. The degree of the polynomial is the highest degree of its terms. For instance, the
degree of the polynomial !2x 3y6 ( 4xy ! x7y 4 is 11 because the sum of the exponents
in the last term is the greatest. The leading coefficient of the polynomial is the
coefficient of the highest-degree term. Expressions are not polynomials if a variable is
underneath a radical or if a polynomial expression (with degree greater than 0) is in the
denominator of a term. The following expressions are not polynomials.
x 3 ! "3x # x 3 ! (3x'1)2
x2 (
5
# x 2 ( 5x!1
x
The exponent “1)2” is not an integer.
The exponent “!1” is not a nonnegative integer.
A28
Appendix A
Review of Fundamental Concepts of Algebra
Operations with Polynomials
You can add and subtract polynomials in much the same way you add and subtract real
numbers. Simply add or subtract the like terms (terms having the same variables to the
same powers) by adding their coefficients. For instance, !3xy 2 and 5xy 2 are like terms
and their sum is
!3xy 2 ( 5xy 2 # (!3 ( 5' xy 2
# 2xy 2.
WARNING / CAUTION
When an expression inside
parentheses is preceded by a
negative sign, remember to
distribute the negative sign to
each term inside the parentheses,
as shown.
! (x 2 ! x ( 3'
# !x 2 ( x ! 3
Example 2
Sums and Differences of Polynomials
a. (5x 3 ! 7x 2 ! 3' ( (x 3 ( 2x 2 ! x ( 8'
# (5x 3 ( x 3' ( (!7x2 ( 2x2' ! x ( (!3 ( 8'
# 6x 3 ! 5x 2 ! x ( 5
b. (7x4 ! x 2 ! 4x ( 2' ! (3x4 ! 4x 2 ( 3x'
# 7x 4 ! x 2 ! 4x ( 2 ! 3x 4 ( 4x 2 ! 3x
#(
!
'((
(
4
2
# 4x ( 3x ! 7x ( 2
7x 4
3x 4
!x2
Group like terms.
Combine like terms.
Distributive Property
' ( (!4x ! 3x' ( 2
4x2
Group like terms.
Combine like terms.
Now try Exercise 41.
To find the product of two polynomials, use the left and right Distributive
Properties. For example, if you treat 5x ( 7 as a single quantity, you can multiply
3x ! 2 by 5x ( 7 as follows.
(3x ! 2'(5x ( 7' # 3x(5x ( 7' ! 2(5x ( 7'
# (3x'(5x' ( (3x'(7' ! (2'(5x' ! (2'(7'
# 15x 2 ( 21x ! 10x ! 14
Product of
First terms
Product of
Outer terms
Product of
Inner terms
Product of
Last terms
# 15x 2 ( 11x ! 14
Note in this FOIL Method (which can only be used to multiply two binomials) that the
outer (O) and inner (I) terms are like terms and can be combined.
Example 3
Finding a Product by the FOIL Method
Use the FOIL Method to find the product of 2x ! 4 and x ( 5.
Solution
F
O
I
L
(2x ! 4'(x ( 5' # 2x 2 ( 10x ! 4x ! 20
# 2x 2 ( 6x ! 20
Now try Exercise 55.
Appendix A.3
Polynomials and Factoring
A29
Special Products
Some binomial products have special forms that occur frequently in algebra. You
do not need to memorize these formulas because you can use the Distributive Property
to multiply. However, becoming familiar with these formulas will enable you to
manipulate the algebra more quickly.
Special Products
Let u and v be real numbers, variables, or algebraic expressions.
Special Product
Sum and Difference of Same Terms
(u ( v'(u ! v' # u 2 ! v 2
Example
(x ( 4'(x ! 4' # x 2 ! 42
# x 2 ! 16
Square of a Binomial
(u ( v' 2 # u 2 ( 2uv ( v 2
(x ( 3' 2 # x 2 ( 2(x'(3' ( 32
# x 2 ( 6x ( 9
(3x ! 2'2 # (3x'2 ! 2(3x'(2' ( 22
# 9x 2 ! 12x ( 4
(u ! v' 2 # u 2 ! 2uv ( v 2
Cube of a Binomial
(u ( v'3 # u 3 ( 3u 2v ( 3uv 2 ( v 3
(x ( 2'3 # x 3 ( 3x 2(2' ( 3x(22' (23
# x 3 ( 6x 2 ( 12x ( 8
(x !1'3 # x 3 !3x 2(1' (3x(12' !13
# x 3 ! 3x 2 ( 3x ! 1
(u ! v'3 # u 3 ! 3u 2v ( 3uv 2 ! v 3
Example 4
Special Products
Find each product.
a. (5x ( 9'(5x ! 9'
b. (x ( y ! 2'(x ( y ( 2'
Solution
a. The product of a sum and a difference of the same two terms has no middle term and
takes the form (u ( v'(u ! v' # u 2 ! v 2.
(5x ( 9'(5x ! 9' # (5x'2 ! 9 2 # 25x 2 ! 81
b. By grouping x ( y in parentheses, you can write the product of the trinomials as a
special product.
Difference
Sum
(x ( y ! 2'(x ( y ( 2' # +(x ( y' ! 2*+(x ( y' ( 2*
# (x ( y' 2 ! 22
2
2
# x ( 2xy ( y ! 4
Now try Exercise 75.
Sum and difference of same terms
A30
Appendix A
Review of Fundamental Concepts of Algebra
Polynomials with Common Factors
The process of writing a polynomial as a product is called factoring. It is an important
tool for solving equations and for simplifying rational expressions.
Unless noted otherwise, when you are asked to factor a polynomial, you can
assume that you are looking for factors with integer coefficients. If a polynomial
cannot be factored using integer coefficients, then it is prime or irreducible over the
integers. For instance, the polynomial x 2 ! 3 is irreducible over the integers. Over the
real numbers, this polynomial can be factored as
x 2 ! 3 # (x ( "3 '(x ! "3 '.
A polynomial is completely factored when each of its factors is prime. For
instance
x 3 ! x 2 ( 4x ! 4 # (x ! 1'(x 2 ( 4'
Completely factored
is completely factored, but
x 3 ! x 2 ! 4x ( 4 # (x ! 1'(x 2 ! 4'
Not completely factored
is not completely factored. Its complete factorization is
x 3 ! x 2 ! 4x ( 4 # (x ! 1'(x ( 2'(x ! 2'.
The simplest type of factoring involves a polynomial that can be written as the
product of a monomial and another polynomial. The technique used here is the
Distributive Property, a(b ( c' # ab ( ac, in the reverse direction.
ab ( ac # a(b ( c'
a is a common factor.
Removing (factoring out) any common factors is the first step in completely factoring
a polynomial.
Example 5
Removing Common Factors
Factor each expression.
a. 6x 3 ! 4x
b. !4x 2 ( 12x ! 16
c. (x ! 2'(2x' ( (x ! 2'(3'
Solution
a. 6x 3 ! 4x # 2x(3x 2' ! 2x(2'
2x is a common factor.
# 2x(3x 2 ! 2'
b. !4x 2 ( 12x ! 16 # !4(x 2' ( (!4'(!3x' ( (!4'4
!4 is a common factor.
# !4(x 2 ! 3x ( 4'
c. (x ! 2'(2x' ( (x ! 2'(3' # (x ! 2'(2x ( 3'
Now try Exercise 99.
(x ! 2' is a common factor.
Appendix A.3
Polynomials and Factoring
A31
Factoring Special Polynomial Forms
Some polynomials have special forms that arise from the special product forms on page
A29. You should learn to recognize these forms so that you can factor such polynomials
easily.
Factoring Special Polynomial Forms
Factored Form
Difference of Two Squares
Example
u 2 ! v 2 # (u ( v'(u ! v'
9x 2 ! 4 # (3x' 2 ! 2 2 # (3x ( 2'(3x ! 2'
Perfect Square Trinomial
u 2 ( 2uv ( v 2 # (u ( v' 2
x 2 ( 6x ( 9 # x 2 ( 2(x'(3' ( 32 # (x ( 3'2
u 2 ! 2uv ( v 2 # (u ! v' 2
x 2 ! 6x ( 9 # x 2 ! 2(x'(3' ( 32 # (x ! 3'2
Sum or Difference of Two Cubes
u 3 ( v 3 # (u ( v'(u 2 ! uv ( v 2'
x 3 ( 8 # x 3 ( 23 # (x ( 2'(x 2 ! 2x ( 4'
u3 ! v3 # (u ! v'(u2 ( uv ( v 2'
27x3 ! 1 # (3x' 3 ! 13 # (3x ! 1'(9x 2 ( 3x ( 1'
One of the easiest special polynomial forms to factor is the difference of two
squares. The factored form is always a set of conjugate pairs.
u 2 ! v 2 # (u ( v'(u ! v'
Difference
Conjugate pairs
Opposite signs
To recognize perfect square terms, look for coefficients that are squares of
integers and variables raised to even powers.
Example 6
In Example 6, note that the first
step in factoring a polynomial
is to check for any common
factors. Once the common
factors are removed, it is often
possible to recognize patterns
that were not immediately
obvious.
Removing a Common Factor First
3 ! 12x 2 # 3(1 ! 4x 2'
3 is a common factor.
# 3+1 ! (2x' *
2
2
# 3(1 ( 2x'(1 ! 2x'
Difference of two squares
Now try Exercise 113.
Example 7
Factoring the Difference of Two Squares
a. (x ( 2'2 ! y 2 # +(x ( 2' ( y*+(x ( 2' ! y*
# (x ( 2 ( y'(x ( 2 ! y'
b. 16x 4 ! 81 # (4x 2'2 ! 92
# (4x 2 ( 9'(4x 2 ! 9'
# (4x2 ( 9'+(2x'2 ! 32*
# (4x2 ( 9'(2x ( 3'(2x ! 3'
Now try Exercise 117.
Difference of two squares
Difference of two squares
A32
Appendix A
Review of Fundamental Concepts of Algebra
A perfect square trinomial is the square of a binomial, and it has the following
form.
u 2 ( 2uv ( v 2 # (u ( v' 2
or
u 2 ! 2uv ( v 2 # (u ! v' 2
Like signs
Like signs
Note that the first and last terms are squares and the middle term is twice the product
of u and v.
Example 8
Factoring Perfect Square Trinomials
Factor each trinomial.
a. x 2 ! 10x ( 25
b. 16x 2 ( 24x ( 9
Solution
a. x 2 ! 10x ( 25 # x 2 ! 2(x'(5' ( 5 2 # (x ! 5'2
b. 16x2 ( 24x ( 9 # (4x'2 ( 2(4x'(3' ( 32 # (4x ( 3'2
Now try Exercise 123.
The next two formulas show the sums and differences of cubes. Pay special
attention to the signs of the terms.
Like signs
Like signs
u 3 ( v 3 # (u ( v'(u 2 ! uv ( v 2'
u 3 ! v 3 # (u ! v'(u 2 ( uv ( v 2'
Unlike signs
Example 9
Unlike signs
Factoring the Difference of Cubes
Factor x 3 ! 27.
Solution
x3 ! 27 # x3 ! 33
Rewrite 27 as 33.
# (x ! 3'(x 2 ( 3x ( 9'
Factor.
Now try Exercise 133.
Example 10
Factoring the Sum of Cubes
a. y 3 ( 8 # y 3 ( 23
Rewrite 8 as 23.
# ( y ( 2'( y 2 ! 2y ( 4'
b. 3(
x3
( 64' # 3(
x3
(
'
43
# 3(x ( 4'(x 2 ! 4x ( 16'
Now try Exercise 135.
Factor.
Rewrite 64 as 43.
Factor.
Appendix A.3
Polynomials and Factoring
A33
Trinomials with Binomial Factors
To factor a trinomial of the form ax 2 ( bx ( c, use the following pattern.
Factors of a
ax2 ( bx ( c # (!x ( !'(!x ( !'
Factors of c
The goal is to find a combination of factors of a and c such that the outer and inner
products add up to the middle term bx. For instance, in the trinomial 6x 2 ( 17x ( 5,
you can write all possible factorizations and determine which one has outer and inner
products that add up to 17x.
(6x ( 5'(x ( 1', (6x ( 1'(x ( 5', (2x ( 1'(3x ( 5', (2x ( 5'(3x ( 1'
You can see that (2x ( 5'(3x ( 1' is the correct factorization because the outer (O) and
inner (I) products add up to 17x.
F
O
I
L
O(I
(2x ( 5'(3x ( 1' # 6x 2 ( 2x ( 15x ( 5 # 6x2 ( 17x ( 5
Example 11
Factoring a Trinomial: Leading Coefficient Is 1
Factor x 2 ! 7x ( 12.
Solution
The possible factorizations are
(x ! 2'(x ! 6', (x ! 1'(x ! 12', and (x ! 3'(x ! 4'.
Testing the middle term, you will find the correct factorization to be
x 2 ! 7x ( 12 # (x ! 3'(x ! 4'.
Now try Exercise 145.
Example 12
Factoring a Trinomial: Leading Coefficient Is Not 1
Factor 2x 2 ( x ! 15.
Solution
Factoring a trinomial can
involve trial and error. However,
once you have produced the
factored form, it is an easy
matter to check your answer.
For instance, you can verify the
factorization in Example 11 by
multiplying out the expression
(x ! 3'(x ! 4' to see that you
obtain the original trinomial,
x2 ! 7x ( 12.
The eight possible factorizations are as follows.
(2x ! 1'(x ( 15'
(2x ( 1'(x ! 15'
(2x ! 3'(x ( 5'
(2x ( 3'(x ! 5'
(2x ! 5'(x ( 3'
(2x ( 5'(x ! 3'
(2x ! 15'(x ( 1'
(2x ( 15'(x ! 1'
Testing the middle term, you will find the correct factorization to be
2x 2 ( x ! 15 # (2x ! 5'(x ( 3'.
Now try Exercise 153.
O ( I # 6x ! 5x # x
A34
Appendix A
Review of Fundamental Concepts of Algebra
Factoring by Grouping
Sometimes polynomials with more than three terms can be factored by a method called
factoring by grouping. It is not always obvious which terms to group, and sometimes
several different groupings will work.
Example 13
Factoring by Grouping
Use factoring by grouping to factor x 3 ! 2x 2 ! 3x ( 6.
Another way to factor the
polynomial in Example 13 is
to group the terms as follows.
x3
!
2x2
Solution
x 3 ! 2x 2 ! 3x ( 6 # (x 3 ! 2x 2' ! (3x ! 6'
! 3x ( 6
# (x3 ! 3x' ! (2x2 ! 6'
# x(x ! 3' ! 2(x ! 3'
2
Group terms.
# x 2(x ! 2' ! 3(x ! 2'
Factor each group.
# (x ! 2'(x 2 ! 3'
Distributive Property
Now try Exercise 161.
2
# (x2 ! 3'(x ! 2'
As you can see, you obtain the
same result as in Example 13.
Factoring a trinomial can involve quite a bit of trial and error. Some of this trial and
error can be lessened by using factoring by grouping. The key to this method of factoring is knowing how to rewrite the middle term. In general, to factor a trinomial
ax2 ( bx ( c by grouping, choose factors of the product ac that add up to b and use
these factors to rewrite the middle term. This technique is illustrated in Example 14.
Example 14
Factoring a Trinomial by Grouping
Use factoring by grouping to factor 2x 2 ( 5x ! 3.
