C H A P T E R 1Equations, Inequalities, and Mathematical Modeling
Section 1.1 Graphs of Equations . . . . . . . . . . . . . . . . . . . . 39
Section 1.2 Linear Equations in One Variable . . . . . . . . . . . . . 43
Section 1.3 Modeling with Linear Equations . . . . . . . . . . . . . . 51
Section 1.4 Quadratic Equations and Applications . . . . . . . . . . . 58
Section 1.5 Complex Numbers . . . . . . . . . . . . . . . . . . . . . 67
Section 1.6 Other Types of Equations . . . . . . . . . . . . . . . . . 71
Section 1.7 Linear Inequalities in One Variable . . . . . . . . . . . . 82
Section 1.8 Other Types of Inequalities . . . . . . . . . . . . . . . . 87
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
39
C H A P T E R 1Equations, Inequalities, and Mathematical Modeling
Section 1.1 Graphs of Equations
■ You should be able to use the point-plotting method of graphing.
■ You should be able to find x- and y-intercepts.
(a) To find the x-intercepts, let and solve for x.
(b) To find the y-intercepts, let and solve for y.
■ You should be able to test for symmetry.
(a) To test for x-axis symmetry, replace y with
(b) To test for y-axis symmetry, replace x with
(c) To test for origin symmetry, replace x with and y with
■ You should know the standard equation of a circle with center and radius r:
�x � h�2 � �y � k�2 � r 2
�h, k�
�y.�x
�x.
�y.
x � 0
y � 0
1.
(a)
Yes, the point is on the graph.
(b)
Yes, the point is on the graph.
3 � �9
�5, 3�: 3 �? �5 � 4
2 � 2
�0, 2�: 2 �? �0 � 4
y � �x � 4 3.
(a)
No, the point is not on the graph.
(b)
Yes, the point is on the graph.
0 � 4 � 4
�6, 0�: 0 �? 4 � �6 � 2�
5 � 4 � 1
�1, 5�: 5 �? 4 � �1 � 2�
y � 4 � �x � 2�
Vocabulary Check
1. solution or solution point 2. graph 3. intercepts
4. axis 5. circle; 6. point-plotting�h, k�; ry-
5.
x541−1−2−3
4
7
2
1
−1
2
3
5
yy � �2x � 5
x 0 1 2
y 7 5 3 1 0
�52, 0��2, 1��1, 3��0, 5���1, 7��x, y�
52�1
40 Chapter 1 Equations, Inequalities, and Mathematical Modeling
7.
1 2
3
4
5
4 5x
−2 −1−1
−2
yy � x2 � 3x
x 0 1 2 3
y 4 0 0
�3, 0��2, �2��1, �2��0, 0���1, 4��x, y�
�2�2
�1
9.
x-intercepts:
y-intercept:
�0, 16�
y � 16 � 4�0�2 � 16
��2, 0�, �2, 0�
x � ±2
x2 � 4
4x2 � 16
0 � 16 � 4x2
y � 16 � 4x2 11.
x-intercepts:
or
y-intercept:
�0, 0�
y � 0
y � 2�0�3 � 4�0�2
�0, 0�, �2, 0�
x � 2 x � 0
0 � 2x2�x � 2�
0 � 2x3 � 4x2
y � 2x3 � 4x2
13. y-axis symmetry
–4 –3 –1 1 3 4
1
2
−2
3
4
x
y
15. Origin symmetry
–4 –3 –2 1 2 3 4
–4
–3
–2
1
2
3
4
x
y
17.
y-axis symmetry
No x-axis symmetry
No origin symmetry ��x�2 � ��y� � 0 ⇒ x2 � y � 0 ⇒
x2 � ��y� � 0 ⇒ x2 � y � 0 ⇒
��x�2 � y � 0 ⇒ x2 � y � 0 ⇒
x2 � y � 0 19.
No y-axis symmetry
No x-axis symmetry
Origin symmetry�y � ��x�3 ⇒ �y � �x3 ⇒ y � x3 ⇒
�y � x3 ⇒ y � �x3 ⇒
y � ��x�3 ⇒ y � �x3 ⇒
y � x3
21.
No y-axis symmetry
No x-axis symmetry
Origin symmetry�y ��x
��x�2 � 1 ⇒ �y �
�xx2 � 1
⇒ y �x
x2 � 1 ⇒
�y �x
x2 � 1 ⇒ y �
�xx2 � 1
⇒
y ��x
��x�2 � 1 ⇒ y �
�xx2 � 1
⇒
y �x
x2 � 1
23.
No y-axis symmetry
x-axis symmetry
No origin symmetry��x���y�2 � 10 � 0 ⇒ �xy2 � 10 � 0 ⇒
x��y�2 � 10 � 0 ⇒ xy2 � 10 � 0 ⇒
��x�y2 � 10 � 0 ⇒ �xy2 � 10 � 0 ⇒
xy2 � 10 � 0
Section 1.1 Graphs of Equations 41
25.
x-intercept:
y-intercept:
No axis or origin symmetry
�0, 1�
�13, 0�
( (x
(0, 1) 13
y
−1−2−3−4 1 2 3 4−1
−2
−3
1
4
5
, 0
y � �3x � 1 27.
Intercepts:
No axis or origin symmetry
�0, 0�, �2, 0�
−1
2
3
4
−1 1 2 3 4x
−2
−2
(0, 0) (2, 0)
yy � x2 � 2x
x 0 1 2 3
y 3 0 0 3�1
�1
29.
Intercepts:
No axis or origin symmetry
−4 −3 −2−1
1
2
4
5
6
7
1 2 3 4x
(0, 3)
−3, 03( (
y�0, 3�, � 3��3, 0�y � x3 � 3
x 0 1 2
y 2 3 4 11�5
�1�2
31.
Domain:
Intercept:
No axis or originsymmetry
�3, 0��3, ��
1 2 3 4 5 6–1
1
2
3
4
5
x(3, 0)
yy � �x � 3
x 3 4 7 12
y 0 1 2 3
33.
Intercepts:
No axis or origin symmetry
�0, 6�, �6, 0�
2−2
2
−2
4
6
8
10
12
4 6 8 10 12x
(6, 0)
(0, 6)
yy � �x � 6�
x 0 2 4 6 8 10
y 8 6 4 2 0 2 4
�2
35.
Intercepts:
x-axis symmetry
–2 1 2 3 4
–3
–2
2
3
x
(0, −1)
(−1, 0) (0, 1)
y
�0, �1�, �0, 1�, ��1, 0�
x � y2 � 1
x 0 3
y 0 ±2±1
�1
37.
Intercepts: �6, 0�, �0, 3�
−10
−10
10
10
y � 3 �12x
39.
Intercepts: �3, 0�, �1, 0�, �0, 3�
−10 10
−10
10
y � x2 � 4x � 3 41.
Intercept: �0, 0�
−10
−10
10
10
y �2x
x � 143.
Intercept: �0, 0�
−10
−10
10
10
y � 3�x
42 Chapter 1 Equations, Inequalities, and Mathematical Modeling
45.
Intercepts: �0, 0�, ��6, 0�
−10
−10
10
10
y � x�x � 6 47.
Intercepts: ��3, 0�, �0, 3�
−10
−10
10
10
y � �x � 3� 49. Center