Solution
In the trinomial 2x 2 ( 5x ! 3, a # 2 and c # !3, which implies that the product ac
is !6. Now, !6 factors as (6'(!1' and 6 ! 1 # 5 # b. So, you can rewrite the
middle term as 5x # 6x ! x. This produces the following.
2x 2 ( 5x ! 3 # 2x 2 ( 6x ! x ! 3
#(
2x 2
( 6x' ! (x ( 3'
Rewrite middle term.
Group terms.
# 2x(x ( 3' ! (x ( 3'
Factor groups.
# (x ( 3'(2x ! 1'
Distributive Property
So, the trinomial factors as 2x 2 ( 5x ! 3 # (x ( 3'(2x ! 1'.
Now try Exercise 167.
Guidelines for Factoring Polynomials
1. Factor out any common factors using the Distributive Property.
2. Factor according to one of the special polynomial forms.
3. Factor as ax2 ( bx ( c # (mx ( r'(nx ( s'.
4. Factor by grouping.
Appendix A.3
A.3
EXERCISES
Polynomials and Factoring
A35
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
1. For the polynomial an x n ( an!1 x n!1 ( . . . ( a1x ( a0, an $ 0, the degree is ________, the leading coefficient is
________, and the constant term is ________.
2. A polynomial in x in standard form is written with ________ powers of x.
3. A polynomial with one term is called a ________, while a polynomial with two terms is called a ________,
and a polynomial with three terms is called a ________.
4. To add or subtract polynomials, add or subtract the ________ ________ by adding their coefficients.
5. The letters in “FOIL” stand for the following.
F ________ O ________ I ________ L ________
6. The process of writing a polynomial as a product is called ________.
7. A polynomial is ________ ________ when each of its factors is prime.
8. A polynomial u2 ( 2uv ( v2 is called a ________ ________ ________.
SKILLS AND APPLICATIONS
In Exercises 9–14, match the polynomial with its description.
[The polynomials are labeled (a), (b), (c), (d), (e), and (f).]
(a)
(c)
(e)
9.
10.
11.
12.
13.
14.
3x2
x3 ( 3x2 ( 3x ( 1
!3x5 ( 2x3 ( x
(b) 1 ! 2x3
(d) 12
2
(f) 3 x4 ( x2 ( 10
A polynomial of degree 0
A trinomial of degree 5
A binomial with leading coefficient !2
A monomial of positive degree
2
A trinomial with leading coefficient 3
A third-degree polynomial with leading coefficient 1
In Exercises 15–18, write a polynomial that fits the description. (There are many correct answers.)
15. A third-degree polynomial with leading coefficient !2
16. A fifth-degree polynomial with leading coefficient 6
17. A fourth-degree binomial with a negative leading
coefficient
18. A third-degree binomial with an even leading coefficient
In Exercises 19–30, (a) write the polynomial in standard
form, (b) identify the degree and leading coefficient of
the polynomial, and (c) state whether the polynomial is a
monomial, a binomial, or a trinomial.
19.
21.
23.
25.
27.
29.
14x ! 12 x 5
x2 ! 4 ! 3x4
3 ! x6
3
1 ( 6x 4 ! 4x 5
4x 3y
20.
22.
24.
26.
28.
30.
2x 2 ! x ( 1
7x
!y ( 25y2 ( 1
!8 ( t2
3 ( 2x
!x 5y ( 2x 2y 2 ( xy 4
In Exercises 31–36, determine whether the expression is a
polynomial. If so, write the polynomial in standard form.
31. 2x ! 3x 3 ( 8
3x ( 4
33.
x
2
35. y ! y 4 ( y 3
32. 5x4 ! 2x2 ( x!2
34.
x 2 ( 2x ! 3
2
36. y4 ! "y
In Exercises 37–54, perform the operation and write the
result in standard form.
37.
38.
39.
40.
41.
42.
43.
44.
45.
47.
49.
51.
53.
(6x ( 5' ! (8x ( 15'
(2x 2 ( 1' ! (x 2 ! 2x ( 1'
! (t3 ! 1' ( (6t3 ! 5t'
! (5x 2 ! 1' ! (!3x 2 ( 5'
(15x 2 ! 6' ! (!8.3x 3 ! 14.7x 2 ! 17'
(15.6w4 ! 14w ! 17.4' ! (16.9w4 ! 9.2w ( 13'
5z ! +3z ! (10z ( 8'*
( y 3 ( 1' ! +( y 2 ( 1' ( (3y ! 7'*
3x(x 2 ! 2x ( 1'
46. y 2(4y 2 ( 2y ! 3'
!5z(3z ! 1'
48. (!3x'(5x ( 2'
3
(1 ! x '(4x'
50. !4x(3 ! x 3'
(1.5t2 ( 5'(!3t'
52. (2 ! 3.5y'(2y 3'
3
!2x(0.1x ( 17'
54. 6y(5 ! 8 y'
In Exercises 55–92, multiply or find the special product.
55.
57.
59.
61.
63.
(x ( 3'(x ( 4'
(3x ! 5'(2x ( 1'
(x ( 10'(x ! 10'
(x ( 2y'(x ! 2y'
(2x ( 3' 2
56.
58.
60.
62.
64.
(x ! 5'(x ( 10'
(7x ! 2'(4x ! 3'
(2x ( 3'(2x ! 3'
(4a ( 5b'(4a ! 5b'
(5 ! 8x' 2
A36
65.
67.
69.
71.
72.
73.
74.
75.
76.
77.
79.
81.
83.
85.
87.
89.
90.
91.
92.
Appendix A
Review of Fundamental Concepts of Algebra
(x ( 1' 3
66. (x ! 2' 3
(2x ! y' 3
68. (3x ( 2y' 3
3
2
(4x ! 3'
70. (8x ( 3'2
(x 2 ! x ( 1'(x 2 ( x ( 1'
(x 2 ( 3x ! 2'(x 2 ! 3x ! 2'
(!x2 ( x ! 5'(3x2 ( 4x ( 1'
(2x2 ! x ( 4'(x2 ( 3x ( 2'
+(m ! 3' ( n*+(m ! 3' ! n*
+(x ! 3y' ( z*+(x ! 3y' ! z*
+(x ! 3' ( y*2
78. +(x ( 1' ! y*2
2
2
(2r ! 5'(2r ( 5'
80. (3a 3 ! 4b2'(3a 3 ( 4b2'
2
2
82. (35 t ( 4'
(14 x ! 5'
84. (3x ( 16 '(3x ! 16 '
(15 x ! 3'(15 x ( 3'
(2.4x ( 3'2
86. (1.8y ! 5'2
(1.5x ! 4'(1.5x ( 4' 88. (2.5y ( 3'(2.5y ! 3'
5x(x ( 1' ! 3x(x ( 1'
(2x ! 1'(x ( 3' ( 3(x ( 3'
(u ( 2'(u ! 2'(u 2 ( 4'
(x ( y'(x ! y'(x 2 ( y 2'
In Exercises 93–96, find the product. (The expressions are not
polynomials, but the formulas can still be used.)
93. ("x ( "y'("x ! "y'
94. (5 ( "x'(5 ! "x'
2
2
95. (x ! "5 '
96. (x ( "3 '
In Exercises 97–104, factor out the common factor.
97.
99.
101.
103.
4x ( 16
98. 5y ! 30
2x 3 ! 6x
100. 3z3 ! 6z2 ( 9z
3x(x ! 5' ( 8(x ! 5' 102. 3x(x ( 2' ! 4(x ( 2'
(x ( 3'2 ! 4(x ( 3' 104. (5x ! 4'2 ( (5x ! 4'
In Exercises 105–110, find the greatest common factor such
that the remaining factors have only integer coefficients.
105. 12 x ( 4
106. 13 y ( 5
107. 12 x 3 ( 2x 2 ! 5x
108. 13 y 4 ! 5y 2 ( 2y
109. 23 x(x ! 3' ! 4(x ! 3' 110. 45 y( y ( 1' ! 2( y ( 1'
In Exercises 111–120, completely factor the difference of two
squares.
111.
113.
115.
117.
119.
x2 ! 81
48y2 ! 27
16x 2 ! 19
(x ! 1' 2 ! 4
9u2 ! 4v 2
112.
114.
116.
118.
120.
x 2 ! 64
50 ! 98z2
4 2
25 y ! 64
25 ! (z ( 5' 2
25x 2 ! 16y 2
In Exercises 121–132, factor the perfect square trinomial.
121.
123.
125.
127.
129.
131.
x 2 ! 4x ( 4
4t 2 ( 4t ( 1
25y 2 ! 10y ( 1
9u2 ( 24uv ( 16v 2
x 2 ! 43x ( 49
4x2 ! 43 x ( 19
122.
124.
126.
128.
130.
132.
x 2 ( 10x ( 25
9x 2 ! 12x ( 4
36y 2 ! 108y ( 81
4x 2 ! 4xy ( y 2
z 2 ( z ( 14
1
9y2 ! 32 y ( 16
In Exercises 133–144, factor the sum or difference of cubes.
133.
135.
137.
139.
141.
143.
x3 ! 8
y 3 ( 64
8
x3 ! 27
8t 3 ! 1
u3 ( 27v 3
(x ( 2'3 ! y3
134.
136.
138.
140.
142.
144.
x 3 ! 27
z 3 ( 216
8
y3 ( 125
27x 3 ( 8
64x 3 ! y 3
(x ! 3y'3 ! 8z3
In Exercises 145–158, factor the trinomial.
145.
147.
149.
151.
153.
155.
157.
x2 ( x ! 2
s 2 ! 5s ( 6
20 ! y ! y 2
x 2 ! 30x ( 200
3x 2 ! 5x ( 2
5x 2 ( 26x ( 5
!9z 2 ( 3z ( 2
146.
148.
150.
152.
154.
156.
158.
x 2 ( 5x ( 6
t2 ! t ! 6
24 ( 5z ! z 2
x 2 ! 13x ( 42
2x 2 ! x ! 1
12x 2 ( 7x ( 1
!5u 2 ! 13u ( 6
In Exercises 159–166, factor by grouping.
159.
161.
163.
165.
x 3 ! x 2 ( 2x ! 2
2x 3 ! x 2 ! 6x ( 3
6 ( 2x ! 3x3 ! x4
6x 3 ! 2x ( 3x 2 ! 1
160.
162.
164.
166.
x 3 ( 5x 2 ! 5x ! 25
5x 3 ! 10x 2 ( 3x ! 6
x 5 ( 2x 3 ( x 2 ( 2
8x 5 ! 6x 2 ( 12x 3 ! 9
In Exercises 167–172, factor the trinomial by grouping.
167. 3x 2 ( 10x ( 8
169. 6x 2 ( x ! 2
171. 15x 2 ! 11x ( 2
168. 2x 2 ( 9x ( 9
170. 6x 2 ! x ! 15
172. 12x2 ! 13x ( 1
In Exercises 173–206, completely factor the expression.
173.
175.
177.
179.
181.
183.
185.
6x 2 ! 54
x3 ! x2
x3 ! 16x
x 2 ! 2x ( 1
1 ! 4x ( 4x 2
2x 2 ( 4x ! 2x 3
2
1 2
81 x ( 9 x ! 8
174.
176.
178.
180.
182.
184.
186.
12x 2 ! 48
x 3 ! 4x 2
x 3 ! 9x
16 ( 6x ! x 2
!9x 2 ( 6x ! 1
13x ( 6 ( 5x 2
1 2
1
1
8 x ! 96 x ! 16
Appendix A.3
3x 3 ( x 2 ( 15x ( 5 188. 5 ! x ( 5x 2 ! x 3
x 4 ! 4x 3 ( x 2 ! 4x 190. 3x3 ( x2 ! 27x ! 9
1 3
3
2
192. 15 x3 ( x2 ! x ! 5
4 x ( 3x ( 4 x ( 9
(t ! 1' 2 ! 49
194. (x 2 ( 1' 2 ! 4x 2
(x 2 ( 8' 2 ! 36x 2
196. 2t 3 ! 16
5x 3 ( 40
198. 4x(2x ! 1' ( (2x ! 1' 2
2
5(3 ! 4x' ! 8(3 ! 4x'(5x ! 1'
2(x ( 1'(x ! 3' 2 ! 3(x ( 1' 2(x ! 3'
7(3x ( 2' 2(1 ! x' 2 ( (3x ( 2'(1 ! x'3
7x(2'(x 2 ( 1'(2x' ! (x2 ( 1' 2(7'
3(x ! 2'2(x ( 1'4 ( (x ! 2' 3(4'(x ( 1' 3
2x(x ! 5' 4 ! x 2(4'(x ! 5' 3
5(x6 ( 1'4(6x5'(3x ( 2'3 ( 3(3x ( 2'2(3'(x6 ( 1'5
x2
206. (x2 ( 1'4 ! (x 2 ( 1'5
2
(a) Write this polynomial in standard form.
(b) Use a calculator to evaluate the polynomial for the
values of r shown in the table.
r
2%
211.
(5x ! 1'(3' ! (3x ( 1'(5'
(5x ! 1'2
212.
(2x ( 3'(4' ! (4x ! 1'(2'
(2x ( 3'2
In Exercises 213–216, find all values of b for which the
trinomial can be factored.
213. x 2 ( bx ! 15
215. x 2 ( bx ( 50
214. x 2 ( bx ! 12
216. x 2 ( bx ( 24
218. 3x 2 ! 10x ( c
220. 2x 2 ( 9x ( c
221. COST, REVENUE, AND PROFIT An electronics
manufacturer can produce and sell x MP3 players per
week. The total cost C (in dollars) of producing x MP3
players is C # 73x ( 25,000, and the total revenue R
(in dollars) is R # 95x.
(a) Find the profit P in terms of x.
(b) Find the profit obtained by selling 5000 MP3 players per week.
4%
412%
(c) What conclusion can you make from the table?
223. VOLUME OF A BOX A take-out fast-food restaurant
is constructing an open box by cutting squares from
the corners of a piece of cardboard that is 18 centimeters
by 26 centimeters (see figure). The edge of each
cut-out square is x centimeters.
x
x
26 − 2x
18 cm
x
x
x
18 − 2x
26 − 2x
26 cm
(a) Find the volume of the box in terms of x.
(b) Find the volume when x # 1, x # 2, and x # 3.
224. VOLUME OF A BOX An overnight shipping company
is designing a closed box by cutting along the solid
lines and folding along the broken lines on the
rectangular piece of corrugated cardboard shown in
the figure. The length and width of the rectangle are
45 centimeters and 15 centimeters, respectively.
45 cm
x
In Exercises 217–220, find two integer values of c such that
the trinomial can be factored. (There are many correct
answers.)
217. 2x 2 ( 5x ( c
219. 3x 2 ! x ( c
312%
15 cm
x4(4'(2x ( 1'3(2x' ( (2x ( 1'4(4x3'
x3(3'(x2 ( 1'2(2x' ( (x2 ( 1'3(3x2'
(2x ! 5'4(3'(5x ! 4'2(5' ( (5x ! 4'3(4'(2x ! 5'3(2'
(x2 ! 5'3(2'(4x ( 3'(4' ( (4x ( 3'2(3'(x2 ! 5'2(x2'
3%
1200(1 ( r'3
In Exercises 207–212, completely factor the expression.
207.
208.
209.
210.
A37
222. COMPOUND INTEREST After 3 years, an investment of $1200 compounded annually at an interest rate
r will yield an amount of 1200(1 ( r'3.
18 − 2x
187.
189.
191.
193.
195.
197.
199.
200.
201.
202.
203.
204.
205.
Polynomials and Factoring
(a) Find the volume of the shipping box in terms of x.
(b) Find the volume when x # 3, x # 5, and x # 7.
225. GEOMETRY Find a polynomial that represents the
total number of square feet for the floor plan shown in
the figure.
x
x
14 ft
22 ft
A38
Appendix A
Review of Fundamental Concepts of Algebra
226. GEOMETRY Find the area of the shaded region in
each figure. Write your result as a polynomial in
standard form.
(a)
(b)
2x + 6
x+4
x
2x
8x
12x
6x
9x
EXPLORATION
GEOMETRIC MODELING In Exercises 227–230, draw a
“geometric factoring model” to represent the factorization.
For instance, a factoring model for
2x2 # 3x # 1 " (2x # 1'(x # 1'
is shown in the figure.
x
x
x
x
x
1
1
1
x
1
x
227.
228.
229.
230.
x
1
x
1
x
1
3x 2 ( 7x ( 2 # (3x ( 1'(x ( 2'
x 2 ( 4x ( 3 # (x ( 3'(x ( 1'
2x 2 ( 7x ( 3 # (2x ( 1'(x ( 3'
x 2 ( 3x ( 2 # (x ( 2'(x ( 1'
GEOMETRY In Exercises 231 and 232, write an expression in
factored form for the area of the shaded portion of the figure.
231.
232.
x+3
r
4
r+2
5
5
(x
4
+ 3)
233. GEOMETRY The cylindrical shell shown in the
figure has a volume of V # "R2h ! "r2h.
R
h
r
(a) Factor the expression for the volume.
(b) From the result of part (a), show that the volume is
2" (average radius'(thickness of the shell'h.
234. CHEMISTRY The rate of change of an autocatalytic
chemical reaction is kQx ! kx 2, where Q is the amount
of the original substance, x is the amount of substance
formed, and k is a constant of proportionality. Factor the
expression.
TRUE OR FALSE? In Exercises 235–238, determine
whether the statement is true or false. Justify your answer.
235. The product of two binomials is always a seconddegree polynomial.
236. The sum of two binomials is always a binomial.
237. The difference of two perfect squares can be factored
as the product of conjugate pairs.
238. The sum of two perfect squares can be factored as the
binomial sum squared.
239. Find the degree of the product of two polynomials of
degrees m and n.
240. Find the degree of the sum of two polynomials of
degrees m and n if m < n.
241. THINK ABOUT IT When the polynomial
!x3 ( 3x2 ( 2x ! 1 is subtracted from an unknown
polynomial, the difference is 5x 2 ( 8. If it is possible,
find the unknown polynomial.
242. LOGICAL REASONING Verify that (x ( y'2 is not
equal to x 2 ( y 2 by letting x # 3 and y # 4 and
evaluating both expressions. Are there any values of x
and y for which (x ( y'2 # x 2 ( y 2 ? Explain.
243. Factor x 2n ! y 2n as completely as possible.
244. Factor x 3n ( y 3n as completely as possible.
245. Give an example of a polynomial that is prime with
respect to the integers.
246. CAPSTONE A third-degree polynomial and a
fourth-degree polynomial are added.
(a) Can the sum be a fourth-degree polynomial?
Explain or give an example.
(b) Can the sum be a second-degree polynomial?
Explain or give an example.
(c) Can the sum be a seventh-degree polynomial?
Explain or give an example.
(d) After adding the two polynomials and factoring
the sum, you obtain a polynomial that is in
factored form. Explain what is meant by saying
that a polynomial is in factored form.
Appendix A.4
Rational Expressions
A39
A.4 RATIONAL EXPRESSIONS
What you should learn
• Find domains of algebraic
expressions.
• Simplify rational expressions.
• Add, subtract, multiply, and divide
rational expressions.
• Simplify complex fractions and
rewrite difference quotients.
Domain of an Algebraic Expression
The set of real numbers for which an algebraic expression is defined is the domain of
the expression. Two algebraic expressions are equivalent if they have the same domain
and yield the same values for all numbers in their domain. For instance,
(x ( 1' ( (x ( 2' and 2x ( 3 are equivalent because
(x ( 1' ( (x ( 2' # x ( 1 ( x ( 2
#x(x(1(2
Why you should learn it
Rational expressions can be used to
solve real-life problems. For instance,
in Exercise 102 on page A48, a rational
expression is used to model the
projected numbers of U.S. households
banking and paying bills online from
2002 through 2007.
# 2x ( 3.
Example 1
Finding the Domain of an Algebraic Expression
a. The domain of the polynomial
2x 3 ( 3x ( 4
is the set of all real numbers. In fact, the domain of any polynomial is the set of all
real numbers, unless the domain is specifically restricted.
b. The domain of the radical expression
"x ! 2
is the set of real numbers greater than or equal to 2, because the square root of a
negative number is not a real number.
c. The domain of the expression
x(2
x!3
is the set of all real numbers except x # 3, which would result in division by zero,
which is undefined.
Now try Exercise 7.
The quotient of two algebraic expressions is a fractional expression. Moreover, the
quotient of two polynomials such as
1
,
x
2x ! 1
,
x(1
or
x2 ! 1
x2 ( 1
is a rational expression.
Simplifying Rational Expressions
Recall that a fraction is in simplest form if its numerator and denominator have no
factors in common aside from ± 1. To write a fraction in simplest form, divide out
common factors.
a
b
* c # a,
*c b
c$0
A40
Appendix A
Review of Fundamental Concepts of Algebra
The key to success in simplifying rational expressions lies in your ability to factor
polynomials. When simplifying rational expressions, be sure to factor each polynomial
completely before concluding that the numerator and denominator have no factors in
common.
Example 2
WARNING / CAUTION
In Example 2, do not make the
mistake of trying to simplify
further by dividing out terms.
x(6 x(6
#x(2
#
3
3
Remember that to simplify
fractions, divide out common
factors, not terms.
Write
Simplifying a Rational Expression
x 2 ( 4x ! 12
in simplest form.
3x ! 6
Solution
x2 ( 4x ! 12 (x ( 6'(x ! 2'
#
3x ! 6
3(x ! 2'
#
x(6
,
3
x$2
Factor completely.
Divide out common factors.
Note that the original expression is undefined when x # 2 (because division by zero is
undefined). To make sure that the simplified expression is equivalent to the original
expression, you must restrict the domain of the simplified expression by excluding the
value x # 2.
Now try Exercise 33.
Sometimes it may be necessary to change the sign of a factor by factoring out (!1'
to simplify a rational expression, as shown in Example 3.
Example 3
Write
Simplifying Rational Expressions
12 ( x ! x2
in simplest form.
2x2 ! 9x ( 4
Solution
12 ( x ! x2
(4 ! x'(3 ( x'
#
2x2 ! 9x ( 4 (2x ! 1'(x ! 4'
#
! (x ! 4'(3 ( x'
(2x ! 1'(x ! 4'
#!
3(x
, x$4
2x ! 1
Factor completely.
(4 ! x' # ! (x ! 4'
Divide out common factors.
Now try Exercise 39.
In this text, when a rational expression is written, the domain is usually
not listed with the expression. It is implied that the real numbers that make
the denominator zero are excluded from the expression. Also, when performing
operations with rational expressions, this text follows the convention of listing
by the simplified expression all values of x that must be specifically excluded from the
domain in order to make the domains of the simplified and original expressions agree.
In Example 3, for instance, the restriction x $ 4 is listed with the simplified expression
1
to make the two domains agree. Note that the value x # 2 is excluded from both
domains, so it is not necessary to list this value.
Appendix A.4
Rational Expressions
A41
Operations with Rational Expressions
To multiply or divide rational expressions, use the properties of fractions
discussed in Appendix A.1. Recall that to divide fractions, you invert the divisor and
multiply.
Example 4
Multiplying Rational Expressions
2x2 ( x ! 6
x2 ( 4x ! 5
*
x3 ! 3x2 ( 2x (2x ! 3'(x ( 2'
#
4x2 ! 6x
(x ( 5'(x ! 1'
#
*
x(x ! 2'(x ! 1'
2x(2x ! 3'
(x ( 2'(x ! 2'
, x $ 0, x $ 1, x $ 32
2(x ( 5'
Now try Exercise 53.
In Example 4, the restrictions x $ 0, x $ 1, and x $ 32 are listed with the simplified
expression in order to make the two domains agree. Note that the value x # !5 is
excluded from both domains, so it is not necessary to list this value.
Example 5
Dividing Rational Expressions
x 3 ! 8 x 2 ( 2x ( 4 x 3 ! 8
)
# 2
x2 ! 4
x3 ( 8
x !4
#
x3 ( 8
* x 2 ( 2x ( 4
Invert and multiply.
(x ! 2'(x2 ( 2x ( 4' (x ( 2'(x2 ! 2x ( 4'
* (x2 ( 2x ( 4'
(x ( 2'(x ! 2'
# x 2 ! 2x ( 4, x $ ± 2
Divide out
common factors.
Now try Exercise 55.
To add or subtract rational expressions, you can use the LCD (least common
denominator) method or the basic definition
a c
ad ± bc
± #
,
b d
bd
b $ 0, d $ 0.
Basic definition
This definition provides an efficient way of adding or subtracting two fractions that
have no common factors in their denominators.
Example 6
WARNING / CAUTION
When subtracting rational
expressions, remember to
distribute the negative sign to
all the terms in the quantity that
is being subtracted.
Subtracting Rational Expressions
x
2
x(3x ( 4' ! 2(x ! 3'
!
#
x ! 3 3x ( 4
(x ! 3'(3x ( 4'
Basic definition
#
3x 2 ( 4x ! 2x ( 6
(x ! 3'(3x ( 4'
Distributive Property
#
3x 2 ( 2x ( 6
(x ! 3'(3x ( 4'
Combine like terms.
Now try Exercise 65.
A42
Appendix A
Review of Fundamental Concepts of Algebra
For three or more fractions, or for fractions with a repeated factor in the denominators, the LCD method works well. Recall that the least common denominator of
several fractions consists of the product of all prime factors in the denominators, with
each factor given the highest power of its occurrence in any denominator. Here is a
numerical example.
1 3 2 1*2 3*3 2*4
( ! #
(
!
6 4 3 6*2 4*3 3*4
#
2
9
8
(
!
12 12 12
#
3
12
#
1
4
The LCD is 12.
Sometimes the numerator of the answer has a factor in common with the
denominator. In such cases the answer should be simplified. For instance, in the
3
example above, 12
was simplified to 14.
Example 7
Combining Rational Expressions: The LCD Method
Perform the operations and simplify.
3
2
x(3
! ( 2
x!1
x
x !1
Solution
Using the factored denominators (x ! 1', x, and (x ( 1'(x ! 1', you can see that the
LCD is x(x ( 1'(x ! 1'.
3
2
x(3
! (
x!1
x
(x ( 1'(x ! 1'
#
3(x'(x ( 1'
2(x ( 1'(x ! 1'
(x ( 3'(x'
!
(
x(x ( 1'(x ! 1'
x(x ( 1'(x ! 1'
x(x ( 1'(x ! 1'
#
3(x'(x ( 1' ! 2(x ( 1'(x ! 1' ( (x ( 3'(x'
x(x ( 1'(x ! 1'
#
3x 2 ( 3x ! 2x 2 ( 2 ( x 2 ( 3x
x(x ( 1'(x ! 1'
Distributive Property
#
3x 2 ! 2x 2 ( x 2 ( 3x ( 3x ( 2
x(x ( 1'(x ! 1'
Group like terms.
#
2x2 ( 6x ( 2
x(x ( 1'(x ! 1'
Combine like terms.
#
2(x 2 ( 3x ( 1'
x(x ( 1'(x ! 1'
Factor.
Now try Exercise 67.
Appendix A.4
Rational Expressions
A43
Complex Fractions and the Difference Quotient
Fractional expressions with separate fractions in the numerator, denominator, or both
are called complex fractions. Here are two examples.
-x.
-x.
1
x2 ( 1
1
and
-x
2
1
(1
.
To simplify a complex fraction, combine the fractions in the numerator into a single
fraction and then combine the fractions in the denominator into a single fraction.
Then invert the denominator and multiply.
Example 8
Simplifying a Complex Fraction
2 ! 3(x'
x
#
1
1(x ! 1' ! 1
1!
x!1
x!1
2
- x ! 3.
-
1
2
. 1
2
- x .
#
x!2
-x ! 1.
Combine fractions.
2 ! 3x
Simplify.
#
2 ! 3x
x
x!1
#
(2 ! 3x'(x ! 1'
, x$1
x(x ! 2'
*x!2
Invert and multiply.
Now try Exercise 73.
Another way to simplify a complex fraction is to multiply its numerator and
denominator by the LCD of all fractions in its numerator and denominator. This method
is applied to the fraction in Example 8 as follows.
- x ! 3.
- x ! 3.
2
-
1
1!
x!1
2
#
. -
1
1!
x!1
x(x ! 1'
.
* x(x ! 1'
-2 !x 3x. * x(x ! 1'
#
-xx !! 21. * x(x ! 1'
#
(2 ! 3x'(x ! 1'
, x$1
x(x ! 2'
LCD is x(x ! 1'.
A44
Appendix A
Review of Fundamental Concepts of Algebra
The next three examples illustrate some methods for simplifying rational
expressions involving negative exponents and radicals. These types of expressions
occur frequently in calculus.
To simplify an expression with negative exponents, one method is to begin by
factoring out the common factor with the smaller exponent. Remember that when
factoring, you subtract exponents. For instance, in 3x!5)2 ( 2x!3)2 the smaller
exponent is ! 52 and the common factor is x!5)2.
3x!5)2 ( 2x!3)2 # x!5)2+3(1' ( 2x!3)2! (!5)2'*
# x!5)2(3 ( 2x1'
#
Example 9
3 ( 2x
x 5)2
Simplifying an Expression
Simplify the following expression containing negative exponents.
x(1 ! 2x'!3)2 ( (1 ! 2x'!1)2
Solution
Begin by factoring out the common factor with the smaller exponent.
x(1 ! 2x'!3)2 ( (1 ! 2x'!1)2 # (1 ! 2x'!3)2+ x ( (1 ! 2x'(!1)2)!(!3)2)*
# (1 ! 2x'!3)2+x ( (1 ! 2x'1*
#
1!x
(1 ! 2x' 3)2
Now try Exercise 81.
A second method for simplifying an expression with negative exponents is shown
in the next example.
Example 10
Simplifying an Expression with Negative Exponents
(4 ! x 2'1)2 ( x 2(4 ! x 2'!1)2
4 ! x2
#
(4 ! x 2'1)2 ( x 2(4 ! x 2'!1)2 (4 ! x 2'1)2
* (4 ! x 2'1)2
4 ! x2
#
(4 ! x 2'1 ( x 2(4 ! x 2' 0
(4 ! x 2' 3)2
#
4 ! x2 ( x2
(4 ! x 2' 3)2
#
4
(4 ! x 2' 3)2
Now try Exercise 83.
Appendix A.4
Example 11
Rational Expressions
A45
Rewriting a Difference Quotient
The following expression from calculus is an example of a difference quotient.
"x ( h ! "x
h
Rewrite this expression by rationalizing its numerator.
Solution
"x ( h ! "x
h
You can review the techniques
for rationalizing a numerator in
Appendix A.2.
#
"x ( h ! "x
h
"x ( h ( "x
* "x ( h ( "x
#
("x ( h '2 ! ("x '2
h("x ( h ( "x '
#
h
h("x ( h ( "x '
#
1
,
"x ( h ( "x
h$0
Notice that the original expression is undefined when h # 0. So, you must exclude
h # 0 from the domain of the simplified expression so that the expressions are
equivalent.
Now try Exercise 89.
Difference quotients, such as that in Example 11, occur frequently in calculus. Often,
they need to be rewritten in an equivalent form that can be evaluated when h # 0. Note
that the equivalent form is not simpler than the original form, but it has the advantage
that it is defined when h # 0.
A.4
EXERCISES
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
1. The set of real numbers for which an algebraic expression is defined is the ________ of the expression.
2. The quotient of two algebraic expressions is a fractional expression and the quotient of
two polynomials is a ________ ________.
3. Fractional expressions with separate fractions in the numerator, denominator, or both are called ________
fractions.
4. To simplify an expression with negative exponents, it is possible to begin by factoring out the common factor
with the ________ exponent.
5. Two algebraic expressions that have the same domain and yield the same values for all numbers
in their domains are called ________.
6. An important rational expression, such as
a ________ ________.
(x ( h'2 ! x2
, that occurs in calculus is called
h
A46
Appendix A
Review of Fundamental Concepts of Algebra
44. Error Analysis
SKILLS AND APPLICATIONS
In Exercises 7–22, find the domain of the expression.
7. 3x 2 ! 4x ( 7
9. 4x 3 ( 3, x & 0
1
11.
3!x
13.
x2
!1
x2 ! 2x ( 1
8. 2x 2 ( 5x ! 2
10. 6x 2 ! 9, x > 0
12.
14.
x(6
3x ( 2
x2
! 5x ( 6
x2 ! 4
x2 ! 2x ! 3
15. 2
x ! 6x ( 9
x2 ! x ! 12
16. 2
x ! 8x ( 16
17. "x ( 7
19. "2x ! 5
18. "4 ! x
20. "4x ( 5
21.
1
22.
"x ! 3
x2
1
"x ( 2
15x 2
10x
3xy
27.
xy ( x
25.
26.
28.
18y 2
60y 5
45.
30.
9x 2 ( 9x
2x ( 2
31.
x!5
10 ! 2x
32.
12 ! 4x
x!3
y 2 ! 16
33.
y(4
35.
37.
39.
41.
34.
x 3 ( 5x 2 ( 6x
x2 ! 4
2
y ! 7y ( 12
y 2 ( 3y ! 18
2 ! x ( 2x 2 ! x 3
x2 ! 4
z3 ! 8
2
z ( 2z ( 4
43. Error Analysis
5x3
2x3 ( 4
#
36.
38.
40.
42.
x 2 ! 25
5!x
x 2 ( 8x ! 20
x 2 ( 11x ( 10
x 2 ! 7x ( 6
x 2 ( 11x ( 10
x2 ! 9
3
x ( x 2 ! 9x ! 9
y 3 ! 2y 2 ! 3y
y3 ( 1
Describe the error.
5x3
5
5
#
#
3
2x ( 4 2 ( 4 6
1
2
3
4
5
6
x(1
46.
0
x
1
2
3
4
5
6
x!3
x2 ! x ! 6
1
x(2
GEOMETRY In Exercises 47 and 48, find the ratio of the
area of the shaded portion of the figure to the total area of
the figure.
47.
48.
x+5
2
r
xy ! y
4y ! 8y 2
10y ! 5
0
x
x2 ! 2x ! 3
x!3
2x 2y
29.
x(x ( 5'(x ! 5' x(x ( 5'
#
(x ! 5'(x ( 3'
x(3
In Exercises 45 and 46, complete the table. What can you conclude?
3
3( ! '
24. #
4 4(x ( 1'
In Exercises 25–42, write the rational expression in simplest
form.
x 3 ( 25x
x(x 2 ( 25'
#
! 2x ! 15 (x ! 5'(x ( 3'
#
In Exercises 23 and 24, find the missing factor in the numerator
such that the two fractions are equivalent.
5
5(!'
23.
#
2x
6x 2
Describe the error.
2x + 3
x+5
2
x+5
In Exercises 49–56, perform the multiplication or division
and simplify.
5
x!1
x ( 13
x(x ! 3'
50. 3
*
*
x ! 1 25(x ! 2'
x (3 ! x'
5
2
r
r
4y ! 16
4!y
51.
)
52.
)
r ! 1 r2 ! 1
5y ( 15 2y ( 6
t2 ! t ! 6
t(3
53. 2
*
t ( 6t ( 9 t 2 ! 4
x 2 ( xy ! 2y 2
x
54.
*
3
2
2
x (x y
x ( 3xy ( 2y 2
49.
55.
x 2 ! 36 x 3 ! 6x 2
) 2
x
x (x
56.
x 2 ! 14x ( 49 3x ! 21
)
x 2 ! 49
x(7
Appendix A.4
In Exercises 57–68, perform the addition or subtraction and
simplify.
57. 6 !
5
x(3
5
x
(
x!1 x!1
3
5
61.
(
x!2 2!x
59.
63.
4
x
!
2x ( 1 x ( 2
64.
3x1)3 ! x!2)3
3x!2)3
!x 3(1 ! x 2'!1)2 ! 2x(1 ! x 2'1)2
84.
x4
83.
1
x
!
x 2 ! x ! 2 x 2 ! 5x ( 6
2
10
66. 2
( 2
x ! x ! 2 x ( 2x ! 8
1
2
1
67. ! ( 2
( 3
x
x (1 x (x
2
2
1
68.
(
(
x ( 1 x ! 1 x2 ! 1
x ( 4 3x ! 8 x ( 4 ! 3x ! 8
!
#
x(2
x(2
x(2
!2x ! 4 !2(x ( 2'
#
#
# !2
x(2
x(2
6!x
x(2
8
70.
(
( 2
2
x(x ( 2'
x
x (x ( 2'
x(6 ! x' ( (x ( 2' 2 ( 8
#
x 2(x ( 2'
6x ! x 2 ( x 2 ( 4 ( 8
#
x 2(x ( 2'
6(x ( 2'
6
# 2
#
x (x ( 2' x 2
69.
In Exercises 71–76, simplify the complex fraction.
x
73.
(x ! 2'
x2
(x ( 1'2
1
2
x
1 (x ( 1' 2
3
75.
-
"x !
(x ! 4'
x
4
!
4
x
2
x !1
x
74.
(x ! 1'2
x
t2
! "t 2 ( 1
"t 2 ( 1
76.
t2
72.
1
2"x
"x
.
In Exercises 85– 88, simplify the difference quotient.
1
1
1
1
!
!
(x ( h' 2 x 2
x(h
x
85.
86.
h
h
x(h
x
1
1
!
!
x(h(1 x(1
x(h!4 x!4
87.
88.
h
h
-
ERROR ANALYSIS In Exercises 69 and 70, describe the
error.
71.
x 5 ! 2x!2
78. x5 ! 5x!3
x 2(x 2 ( 1'!5 ! (x 2 ( 1'!4
2x(x ! 5'!3 ! 4x 2(x ! 5'!4
2x 2(x ! 1'1)2 ! 5(x ! 1'!1)2
4x 3(2x ! 1'3)2 ! 2x(2x ! 1'!1)2
In Exercises 83 and 84, simplify the expression.
2
5x
(
x ! 3 3x ( 4
65.
- 2 ! 1.
In Exercises 77–82, factor the expression by removing the
common factor with the smaller exponent.
77.
79.
80.
81.
82.
3
!5
x!1
2x ! 1 1 ! x
60.
(
x(3
x(3
2x
5
62.
!
x!5 5!x
58.
-
1
-
A47
Rational Expressions
.
.
2
.
1
.
-
2
-
.
.
In Exercises 89–94, simplify the difference quotient by
rationalizing the numerator.
89.
91.
93.
94.
"x ( 2 ! "x
90.
2
"t ( 3 ! "3
92.
t
"z ! 3 ! "z
3
"x ( 5 ! "5
x
"x ( h ( 1 ! "x ( 1
h
"x ( h ! 2 ! "x ! 2
h
PROBABILITY In Exercises 95 and 96, consider an experiment in which a marble is tossed into a box whose base is
shown in the figure. The probability that the marble will
come to rest in the shaded portion of the box is equal to the
ratio of the shaded area to the total area of the figure. Find
the probability.
95.
96.
x
2x + 1
x
2
x
x+4
x
x+2
4
x
(x + 2)
97. RATE A digital copier copies in color at a rate of
50 pages per minute.
(a) Find the time required to copy one page.
A48
Appendix A
Review of Fundamental Concepts of Algebra
(b) Find the time required to copy x pages.
(c) Find the time required to copy 120 pages.
98. RATE After working together for t hours on a
common task, two workers have done fractional parts
of the job equal to t)3 and t)5, respectively. What
fractional part of the task has been completed?
102. INTERACTIVE MONEY MANAGEMENT The table
shows the projected numbers of U.S. households (in
millions) banking online and paying bills online from
2002 through 2007. (Source: eMarketer; Forrester
Research)
FINANCE In Exercises 99 and 100, the formula that approximates the annual interest rate r of a monthly installment
loan is given by
24(NM ! P'
[
]
N
r"
3P # NM
12 4
Year
Banking
Paying Bills
2002
2003
2004
2005
2006
2007
21.9
26.8
31.5
35.0
40.0
45.0
13.7
17.4
20.9
23.9
26.7
29.1
where N is the total number of payments, M is the monthly
payment, and P is the amount financed.
Mathematical models for these data are
99. (a) Approximate the annual interest rate for a
four-year car loan of $20,000 that has monthly
payments of $475.
(b) Simplify the expression for the annual interest rate
r, and then rework part (a).
100. (a) Approximate the annual interest rate for a fiveyear car loan of $28,000 that has monthly
payments of $525.
(b) Simplify the expression for the annual interest rate
r, and then rework part (a).
101. REFRIGERATION When food (at room temperature)
is placed in a refrigerator, the time required for the
food to cool depends on the amount of food, the air
circulation in the refrigerator, the original temperature
of the food, and the temperature of the refrigerator.
The model that gives the temperature of food that has
an original temperature of 75,F and is placed in a
40,F refrigerator is
Number banking online #
T # 10
-t
4t 2 ( 16t ( 75
2 ( 4t ( 10
.
0
2
4
6
8
10
12
T
t
14
16
18
and
Number paying bills online #
4.39t ( 5.5
0.002t2 ( 0.01t ( 1.0
where t represents the year, with t # 2 corresponding
to 2002.
(a) Using the models, create a table to estimate the
projected numbers of households banking online
and the projected numbers of households paying
bills online for the given years.
(b) Compare the values given by the models with the
actual data.
(c) Determine a model for the ratio of the projected
number of households paying bills online to the
projected number of households banking online.
(d) Use the model from part (c) to find the ratios for
the given years. Interpret your results.
EXPLORATION
where T is the temperature (in degrees Fahrenheit) and
t is the time (in hours).
(a) Complete the table.
t
!0.728t2 ( 23.81t ! 0.3
!0.049t2 ( 0.61t ( 1.0
20
22
T
(b) What value of T does the mathematical model
appear to be approaching?
TRUE OR FALSE? In Exercises 103 and 104, determine
whether the statement is true or false. Justify your answer.
103.
x 2n ! 12n
# x n ( 1n
x n ! 1n
104.
x 2 ! 3x ( 2
# x ! 2, for all values of x
x!1
105. THINK ABOUT IT How do you determine whether a
rational expression is in simplest form?
106. CAPSTONE In your own words, explain how to
divide rational expressions.
Appendix A.5
Solving Equations
A49
A.5 SOLVING EQUATIONS
What you should learn
• Identify different types of equations.
• Solve linear equations in one variable
and equations that lead to linear
equations.
• Solve quadratic equations by factoring,
extracting square roots, completing
the square, and using the Quadratic
Formula.
• Solve polynomial equations of
degree three or greater.
• Solve equations involving radicals.
• Solve equations with absolute values.
• Use common formulas to solve reallife problems.
Why you should learn it
Linear equations are used in many
real-life applications. For example, in
Exercises 155 and 156 on pages A61
and A62, linear equations can be used
to model the relationship between the
length of a thigh bone and the height
of a person, helping researchers learn
about ancient cultures.
Equations and Solutions of Equations
An equation in x is a statement that two algebraic expressions are equal. For example
3x ! 5 # 7, x 2 ! x ! 6 # 0, and "2x # 4
are equations. To solve an equation in x means to find all values of x for which the equation is true. Such values are solutions. For instance, x # 4 is a solution of the equation
3x ! 5 # 7
because 3(4' ! 5 # 7 is a true statement.
The solutions of an equation depend on the kinds of numbers being considered. For
instance, in the set of rational numbers, x 2 # 10 has no solution because there is no
rational number whose square is 10. However, in the set of real numbers, the equation
has the two solutions x # "10 and x # ! "10.
An equation that is true for every real number in the domain of the variable is called
an identity. The domain is the set of all real numbers for which the equation is defined.
For example
x2 ! 9 # (x ( 3'(x ! 3'
Identity
is an identity because it is a true statement for any real value of x. The equation
x
1
#
3x2
3x
Identity
where x $ 0, is an identity because it is true for any nonzero real value of x.
An equation that is true for just some (or even none) of the real numbers in
the domain of the variable is called a conditional equation. For example, the equation
x2 ! 9 # 0
Conditional equation
is conditional because x # 3 and x # !3 are the only values in the domain that
satisfy the equation. The equation 2x ! 4 # 2x ( 1 is conditional because there are no
real values of x for which the equation is true.
Linear Equations in One Variable
Definition of a Linear Equation
A linear equation in one variable x is an equation that can be written in the
standard form
ax ( b # 0
where a and b are real numbers with a $ 0.
A50
Appendix A
Review of Fundamental Concepts of Algebra
A linear equation in one variable, written in standard form, always has exactly one
solution. To see this, consider the following steps.
Note that some linear equations
in nonstandard form have no
solution or infinitely many
solutions. For instance,
x#x(1
has no solution because it is not
true for any value of x. Because
5x ( 10 # 5(x ( 2'
is true for any value of x, the
equation has infinitely many
solutions.
ax ( b # 0
Original equation, with a $ 0
ax # !b
x#!
b
a
Subtract b from each side.
Divide each side by a.
To solve a conditional equation in x, isolate x on one side of the equation
by a sequence of equivalent (and usually simpler) equations, each having the same
solution(s) as the original equation. The operations that yield equivalent equations come
from the Substitution Principle (see Appendix A.1) and simplification techniques.
Generating Equivalent Equations
An equation can be transformed into an equivalent equation by one or more of
the following steps.
Given Equation
2x ! x # 4
Equivalent
Equation
x#4
2. Add (or subtract) the same
quantity to (from) each side
of the equation.
x(1#6
x#5
3. Multiply (or divide) each
side of the equation by the
same nonzero quantity.
2x # 6
x#3
4. Interchange the two sides of
the equation.
2#x
x#2
1. Remove symbols of grouping,
combine like terms, or simplify
fractions on one or both sides
of the equation.
After solving an equation, you
should check each solution in the
original equation. For instance,
you can check the solution of
Example 1(a) as follows.
3x ! 6 # 0
?
3(2' ! 6 # 0
0#0
Write original
equation.
Try checking the solution of
Example 1(b).
Solving a Linear Equation
a. 3x ! 6 # 0
Original equation
3x # 6
Add 6 to each side.
x#2
Divide each side by 3.
b. 5x ( 4 # 3x ! 8
Substitute 2 for x.
Solution checks.
Example 1
✓
2x ( 4 # !8
2x # !12
x # !6
Original equation
Subtract 3x from each side.
Subtract 4 from each side.
Divide each side by 2.
Now try Exercise 15.
Appendix A.5
An equation with a single
fraction on each side can be
cleared of denominators by
cross multiplying. To do this,
multiply the left numerator by
the right denominator and the
right numerator by the left
denominator as follows.
a
c
#
b d
ad # cb
A51
To solve an equation involving fractional expressions, find the least common
denominator (LCD) of all terms and multiply every term by the LCD. This process will
clear the original equation of fractions and produce a simpler equation.
Example 2
Solve
An Equation Involving Fractional Expressions
x
3x
(
# 2.
3
4
Solution
3x
x
(
#2
3
4
Original equation
Cross multiply.
Solving Equations
Write original equation.
x
3x
(12' ( (12' # (12'2
3
4
Multiply each term by the LCD of 12.
4x ( 9x # 24
Divide out and multiply.
13x # 24
x#
Combine like terms.
24
13
Divide each side by 13.
The solution is x # 24
13 . Check this in the original equation.
Now try Exercise 23.
When multiplying or dividing an equation by a variable quantity, it is possible to
introduce an extraneous solution. An extraneous solution is one that does not satisfy
the original equation. Therefore, it is essential that you check your solutions.
Example 3
Solve
An Equation with an Extraneous Solution
1
3
6x .
#
!
x ! 2 x ( 2 x2 ! 4
Solution
The LCD is x 2 ! 4, or (x ( 2'(x ! 2'. Multiply each term by this LCD.
Recall that the least common
denominator of two or more
fractions consists of the product
of all prime factors in the
denominators, with each factor
given the highest power of its
occurrence in any denominator.
For instance, in Example 3, by
factoring each denominator you
can determine that the LCD is
(x ( 2'(x ! 2'.
1
3
6x
(x ( 2'(x ! 2' #
(x ( 2'(x ! 2' ! 2
(x ( 2'(x ! 2'
x!2
x(2
x !4
x ( 2 # 3(x ! 2' ! 6x,
x $ ±2
x ( 2 # 3x ! 6 ! 6x
x ( 2 # !3x ! 6
4x # !8
x # !2
Extraneous solution
In the original equation, x # !2 yields a denominator of zero. So, x # !2 is an
extraneous solution, and the original equation has no solution.
Now try Exercise 35.
A52
Appendix A
Review of Fundamental Concepts of Algebra
Quadratic Equations
A quadratic equation in x is an equation that can be written in the general form
ax2 ( bx ( c # 0
where a, b, and c are real numbers, with a $ 0. A quadratic equation in x is also known
as a second-degree polynomial equation in x.
You should be familiar with the following four methods of solving quadratic
equations.
Solving a Quadratic Equation
Factoring: If ab # 0, then a # 0 or b # 0.
x2 ! x ! 6 # 0
Example:
(x ! 3'(x ( 2' # 0
x!3#0
x#3
x(2#0
x # !2
Square Root Principle: If u 2 # c, where c > 0, then u # ± "c.
(x ( 3'2 # 16
Example:
The Square Root Principle is
also referred to as extracting
square roots.
x ( 3 # ±4
x # !3 ± 4
or
x#1
x # !7
Completing the Square: If x 2 ( bx # c, then
-2.
2
-x ( 2 .
2
x 2 ( bx (
b
b
Example:
#c(
-2.
#c(
b2
.
4
b
2
Add
-b2.
2
Add
-62.
2
to each side.
x 2 ( 6x # 5
x 2 ( 6x ( 32 # 5 ( 32
to each side.
(x ( 3' # 14
2
x ( 3 # ± "14
x # !3 ± "14
Quadratic Formula: If ax 2 ( bx ( c # 0, then x #
You can solve every quadratic
equation by completing the
square or using the Quadratic
Formula.
Example:
!b ± "b2 ! 4ac
.
2a
2x 2 ( 3x ! 1 # 0
x#
#
!3 ± "32 ! 4(2'(!1'
2(2'
!3 ± "17
4
Appendix A.5
Example 4
a.
Solving Equations
A53
Solving a Quadratic Equation by Factoring
2x 2 ( 9x ( 7 # 3
Original equation
2x2
Write in general form.
( 9x ( 4 # 0
(2x ( 1'(x ( 4' # 0
Factor.
1
2
2x ( 1 # 0
x#!
x(4#0
x # !4
Set 1st factor equal to 0.
Set 2nd factor equal to 0.
The solutions are x # ! 12 and x # !4. Check these in the original equation.
b.
6x 2 ! 3x # 0
Original equation
3x(2x ! 1' # 0
3x # 0
2x ! 1 # 0
Factor.
x#0
x#
1
2
Set 1st factor equal to 0.
Set 2nd factor equal to 0.
The solutions are x # 0 and x # 12. Check these in the original equation.
Now try Exercise 49.
Note that the method of solution in Example 4 is based on the Zero-Factor Property
from Appendix A.1. Be sure you see that this property works only for equations written
in general form (in which the right side of the equation is zero). So, all terms must
be collected on one side before factoring. For instance, in the equation
(x ! 5'(x ( 2' # 8, it is incorrect to set each factor equal to 8. Try to solve this
equation correctly.
Example 5
Extracting Square Roots
Solve each equation by extracting square roots.
b. (x ! 3'2 # 7
a. 4x 2 # 12
Solution
a. 4x 2 # 12
x2
Write original equation.
#3
Divide each side by 4.
x # ± "3
Extract square roots.
When you take the square root of a variable expression, you must account for both
positive and negative solutions. So, the solutions are x # "3 and x # ! "3. Check
these in the original equation.
b. (x ! 3'2 # 7
x ! 3 # ± "7
x # 3 ± "7
Write original equation.
Extract square roots.
Add 3 to each side.
The solutions are x # 3 ± "7. Check these in the original equation.
Now try Exercise 65.
A54
Appendix A
Review of Fundamental Concepts of Algebra
When solving quadratic equations by completing the square, you must add (b)2' 2
to each side in order to maintain equality. If the leading coefficient is not 1, you must
divide each side of the equation by the leading coefficient before completing the square,
as shown in Example 7.
Example 6
Completing the Square: Leading Coefficient Is 1
Solve x 2 ( 2x ! 6 # 0 by completing the square.
Solution
x 2 ( 2x ! 6 # 0
Write original equation.
x 2 ( 2x # 6
2
Add 6 to each side.
2
2
x ( 2x ( 1 # 6 ( 1
Add 12 to each side.
2
(half of 2'
(x ( 1'2 # 7
Simplify.
x ( 1 # ± "7
x # !1 ± "7
Take square root of each side.
Subtract 1 from each side.
The solutions are x # !1 ± "7. Check these in the original equation.
Now try Exercise 73.
Example 7
Completing the Square: Leading Coefficient Is Not 1
3x2 ! 4x ! 5 # 0
Original equation
3x2 ! 4x # 5
Add 5 to each side.
4
5
x2 ! x #
3
3
- .
4
2
x2 ! x ( !
3
3
2
#
Divide each side by 3.
- .
5
2
( !
3
3
2
Add (! 3 ' to each side.
2 2
(half of ! 43 '2
4
4 19
x2 ! x ( #
3
9
9
-
x!
2
3
x!
.
2
#
19
9
"19
2
# ±
3
3
x#
"19
2
±
3
3
Now try Exercise 77.
Simplify.
Perfect square trinomial
Extract square roots.
Solutions
Appendix A.5
WARNING / CAUTION
When using the Quadratic
Formula, remember that before
the formula can be applied, you
must first write the quadratic
equation in general form.
Example 8
Solving Equations
A55
The Quadratic Formula: Two Distinct Solutions
Use the Quadratic Formula to solve x 2 ( 3x # 9.
Solution
x2 ( 3x # 9
Write original equation.
2
x ( 3x ! 9 # 0
Write in general form.
x#
!b ± "b2 ! 4ac
2a
Quadratic Formula
x#
!3 ± "(3'2 ! 4(1'(!9'
2(1'
Substitute a # 1,
b # 3, and c # !9.
x#
!3 ± "45
2
Simplify.
x#
!3 ± 3"5
2
Simplify.
The equation has two solutions:
x#
!3 ( 3"5
2
and
x#
!3 ! 3"5
.
2
Check these in the original equation.
Now try Exercise 87.
Example 9
The Quadratic Formula: One Solution
Use the Quadratic Formula to solve 8x 2 ! 24x ( 18 # 0.
Solution
8x2 ! 24x ( 18 # 0
4x2 ! 12x ( 9 # 0
Write original equation.
Divide out common
factor of 2.
x#
!b ± "b2 ! 4ac
2a
Quadratic Formula
x#
! (!12' ± "(!12'2 ! 4(4'(9'
2(4'
Substitute a # 4,
b # !12, and c # 9.
x#
12 ± "0 3
#
8
2
Simplify.
This quadratic equation has only one solution: x # 32. Check this in the original
equation.
Now try Exercise 91.
Note that Example 9 could have been solved without first dividing out a common
factor of 2. Substituting a # 8, b # !24, and c # 18 into the Quadratic Formula
produces the same result.
A56
Appendix A
Review of Fundamental Concepts of Algebra
Polynomial Equations of Higher Degree
WARNING / CAUTION
A common mistake that is made
in solving equations such as the
equation in Example 10 is to
divide each side of the equation
by the variable factor x 2. This
loses the solution x # 0. When
solving an equation, always
write the equation in general
form, then factor the equation
and set each factor equal to
zero. Do not divide each side of
an equation by a variable factor
in an attempt to simplify the
equation.
The methods used to solve quadratic equations can sometimes be extended to solve
polynomial equations of higher degree.
Example 10
Solving a Polynomial Equation by Factoring
Solve 3x 4 # 48x 2.
Solution
First write the polynomial equation in general form with zero on one side, factor the
other side, and then set each factor equal to zero and solve.
3x 4 # 48x 2
3x 4
!
48x 2
Write original equation.
#0
Write in general form.
3x 2(x 2 ! 16' # 0
Factor out common factor.
3x 2(x ( 4'(x ! 4' # 0
3x 2 # 0
Write in factored form.
x#0
Set 1st factor equal to 0.
x(4#0
x # !4
Set 2nd factor equal to 0.
x!4#0
x#4
Set 3rd factor equal to 0.
You can check these solutions by substituting in the original equation, as follows.
Check
3(0'4 # 48(0' 2
3(!4' # 48(!4'
4
0 checks.
2
✓
!4 checks.
3(4'4 # 48(4' 2
4 checks.
✓
✓
So, you can conclude that the solutions are x # 0, x # !4, and x # 4.
Now try Exercise 113.
Example 11
Solving a Polynomial Equation by Factoring
Solve x 3 ! 3x 2 ! 3x ( 9 # 0.
Solution
x3 ! 3x 2 ! 3x ( 9 # 0
Write original equation.
x2(x ! 3' ! 3(x ! 3' # 0
Factor by grouping.
(x ! 3'(x 2 ! 3' # 0
x!3#0
x2 ! 3 # 0
Distributive Property
x#3
Set 1st factor equal to 0.
x # ± "3
Set 2nd factor equal to 0.
The solutions are x # 3, x # "3, and x # ! "3. Check these in the original
equation.
Now try Exercise 119.
Appendix A.5
Solving Equations
A57
Equations Involving Radicals
Operations such as squaring each side of an equation, raising each side of an equation
to a rational power, and multiplying each side of an equation by a variable quantity all
can introduce extraneous solutions. So, when you use any of these operations, checking
your solutions is crucial.
Example 12
Solving Equations Involving Radicals
a. "2x ( 7 ! x # 2
Original equation
"2x ( 7 # x ( 2
2x ( 7 #
x2
Isolate radical.
( 4x ( 4
Square each side.
0 # x 2 ( 2x ! 3
Write in general form.
0 # (x ( 3'(x ! 1'
Factor.
x(3#0
x # !3
Set 1st factor equal to 0.
x!1#0
x#1
Set 2nd factor equal to 0.
By checking these values, you can determine that the only solution is x # 1.
b. "2x ! 5 ! "x ! 3 # 1
Original equation
"2x ! 5 # "x ! 3 ( 1
When an equation contains two
radicals, it may not be possible
to isolate both. In such cases,
you may have to raise each side
of the equation to a power at two
different stages in the solution,
as shown in Example 12(b).
Isolate "2x ! 5.
2x ! 5 # x ! 3 ( 2"x ! 3 ( 1
Square each side.
2x ! 5 # x ! 2 ( 2"x ! 3
Combine like terms.
x ! 3 # 2"x ! 3
Isolate 2"x ! 3.
x 2 ! 6x ( 9 # 4(x ! 3'
Square each side.
x 2 ! 10x ( 21 # 0
Write in general form.
(x ! 3'(x ! 7' # 0
Factor.
x!3#0
x#3
Set 1st factor equal to 0.
x!7#0
x#7
Set 2nd factor equal to 0.
The solutions are x # 3 and x # 7. Check these in the original equation.
Now try Exercise 129.
Example 13
Solving an Equation Involving a Rational Exponent
(x ! 4'2)3 # 25
3
"
(x ! 4'2 # 25
(x ! 4' # 15,625
2
x ! 4 # ± 125
x # 129, x # !121
Now try Exercise 137.
Original equation
Rewrite in radical form.
Cube each side.
Take square root of each side.
Add 4 to each side.
A58
Appendix A
Review of Fundamental Concepts of Algebra
Equations with Absolute Values
To solve an equation involving an absolute value, remember that the expression inside
the absolute value signs can be positive or negative. This results in two separate
equations, each of which must be solved. For instance, the equation
,x ! 2, # 3
results in the two equations x ! 2 # 3 and ! (x ! 2' # 3, which implies that the equation has two solutions: x # 5 and x # !1.
Example 14
,
Solving an Equation Involving Absolute Value
,
Solve x 2 ! 3x # !4x ( 6.
Solution
Because the variable expression inside the absolute value signs can be positive or
negative, you must solve the following two equations.
First Equation
x 2 ! 3x # !4x ( 6
Use positive expression.
2
x (x!6#0
Write in general form.
(x ( 3'(x ! 2' # 0
Factor.
x(3#0
x # !3
Set 1st factor equal to 0.
x!2#0
x#2
Set 2nd factor equal to 0.
Second Equation
! (x 2 ! 3x' # !4x ( 6
Use negative expression.
x 2 ! 7x ( 6 # 0
Write in general form.
(x ! 1'(x ! 6' # 0
Factor.
x!1#0
x#1
Set 1st factor equal to 0.
x!6#0
x#6
Set 2nd factor equal to 0.
Check
?
,(!3'2 ! 3(!3', # !4(!3' ( 6
Substitute !3 for x.
✓
18 # 18
?
(2'2 ! 3(2' # !4(2' ( 6
!3 checks.
2 $ !2
?
(1'2 ! 3(1' # !4(1' ( 6
2 does not check.
2#2
?
(6'2 ! 3(6' # !4(6' ( 6
1 checks.
,
,
,
,
,
,
18 $ !18
The solutions are x # !3 and x # 1.
Now try Exercise 151.
Substitute 2 for x.
Substitute 1 for x.
✓
Substitute 6 for x.
6 does not check.
Appendix A.5
Solving Equations
A59
Common Formulas
The following geometric formulas are used at various times throughout this course. For
your convenience, some of these formulas along with several others are also provided
on the inside cover of this text.
Common Formulas for Area A, Perimeter P, Circumference C, and Volume V
Rectangle
Circle
A # lw
A # " r2
P # 2l ( 2w
C # 2" r
w
Triangle
1
A # bh
2
P#a(b(c
l
V # lwh
V # " r 2h
4
V # " r3
3
r
h
w
r
l
h
Using a Geometric Formula
A cylindrical can has a volume of 200 cubic centimeters (cm3' and a radius of
4 centimeters (cm), as shown in Figure A.13. Find the height of the can.
4 cm
h
A.13
Sphere
b
Example 15
FIGURE
Circular Cylinder
c
a h
r
Rectangular Solid
Solution
The formula for the volume of a cylinder is V # " r 2h. To find the height of the can,
solve for h.
h#
V
"r2
Then, using V # 200 and r # 4, find the height.
h#
#
200
" (4' 2
Substitute 200 for V and 4 for r.
200
16"
Simplify denominator.
& 3.98
Use a calculator.
Because the value of h was rounded in the solution, a check of the solution will not
result in an equality. If the solution is valid, the expressions on each side of the equal
sign will be approximately equal to each other.
V # "r2 h
?
200 & "(4'2(3.98'
Write original equation.
200 & 200.06
Solution checks.
Substitute 200 for V, 4 for r, and 3.98 for h.
✓
You can also use unit analysis to check that your answer is reasonable.
200 cm3
& 3.98 cm
16" cm2
Now try Exercise 157.
A60
A.5
Appendix A
Review of Fundamental Concepts of Algebra
EXERCISES
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
1. A(n) ________ is a statement that equates two algebraic expressions.
2. A linear equation in one variable is an equation that can be written in the standard form ________.
3. When solving an equation, it is possible to introduce an ________ solution, which is a value that does not
satisfy the original equation.
4. The four methods that can be used to solve a quadratic equation are ________, ________, ________, and the ________.
SKILLS AND APPLICATIONS
In Exercises 5–12, determine whether the equation is an
identity or a conditional equation.
4(x ( 1' # 4x ( 4
2(x ! 3' # 7x ! 1
!6(x ! 3' ( 5 # !2x ( 10
3(x ( 2' ! 5 # 3x ( 1
4(x ( 1' ! 2x # 2(x ( 2'
x 2 ( 2(3x ! 2' # x 2 ( 6x ! 4
1
4x
5 3
11. 3 (
#
12. ( # 24
x(1 x(1
x
x
5.
6.
7.
8.
9.
10.
In Exercises 13–26, solve the equation and check your
solution.
13.
15.
17.
19.
20.
21.
22.
x ( 11 # 15
14. 7 ! x # 19
7 ! 2x # 25
16. 7x ( 2 # 23
8x ! 5 # 3x ( 20
18. 7x ( 3 # 3x ! 17
4y ( 2 ! 5y # 7 ! 6y
3(x ( 3' # 5(1 ! x' ! 1
x ! 3(2x ( 3' # 8 ! 5x
9x ! 10 # 5x ( 2(2x ! 5'
x
x
3x
3x 4x
23.
!
#4
24. ! # 3 (
8
3
5 2
10
3
1
25. 2(z ( 5' ! 4(z ( 24' # 0
26. 0.60x ( 0.40(100 ! x' # 50
In Exercises 27–42, solve the equation and check your
solution. (If not possible, explain why.)
27. x ( 8 # 2(x ! 2' ! x
28. 8(x ( 2' ! 3(2x ( 1' # 2(x ( 5'
100 ! 4x 5x ( 6
29.
#
(6
3
4
17 ( y 32 ( y
30.
(
# 100
y
y
5x ! 4 2
6
15
31.
#
32.
!4# (3
5x ( 4 3
x
x
1
2
34. (
z(2
x
x
x
4
(
(2#0
x(4 x(4
8x
7
!
# !4
2x ( 1 2x ! 1
2
1
2
#
(
(x ! 4'(x ! 2' x ! 4 x ! 2
1
3
4
(
#
x ! 2 x ( 3 x2 ( x ! 6
3
4
1
6
40. !
( #
2
x ! 3x
x
x!3
x
x
(x ( 2'2 ( 5 # (x ( 3'2
(2x ( 1'2 # 4(x 2 ( x ( 1'
33. 3 # 2 (
35.
36.
37.
38.
39.
41.
42.
2
#0
!5
2
3(x ( 5'
# 2
(3
x ( 3x
In Exercises 43– 46, write the quadratic equation in general
form.
43. 2x 2 # 3 ! 8x
45. 15(3x 2 ! 10' # 18x
44. 13 ! 3(x ( 7'2 # 0
46. x(x ( 2' # 5x 2 ( 1
In Exercises 47– 58, solve the quadratic equation by factoring.
47.
49.
51.
53.
55.
57.
58.
48. 9x 2 ! 4 # 0
6x 2 ( 3x # 0
50. x 2 ! 10x ( 9 # 0
x 2 ! 2x ! 8 # 0
52. 4x 2 ( 12x ( 9 # 0
x2 ! 12x ( 35 # 0
2
54. 2x 2 # 19x ( 33
3 ( 5x ! 2x # 0
56. 18 x 2 ! x ! 16 # 0
x 2 ( 4x # 12
x 2 ( 2ax ( a 2 # 0, a is a real number
(x ( a'2 ! b 2 # 0, a and b are real numbers
In Exercises 59–70, solve the equation by extracting square roots.
59.
61.
63.
65.
67.
69.
x 2 # 49
3x 2 # 81
(x ! 12'2 # 16
(x ( 2' 2 # 14
(2x ! 1'2 # 18
(x ! 7'2 # (x ( 3' 2
60.
62.
64.
66.
68.
70.
x 2 # 32
9x 2 # 36
(x ( 13'2 # 25
(x ! 5'2 # 30
(2x ( 3'2 ! 27 # 0
(x ( 5'2 # (x ( 4' 2
Appendix A.5
In Exercises 71– 80, solve the quadratic equation by
completing the square.
71.
73.
75.
77.
79.
x 2 ( 4x ! 32 # 0
x 2 ( 12x ( 25 # 0
8 ( 4x ! x 2 # 0
2x 2 ( 5x ! 8 # 0
5x2 ! 15x ( 7 # 0
72.
74.
76.
78.
80.
x2 ( 6x ( 2 # 0
x 2 ( 8x ( 14 # 0
9x 2 ! 12x # 14
4x 2 ! 4x ! 99 # 0
3x2 ( 9x ( 5 # 0
In Exercises 81–98, use the Quadratic Formula to solve the
equation.
81.
83.
85.
87.
89.
91.
93.
95.
97.
2x 2 ( x ! 1 # 0
2 ( 2x ! x 2 # 0
x 2 ( 14x ( 44 # 0
x 2 ( 8x ! 4 # 0
12x ! 9x 2 # !3
9x2 ( 24x ( 16 # 0
28x ! 49x 2 # 4
8t # 5 ( 2t 2
( y ! 5'2 # 2y
82.
84.
86.
88.
90.
92.
94.
96.
98.
25x 2 ! 20x ( 3 # 0
x 2 ! 10x ( 22 # 0
6x # 4 ! x 2
4x 2 ! 4x ! 4 # 0
16x 2 ( 22 # 40x
16x 2 ! 40x ( 5 # 0
3x ( x 2 ! 1 # 0
25h2 ( 80h ( 61 # 0
(57x ! 14'2 # 8x
In Exercises 99–104, use the Quadratic Formula to solve the
equation. (Round your answer to three decimal places.)
99.
100.
101.
102.
103.
104.
0.1x2 ( 0.2x ! 0.5 # 0
2x 2 ! 2.50x ! 0.42 # 0
!0.067x 2 ! 0.852x ( 1.277 # 0
!0.005x 2 ( 0.101x ! 0.193 # 0
422x 2 ! 506x ! 347 # 0
!3.22x 2 ! 0.08x ( 28.651 # 0
In Exercises 105–112, solve the equation using any
convenient method.
105.
107.
109.
111.
112.
! 2x ! 1 # 0
106.
( 33x # 0
2
2
(x ( 3' # 81
108. x ! 14x ( 49 # 0
11
2
x !x! 4 #0
110. 3x ( 4 # 2x2 ! 7
4x 2 ( 2x ( 4 # 2x ( 8
a 2x 2 ! b 2 # 0, a and b are real numbers, a $ 0
x2
11x 2
In Exercises 113–126, find all real solutions of the equation.
Check your solutions in the original equation.
113.
115.
117.
118.
119.
120.
2x4 ! 50x2 # 0
114. 20x3 ! 125x # 0
x 4 ! 81 # 0
116. x6 ! 64 # 0
3
x ( 216 # 0
9x4 ! 24x3 ( 16x 2 # 0
x3 ! 3x2 ! x ( 3 # 0
x3 ( 2x2 ( 3x ( 6 # 0
121.
122.
123.
125.
x4
x4
x4
x6
(
!
!
(
Solving Equations
A61
x # x3 ( 1
2x3 # 16 ( 8x ! 4x3
124. 36t 4 ( 29t 2 ! 7 # 0
4x2 ( 3 # 0
126. x6 ( 3x3 ( 2 # 0
7x3 ! 8 # 0
In Exercises 127–154, find all solutions of the equation.
Check your solutions in the original equation.
127.
129.
131.
133.
135.
137.
139.
141.
142.
143.
145.
147.
149.
151.
153.
128.
130.
132.
134.
136.
138.
140.
"2x ! 10 # 0
7"x ! 6 # 0
"x ! 10 ! 4 # 0
"5 ! x ! 3 # 0
"2x ( 5 ( 3 # 0
"3 ! 2x ! 2 # 0
3
3
"
"
2x ( 1 ( 8 # 0
4x ! 3 ( 2 # 0
"5x ! 26 ( 4 # x
"x ( 5 # "2x ! 5
(x ! 6'3)2 # 8
(x ( 3'3)2 # 8
(x ( 3'2)3 # 5
(x2 ! x ! 22'4)3 # 16
3x(x ! 1'1)2 ( 2(x ! 1'3)2 # 0
4x2(x ! 1'1)3 ( 6x(x ! 1'4)3 # 0
3 1
4
3
144.
x# (
!
#1
x
2
x(1 x(2
20 ! x
3
146. 4x ( 1 #
#x
x
x
x
1
x(1 x(1
(
# 3 148.
!
#0
x2 ! 4 x ( 2
3
x(2
150. 13x ( 1 # 12
2x ! 1 # 5
2
152. x 2 ( 6x # 3x ( 18
x #x (x!3
154. x ! 10 # x 2 ! 10x
x ( 1 # x2 ! 5
,
,
,,
, ,
,
,
,
,
,
,
ANTHROPOLOGY In Exercises 155 and 156, use the
following information. The relationship between the length
of an adult’s femur (thigh bone) and the height of the adult
can be approximated by the linear equations
y " 0.432 x ! 10.44
Female
y " 0.449x ! 12.15
Male
where y is the length of the femur in inches and x is the
height of the adult in inches (see figure).
x in.
y in.
femur
155. An anthropologist discovers a femur belonging to an
adult human female. The bone is 16 inches long.
Estimate the height of the female.
A62
Appendix A
Review of Fundamental Concepts of Algebra
156. From the foot bones of an adult human male, an
anthropologist estimates that the person’s height was
69 inches. A few feet away from the site where the
foot bones were discovered, the anthropologist
discovers a male adult femur that is 19 inches long. Is
it likely that both the foot bones and the thigh bone
came from the same person?
157. VOLUME OF A BILLIARD BALL A billiard ball has
a volume of 5.96 cubic inches. Find the radius of a
billiard ball.
158. LENGTH OF A TANK The diameter of a cylindrical
propane gas tank is 4 feet. The total volume of the tank
is 603.2 cubic feet. Find the length of the tank.
159. GEOMETRY A “Slow Moving Vehicle” sign has the
shape of an equilateral triangle. The sign has a
perimeter of 129 centimeters. Find the length of each
side of the sign. Find the area of the sign.
160. GEOMETRY The radius of a traffic cone is
14 centimeters and the lateral surface area of the cone is
1617 square centimeters. Find the height of the cone.
161. VOTING POPULATION The total voting-age
population P (in millions) in the United States from
1990 through 2006 can be modeled by
P#
182.17 ! 1.542t
, 0 % t % 16
1 ! 0.018t
where t represents the year, with t # 0 corresponding
to 1990. (Source: U.S. Census Bureau)
(a) In which year did the total voting-age population
reach 200 million?
(b) Use the model to predict the year in which the
total voting-age population will reach 241 million.
Is this prediction reasonable? Explain.
162. AIRLINE PASSENGERS An airline offers daily
flights between Chicago and Denver. The total monthly
cost C (in millions of dollars) of these flights is
C # "0.2x ( 1, where x is the number of passengers
(in thousands). The total cost of the flights for June is
2.5 million dollars. How many passengers flew in
June?
EXPLORATION
TRUE OR FALSE? In Exercises 163 and 164, determine
whether the statement is true or false. Justify your answer.
163. An equation can never have more than one extraneous
solution.
164. When solving an absolute value equation, you will
always have to check more than one solution.
165. THINK ABOUT IT What is meant by equivalent
equations? Give an example of two equivalent
equations.
166. Solve 3(x ( 4'2 ( (x ( 4' ! 2 # 0 in two ways.
(a) Let u # x ( 4, and solve the resulting equation
for u. Then solve the u-solution for x.
(b) Expand and collect like terms in the equation, and
solve the resulting equation for x.
(c) Which method is easier? Explain.
THINK ABOUT IT In Exercises 167–170, write a quadratic
equation that has the given solutions. (There are many
correct answers.)
167.
168.
169.
170.
!3 and 6
!4 and !11
1 ( "2 and 1 ! "2
!3 ( "5 and !3 ! "5
In Exercises 171 and 172, consider an equation of the form
x # x ! a " b, where a and b are constants.
,
,
171. Find a and b when the solution of the equation is
x # 9. (There are many correct answers.)
172. WRITING Write a short paragraph listing the steps
required to solve this equation involving absolute values,
and explain why it is important to check your solutions.
In Exercises 173 and 174, consider an equation of the form
x # "x ! a " b, where a and b are constants.
173. Find a and b when the solution of the equation is
x # 20. (There are many correct answers.)
174. WRITING Write a short paragraph listing the steps
required to solve this equation involving radicals, and
explain why it is important to check your solutions.
175. Solve each equation, given that a and b are not zero.
(a) ax 2 ( bx # 0
(b) ax 2 ! ax # 0
176. CAPSTONE
(a) Explain the difference between a conditional
equation and an identity.
(b) Give an example of an absolute value equation
that has only one solution.
(c) State the Quadratic Formula in words.
(d) Does raising each side of an equation to the nth
power always yield an equivalent equation? Explain.
Appendix A.6
Linear Inequalities in One Variable
A63
A.6 LINEAR INEQUALITIES IN ONE VARIABLE
What you should learn
• Represent solutions of linear
inequalities in one variable.
• Use properties of inequalities to
create equivalent inequalities.
• Solve linear inequalities in one
variable.
• Solve inequalities involving absolute
values.
• Use inequalities to model and solve
real-life problems.
Why you should learn it
Inequalities can be used to model and
solve real-life problems. For instance,
in Exercise 121 on page A71, you will
use a linear inequality to analyze the
average salary for elementary school
teachers.
Introduction
Simple inequalities were discussed in Appendix A.1. There, you used the inequality
symbols <, ≤, >, and ≥ to compare two numbers and to denote subsets of real numbers.
For instance, the simple inequality
x & 3
denotes all real numbers x that are greater than or equal to 3.
Now, you will expand your work with inequalities to include more involved
statements such as
5x ! 7 < 3x ( 9
and
!3 % 6x ! 1 < 3.
As with an equation, you solve an inequality in the variable x by finding all values of
x for which the inequality is true. Such values are solutions and are said to satisfy the
inequality. The set of all real numbers that are solutions of an inequality is the solution
set of the inequality. For instance, the solution set of
x(1 < 4
is all real numbers that are less than 3.
The set of all points on the real number line that represents the solution set is the
graph of the inequality. Graphs of many types of inequalities consist of intervals on
the real number line. See Appendix A.1 to review the nine basic types of intervals on
the real number line. Note that each type of interval can be classified as bounded or
unbounded.
Example 1
Intervals and Inequalities
Write an inequality to represent each interval, and state whether the interval is bounded
or unbounded.
a. (!3, 5*
b. (!3, ''
c. +0, 2*
d. (! ', ''
Solution
a. (!3, 5* corresponds to !3 < x % 5.
b. (!3, '' corresponds to !3 < x.
c. +0, 2* corresponds to 0 % x % 2.
d. (! ', '' corresponds to ! ' < x < '.
Now try Exercise 9.
Bounded
Unbounded
Bounded
Unbounded
A64
Appendix A
Review of Fundamental Concepts of Algebra
Properties of Inequalities
The procedures for solving linear inequalities in one variable are much like those for
solving linear equations. To isolate the variable, you can make use of the Properties of
Inequalities. These properties are similar to the properties of equality, but there are two
important exceptions. When each side of an inequality is multiplied or divided by a
negative number, the direction of the inequality symbol must be reversed. Here is an
example.
!2 < 5
Original inequality
(!3'(!2' > (!3'(5'
Multiply each side by !3 and reverse inequality.
6 > !15
Simplify.
Notice that if the inequality was not reversed, you would obtain the false statement
6 < !15.
Two inequalities that have the same solution set are equivalent. For instance, the
inequalities
x(2 < 5
and
x < 3
are equivalent. To obtain the second inequality from the first, you can subtract 2 from
each side of the inequality. The following list describes the operations that can be used
to create equivalent inequalities.
Properties of Inequalities
Let a, b, c, and d be real numbers.
1. Transitive Property
a < b and b < c
a < c
2. Addition of Inequalities
a < b and c < d
a(c < b(d
3. Addition of a Constant
a < b
a(c < b(c
4. Multiplication by a Constant
For c > 0, a < b
ac < bc
For c < 0, a < b
ac > bc
Reverse the inequality.
Each of the properties above is true if the symbol < is replaced by % and
the symbol > is replaced by ≥. For instance, another form of the multiplication property
would be as follows.
For c > 0, a % b
ac % bc
For c < 0, a % b
ac & bc
Appendix A.6
Linear Inequalities in One Variable
A65
Solving a Linear Inequality in One Variable
The simplest type of inequality is a linear inequality in one variable. For instance,
2x ( 3 > 4 is a linear inequality in x.
In the following examples, pay special attention to the steps in which the inequality
symbol is reversed. Remember that when you multiply or divide by a negative number,
you must reverse the inequality symbol.
Example 2
Solving a Linear Inequality
Solve 5x ! 7 > 3x ( 9.
Solution
Checking the solution set of an
inequality is not as simple as
checking the solutions of an
equation. You can, however, get
an indication of the validity of
a solution set by substituting a
few convenient values of x. For
instance, in Example 2, try
substituting x # 5 and x # 10
into the original inequality.
5x ! 7 > 3x ( 9
Write original inequality.
2x ! 7 > 9
Subtract 3x from each side.
2x > 16
Add 7 to each side.
x > 8
Divide each side by 2.
The solution set is all real numbers that are greater than 8, which is denoted by (8, ''.
The graph of this solution set is shown in Figure A.14. Note that a parenthesis at 8 on
the real number line indicates that 8 is not part of the solution set.
x
6
7
8
9
10
Solution interval: (8, ''
FIGURE A.14
Now try Exercise 35.
Example 3
Solving a Linear Inequality
Solve 1 ! 32 x & x ! 4.
Graphical Solution
Algebraic Solution
3x
1!
& x!4
2
Write original inequality.
2 ! 3x & 2x ! 8
Multiply each side by 2.
2 ! 5x & !8
Subtract 2x from each side.
!5x & !10
x % 2
Use a graphing utility to graph y1 # 1 ! 32 x and
y2 # x ! 4 in the same viewing window. In Figure A.16,
you can see that the graphs appear to intersect at the
point (2, !2'. Use the intersect feature of the graphing
utility to confirm this. The graph of y1 lies above the
graph of y2 to the left of their point of intersection, which
implies that y1 & y2 for all x % 2.
Subtract 2 from each side.
Divide each side by !5 and reverse the inequality.
The solution set is all real numbers that are less than or equal to 2,
which is denoted by (! ', 2*. The graph of this solution set is
shown in Figure A.15. Note that a bracket at 2 on the real number
line indicates that 2 is part of the solution set.
2
−5
7
y1 = 1 − 32 x
x
0
1
2
Solution interval: (! ', 2*
FIGURE A.15
Now try Exercise 37.
3
y2 = x − 4
−6
4
FIGURE
A.16
A66
Appendix A
Review of Fundamental Concepts of Algebra
Sometimes it is possible to write two inequalities as a double inequality. For
instance, you can write the two inequalities !4 % 5x ! 2 and 5x ! 2 < 7 more
simply as
!4 % 5x ! 2 < 7.
Double inequality
This form allows you to solve the two inequalities together, as demonstrated in
Example 4.
Example 4
Solving a Double Inequality
To solve a double inequality, you can isolate x as the middle term.
!3 % 6x ! 1 < 3
Original inequality
!3 ( 1 % 6x ! 1 ( 1 < 3 ( 1
Add 1 to each part.
!2 % 6x < 4
Simplify.
!2
6x 4
<
%
6
6
6
Divide each part by 6.
!
1
2
% x<
3
3
Simplify.
The solution set is all real numbers that are greater than or equal to ! 13 and less than
2
1 2
3 , which is denoted by +! 3 , 3 '. The graph of this solution set is shown in Figure A.17.
− 13
2
3
x
−1
0
1
Solution interval: +! 13, 23 '
FIGURE A.17
Now try Exercise 47.
The double inequality in Example 4 could have been solved in two parts, as
follows.
!3 % 6x ! 1
!2 % 6x
1
! % x
3
and
6x ! 1 < 3
6x < 4
x <
2
3
The solution set consists of all real numbers that satisfy both inequalities. In other
words, the solution set is the set of all values of x for which
!
2
1
%x< .
3
3
When combining two inequalities to form a double inequality, be sure that the
inequalities satisfy the Transitive Property. For instance, it is incorrect to combine the
inequalities 3 < x and x % !1 as 3 < x % !1. This “inequality” is wrong because
3 is not less than !1.
Appendix A.6
A67
Linear Inequalities in One Variable
Inequalities Involving Absolute Values
Solving an Absolute Value Inequality
T E C H N O LO G Y
Let x be a variable or an algebraic expression and let a be a real number such
that a & 0.
A graphing utility can be used
to identify the solution set of
the graph of an inequality. For
instance, to find the solution set
of x ! 5 < 2 (see Example 5a),
rewrite the inequality as
x ! 5 ! 2 < 0, enter
,
,
,,
!a < x < a.
Double inequality
,,
2. The solutions of x > a are all values of x that are less than !a or greater
than a.
,
,x, > a
Y1 " abs / X ! 50 ! 2
6
if and only if
x < a
,
and press the graph key. The
graph should look like the one
shown below.
,,
1. The solutions of x < a are all values of x that lie between !a and a.
if and only if
x < !a or
x > a.
Compound inequality
These rules are also valid if < is replaced by ≤ and > is replaced by ≥.
Example 5
Solving an Absolute Value Inequality
Solve each inequality.
−1
10
,
,
,
a. x ! 5 < 2
,
b. x ( 3 & 7
Solution
a.
−4
,x ! 5, < 2
Write original inequality.
!2 < x ! 5 < 2
Notice that the graph is below
the x-axis on the interval /3, 70.
Write equivalent inequalities.
!2 ( 5 < x ! 5 ( 5 < 2 ( 5
Add 5 to each part.
3 < x < 7
Simplify.
The solution set is all real numbers that are greater than 3 and less than 7, which is
denoted by (3, 7'. The graph of this solution set is shown in Figure A.18.
b.
,x ( 3, &
7
Write original inequality.
x ( 3 % !7
or
x ( 3 ! 3 % !7 ! 3
x(3 & 7
x ( 3!3 & 7!3
,
,
Subtract 3 from each side.
x & 4
x % !10
Note that the graph of the
inequality x ! 5 < 2 can be
described as all real numbers
within two units of 5, as shown
in Figure A.18.
Write equivalent inequalities.
Simplify.
The solution set is all real numbers that are less than or equal to !10 or greater than
or equal to 4. The interval notation for this solution set is (! ', !10* % +4, ''. The
symbol % is called a union symbol and is used to denote the combining of two sets.
The graph of this solution set is shown in Figure A.19.
2 units
2 units
7 units
7 units
x
x
2
3
4
5
6
7
8
,x ! 5, < 2: Solutions lie inside (3, 7'.
FIGURE
A.18
Now try Exercise 61.
−12 −10 −8 −6 −4 −2
0
2
4
6
,x ( 3, & 7: Solutions lie outside (!10, 4'.
FIGURE
A.19
A68
Appendix A
Review of Fundamental Concepts of Algebra
Applications
A problem-solving plan can be used to model and solve real-life problems that involve
inequalities, as illustrated in Example 6.
Example 6
Comparative Shopping
You are choosing between two different cell phone plans. Plan A costs $49.99 per month
for 500 minutes plus $0.40 for each additional minute. Plan B costs $45.99 per month
for 500 minutes plus $0.45 for each additional minute. How many additional minutes
must you use in one month for plan B to cost more than plan A?
Solution
Verbal
Model:
Labels:
Monthly cost for plan B
>
Monthly cost for plan A
Minutes used (over 500) in one month # m
Monthly cost for plan A # 0.40m ( 49.99
Monthly cost for plan B # 0.45m ( 45.99
(minutes)
(dollars)
(dollars)
Inequality: 0.45m ( 45.99 > 0.40m ( 49.99
0.05m > 4
m > 80 minutes
Plan B costs more if you use more than 80 additional minutes in one month.
Now try Exercise 111.
Example 7
Accuracy of a Measurement
You go to a candy store to buy chocolates that cost $9.89 per pound. The scale that
is used in the store has a state seal of approval that indicates the scale is accurate to
1
within half an ounce (or 32
of a pound). According to the scale, your purchase weighs
one-half pound and costs $4.95. How much might you have been undercharged or
overcharged as a result of inaccuracy in the scale?
Solution
Let x represent the true weight of the candy. Because the scale is accurate
1
to within half an ounce (or 32
of a pound), the difference between the exact weight
, ,
(x' and the scale weight ( 12 ' is less than or equal to 321 of a pound. That is, x ! 12 ≤
You can solve this inequality as follows.
1
1
! 32
% x!2 %
15
32
% x %
1
32 .
1
32
17
32
0.46875 % x % 0.53125
In other words, your “one-half pound” of candy could have weighed as little as
0.46875 pound (which would have cost $4.64) or as much as 0.53125 pound (which
would have cost $5.25). So, you could have been overcharged by as much as $0.31 or
undercharged by as much as $0.30.
Now try Exercise 125.
Appendix A.6
A.6
EXERCISES
Linear Inequalities in One Variable
A69
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
VOCABULARY: Fill in the blanks.
1. The set of all real numbers that are solutions of an inequality is the ________ ________ of the inequality.
2. The set of all points on the real number line that represents the solution set of an inequality is the ________ of the inequality.
3. To solve a linear inequality in one variable, you can use the properties of inequalities, which are identical
to those used to solve equations, with the exception of multiplying or dividing each side by a ________ number.
4. Two inequalities that have the same solution set are ________.
5. It is sometimes possible to write two inequalities as one inequality, called a ________ inequality.
6. The symbol % is called a ________ symbol and is used to denote the combining of two sets.
SKILLS AND APPLICATIONS
In Exercises 7–14, (a) write an inequality that represents the
interval and (b) state whether the interval is bounded or
unbounded.
7.
9.
11.
13.
+0, 9'
+!1, 5*
(11, ''
(! ', !2'
(!7, 4'
(2, 10*
+!5, ''
(! ', 7*
8.
10.
12.
14.
(a)
x
−4
−3
−2
−1
0
1
2
3
4
x
3
4
5
x
−2
−1
0
0
1
1
2
3
4
5
x
2
3
4
5
(e)
x
−3
−2
−1
0
1
2
3
4
5
6
(f)
x
−5
−4
−3
−2
−1
0
1
2
3
4
5
(g)
x
−3
−2
−1
0
1
2
3
4
5
6
(h)
x
4
15.
17.
19.
21.
5
x < 3
!3 < x % 4
x < 3
!1 % x % 52
,,
6
7
16.
18.
20.
22.
8
x & 5
0 % x % 92
x > 4
!1 < x < 52
,,
,
,
,
,
27. x ! 10 & 3
28. 2x ! 3 < 15
6
(d)
−1
x!2
< 2
4
26. !5 < 2x ! 1 % 1
6
(c)
−3
25. 0 <
5
(b)
2
Inequality
23. 5x ! 12 > 0
24. 2x ( 1 < !3
In Exercises 15–22, match the inequality with its graph. [The
graphs are labeled (a)–(h).]
−5
In Exercises 23–28, determine whether each value of x is a
solution of the inequality.
(a)
(c)
(a)
(c)
(a)
(c)
(a)
(c)
(a)
(c)
(a)
(c)
Values
x#3
(b)
5
x#2
(d)
x#0
(b)
x # !4
(d)
x#4
(b)
x#0
(d)
1
x # !2
(b)
4
x#3
(d)
x # 13
(b)
x # 14
(d)
x # !6
(b)
x # 12
(d)
x # !3
x # 32
x # ! 14
x # ! 32
x # 10
x # 72
x # ! 52
x#0
x # !1
x#9
x#0
x#7
In Exercises 29–56, solve the inequality and sketch the
solution on the real number line. (Some inequalities have no
solutions.)
29. 4x < 12
31. !2x > !3
33.
35.
37.
39.
41.
43.
45.
47.
48.
49.
50.
30. 10x < ! 40
32. !6x > 15
x!5 & 7
34.
2x ( 7 < 3 ( 4x
36.
2x ! 1 & 1 ! 5x
38.
4 ! 2x < 3(3 ! x'
40.
3
42.
4x ! 6 % x ! 7
1
5
(
8x
(
1
'
3x
(
44.
&
2
2
3.6x ( 11 & !3.4
46.
1 < 2x ( 3 < 9
!8 % ! (3x ( 5' < 13
!8 % 1 ! 3(x ! 2' < 13
0 % 2 ! 3(x ( 1' < 20
x ( 7 % 12
3x ( 1 & 2 ( x
6x ! 4 % 2 ( 8x
4(x ( 1' < 2x ( 3
3 ( 27 x > x ! 2
9x ! 1 < 34(16x ! 2'
15.6 ! 1.3x < !5.2
A70
Appendix A
Review of Fundamental Concepts of Algebra
2x ! 3
< 4
3
3
1
53. > x ( 1 >
4
4
51. !4 <
55. 3.2 % 0.4x ! 1 % 4.4
52. 0 %
,
x(3
< 5
2
x
< 1
3
1.5x ( 6
56. 4.5 >
> 10.5
2
54. !1 < 2 !
In Exercises 57–72, solve the inequality and sketch the
solution on the real number line. (Some inequalities have no
solution.)
,,
57. x < 5
x
59.
> 1
2
61. x ! 5 < !1
63. x ! 20 % 6
65. 3 ! 4x & 9
x!3
67.
&4
2
69. 9 ! 2x ! 2 < !1
71. 2 x ( 10 & 9
,, ,
,
,
,
,
,
,, , ,
,
,
,,
58. x & 8
x
60.
> 3
5
,, ,
,
,
6x > 12
5 ! 2x & 1
4(x ! 3' % 8 ! x
x ! 8 % 14
2 x ( 7 & 13
,
,
,
,
,
,
83.
84.
85.
86.
87.
88. y #
,
,
1
2x
,
,
(1
(a)
(a)
(a)
(a)
(a)
x
−3
−2
−1
0
1
2
3
2
3
x
,
,
,
Inequalities
y & 1
(b)
y % 5
(b)
0 % y % 3
(b)
!1 % y % 3 (b)
y % 2
(b)
(a) y % 4
y
y
y
y
y
%
&
&
%
&
90. "x ! 10
92. "3 ! x
4 6x ( 15
94. "
−2
−1
0
1
x
5
6
7
8
9
10
11
12
13
14
0
1
2
3
x
−7
0
0
0
0
4
(b) y & 1
In Exercises 89–94, find the interval(s) on the real number
line for which the radicand is nonnegative.
89. "x ! 5
91. "x ( 3
4 7 ! 2x
93. "
97.
100.
GRAPHICAL ANALYSIS In Exercises 83– 88, use a graphing
utility to graph the equation. Use the graph to approximate
the values of x that satisfy each inequality.
Equation
y # 2x ! 3
y # 23x ( 1
y # ! 12x ( 2
y # !3x ( 8
y# x!3
,
In Exercises 97–104, use absolute value notation to define the
interval (or pair of intervals) on the real number line.
4
3x ! 1 % 5
20 < 6x ! 1
3(x ( 1' < x ( 7
2x ( 9 > 13
1
2 x ( 1 % 3
,
,
99.
,
,, ,,
74.
76.
78.
80.
82.
,
,
,
−3
GRAPHICAL ANALYSIS In Exercises 73 – 82, use a
graphing utility to graph the inequality and identify the
solution set.
73.
75.
77.
79.
81.
,
98.
62. x ! 7 < !5
64. x ! 8 & 0
66. 1 ! 2x < 5
2x
68. 1 !
< 1
3
70. x ( 14 ( 3 > 17
72. 3 4 ! 5x % 9
,
,
,
95. THINK ABOUT IT The graph of x ! 5 < 3 can be
described as all real numbers within three units of 5.
Give a similar description of x ! 10 < 8.
96. THINK ABOUT IT The graph of x ! 2 > 5 can be
described as all real numbers more than five units
from 2. Give a similar description of x ! 8 > 4.
101.
102.
103.
104.
−6
−5
−4
−3
−2
−1
All real numbers within 10 units of 12
All real numbers at least five units from 8
All real numbers more than four units from !3
All real numbers no more than seven units from !6
In Exercises 105–108, use inequality notation to describe the
subset of real numbers.
105. A company expects its earnings per share E for the next
quarter to be no less than $4.10 and no more than $4.25.
106. The estimated daily oil production p at a refinery is
greater than 2 million barrels but less than 2.4 million
barrels.
107. According to a survey, the percent p of U.S. citizens
that now conduct most of their banking transactions
online is no more than 45%.
108. The net income I of a company is expected to be no
less than $239 million.
PHYSIOLOGY In Exercises 109 and 110, use the following
information. The maximum heart rate of a person in normal
health is related to the person’s age by the equation
r " 220 ! A, where r is the maximum heart rate in beats per
minute and A is the person’s age in years. Some physiologists
recommend that during physical activity a sedentary person
should strive to increase his or her heart rate to at least 50%
of the maximum heart rate, and a highly fit person should
strive to increase his or her heart rate to at most 85% of the
maximum heart rate. (Source: American Heart Association)
109. Express as an interval the range of the target heart rate
for a 20-year-old.
110. Express as an interval the range of the target heart rate
for a 40-year-old.
Appendix A.6
111. JOB OFFERS You are considering two job offers.
The first job pays $13.50 per hour. The second job
pays $9.00 per hour plus $0.75 per unit produced per
hour. Write an inequality yielding the number of units x
that must be produced per hour to make the second job
pay the greater hourly wage. Solve the inequality.
112. JOB OFFERS You are considering two job offers.
The first job pays $3000 per month. The second job
pays $1000 per month plus a commission of 4% of
your gross sales. Write an inequality yielding the gross
sales x per month for which the second job will pay the
greater monthly wage. Solve the inequality.
113. INVESTMENT In order for an investment of $1000 to
grow to more than $1062.50 in 2 years, what must the
annual interest rate be? +A # P(1 ( rt'*
114. INVESTMENT In order for an investment of $750 to
grow to more than $825 in 2 years, what must the
annual interest rate be? +A # P(1 ( rt'*
115. COST, REVENUE, AND PROFIT The revenue from
selling x units of a product is R # 115.95x. The cost of
producing x units is C # 95x ( 750. To obtain a
profit, the revenue must be greater than the cost. For
what values of x will this product return a profit?
116. COST, REVENUE, AND PROFIT The revenue from
selling x units of a product is R # 24.55x. The cost of
producing x units is C # 15.4x ( 150,000. To obtain
a profit, the revenue must be greater than the cost. For
what values of x will this product return a profit?
117. DAILY SALES A doughnut shop sells a dozen
doughnuts for $4.50. Beyond the fixed costs (rent,
utilities, and insurance) of $220 per day, it costs $2.75
for enough materials (flour, sugar, and so on) and labor
to produce a dozen doughnuts. The daily profit from
doughnut sales varies from $60 to $270. Between what
levels (in dozens) do the daily sales vary?
118. WEIGHT LOSS PROGRAM A person enrolls in a
diet and exercise program that guarantees a loss of at
least 112 pounds per week. The person’s weight at the
beginning of the program is 164 pounds. Find the
maximum number of weeks before the person attains a
goal weight of 128 pounds.
119. DATA ANALYSIS: IQ SCORES AND GPA The
admissions office of a college wants to determine
whether there is a relationship between IQ scores x and
grade-point averages y after the first year of school. An
equation that models the data the admissions office
obtained is y # 0.067x ! 5.638.
(a) Use a graphing utility to graph the model.
(b) Use the graph to estimate the values of x that
predict a grade-point average of at least 3.0.
Linear Inequalities in One Variable
A71
120. DATA ANALYSIS: WEIGHTLIFTING You want to
determine whether there is a relationship between an
athlete’s weight x (in pounds) and the athlete’s
maximum bench-press weight y (in pounds). The table
shows a sample of data from 12 athletes.
Athlete’s
weight, x
Bench-press
weight, y
165
184
150
210
196
240
202
170
185
190
230
160
170
185
200
255
205
295
190
175
195
185
250
155
(a) Use a graphing utility to plot the data.
(b) A model for the data is y # 1.3x ! 36. Use a
graphing utility to graph the model in the same
viewing window used in part (a).
(c) Use the graph to estimate the values of x that
predict a maximum bench-press weight of at least
200 pounds.
(d) Verify your estimate from part (c) algebraically.
(e) Use the graph to write a statement about the
accuracy of the model. If you think the graph
indicates that an athlete’s weight is not a
particularly good indicator of the athlete’s
maximum bench-press weight, list other factors
that might influence an individual’s maximum
bench-press weight.
121. TEACHERS’ SALARIES The average salaries S (in
thousands of dollars) for elementary school teachers
in the United States from 1990 through 2005 are
approximated by the model
S # 1.09t ( 30.9, 0 % t % 15
where t represents the year, with t # 0 corresponding
to 1990. (Source: National Education Association)
(a) According to this model, when was the average
salary at least $32,500, but not more than $42,000?
(b) According to this model, when will the average
salary exceed $54,000?
Appendix A
Review of Fundamental Concepts of Algebra
122. EGG PRODUCTION The numbers of eggs E (in
billions) produced in the United States from 1990
through 2006 can be modeled by
E # 1.52t ( 68.0, 0 % t % 16
123.
124.
125.
126.
127.
where t represents the year, with t # 0 corresponding
to 1990. (Source: U.S. Department of Agriculture)
(a) According to this model, when was the annual egg
production 70 billion, but no more than 80 billion?
(b) According to this model, when will the annual egg
production exceed 100 billion?
GEOMETRY The side of a square is measured as
1
10.4 inches with a possible error of 16
inch. Using
these measurements, determine the interval containing
the possible areas of the square.
GEOMETRY The side of a square is measured
as 24.2 centimeters with a possible error of
0.25 centimeter. Using these measurements, determine
the interval containing the possible areas of the square.
ACCURACY OF MEASUREMENT You stop at a
self-service gas station to buy 15 gallons of 87-octane
gasoline at $2.09 a gallon. The gas pump is accurate to
1
within 10
of a gallon. How much might you be
undercharged or overcharged?
ACCURACY OF MEASUREMENT You buy six
T-bone steaks that cost $14.99 per pound. The weight
that is listed on the package is 5.72 pounds. The scale
that weighed the package is accurate to within 12 ounce.
How much might you be undercharged or overcharged?
TIME STUDY A time study was conducted to
determine the length of time required to perform a
particular task in a manufacturing process. The times
required by approximately two-thirds of the workers in
the study satisfied the inequality
,
,
t ! 15.6
< 1
1.9
where t is time in minutes. Determine the interval on
the real number line in which these times lie.
128. HEIGHT The heights h of two-thirds of the members
of a population satisfy the inequality
,
,
130. MUSIC Michael Kasha of Florida State University
used physics and mathematics to design a new
classical guitar. The model he used for the frequency
of the vibrations on a circular plate was
v # (2.6t)d 2'"E)-, where v is the frequency (in
vibrations per second), t is the plate thickness (in
millimeters), d is the diameter of the plate, E is the
elasticity of the plate material, and - is the density of
the plate material. For fixed values of d, E, and -, the
graph of the equation is a line (see figure).
Frequency
(vibrations per second)
A72
v
700
600
500
400
300
200
100
1
2
(a) Estimate the frequency when the plate thickness is
2 millimeters.
(b) Estimate the plate thickness when the frequency is
600 vibrations per second.
(c) Approximate the interval for the plate thickness
when the frequency is between 200 and 400
vibrations per second.
(d) Approximate the interval for the frequency when
the plate thickness is less than 3 millimeters.
EXPLORATION
TRUE OR FALSE? In Exercises 131 and 132, determine
whether the statement is true or false. Justify your answer.
131. If a, b, and c are real numbers, and a % b, then
ac % bc.
132. If !10 % x % 8, then !10 & !x and !x & !8.
,
,
,
,
133. Identify the graph of the inequality x ! a & 2.
(a)
(b)
x
a−2
2
x
a−2
a+2
a
x
2−a
where h is measured in inches. Determine the interval
on the real number line in which these heights lie.
129. METEOROLOGY An electronic device is to be
operated in an environment with relative humidity h in
the interval defined by h ! 50 % 30. What are the
minimum and maximum relative humidities for the
operation of this device?
t
4
Plate thickness (in millimeters)
(c)
h ! 68.5
%1
2.7
3
a
a+2
(d)
x
2−a
2+a
2
2+a
134. Find sets of values of a, b, and c such that 0 % x % 10
is a solution of the inequality ax ! b % c.
135. Give an example of an inequality with an unbounded
solution set.
,
,
136. CAPSTONE Describe any differences between
properties of equalities and properties of inequalities.
PreCalculus Practice Test 2014-2015 (Due August 14, 2014)
1. The length of a rectangle is (
) centimeters and the width is (
the area, in square centimeters, of the rectangle.
) centimeters. Which expression represents
2. Find the area of the given triangle.
23
4
10
9
3. The area of a triangle is 78 square feet. Find the length of the base if the height is 13 feet.
4. Find the area, in mm, of the shaded region comprised of a rectangle and a right triangle.
8
4
6
10
5. Given:
1
q
p
2
m
What is
n
?
6. In the figure below, line m is parallel to line n.
1
m
2
3
n
5
4
Determine whether the folloing statements are true or false.
True or False
True or False
True or False
True or False
True or False
7. In the accompanying diagram, parallel lines l and m are cut by a transversal t.
t
1
l
2
m
Determine whether the following statements are true or false.
is the complement of
True or False
.
and
are right
angles.
is the supplement of
.
True or False
True or False
True or False
True or False
8. If the area of a circle is 49
9. In the triangles below,
inches , find the circumference of the circle.
.
R
(3x)°
Q
S
U
(5x–8)°
(8x+12)°
T
V
What is the value of x?
10. A right triangle’s hypotenuse has length 6. If one leg has length 5, what is the length of the other leg?
11. Find the value of x. Write your answer in simplest radical form.
10
45º
x
12. If
in the right triangle below, what is the values of b?
a
b
c
30 º
13. What is the area, in square inches (in), of the triangle below?
16 in.
16 in.
16 in.
14. Simplify:
(x + 2)(5x – 2) – (–3x – 8)
15. Simplify:
16. Evaluate.
17. Given the polynomial, 7x + 5x3 – 35x2 – 49, identify the leading coefficient, degree and constant.
18. Evaluate.
19. Simplify.
20. Simplify.
21. Rationalize and simplify.
22. Write the rational expression in simplest form.
23. Factor the sum of two cubes.
24. Factor completely.
25. Factor by grouping.
20x3 – 25x2 – 16x + 20
26. Multiply.
27. Simplify.
28. Solve.
29. Solve.
30. Solve.
31. Write the interval [–9, 1) in inquality notation, sketch the subset on a real number line, and state whether the
interval is bound or unbound.
32. What is the complete solution to the equation
33. Perform the opertation.
34. Write the rational expression in simplest form.
35. Solve the equation.
36. Solve.
37. Solve.
38. Solve.
39. Simplify
to lowest terms.
40. Which is the factored form of
?
PreCalculus Practice Test 2014-2015 (Due August 14, 2014)
Answer Section
1.
2.
3.
4.
5.
6.
7.
8.
9.
18 units2
12 feet
68
113º
T,F,T,T,T
F,F,F,T,T
C = 14 inches
11
10.
11
11. 10
12. 11
13. 64
14.
15.
16. 3
17. 5, 3, –49
18. 64
19.
20.
21.
22.
23.
24.
25. (5x 2 – 4)(4x – 5)
26.
27.
28. 13x = 17
29. x = 0, +
3
2
,
30. x = 128
31.
–9
–8
–7
–6
32. x = –2; x = –4
–5
–4
–3
–2
–1
0
1
33.
34.
35.
, –6
36. x = 21
37. x > 20 or x < –14
38. x = –5, 2
39.
40. 2(a – 3b)(a – 3b)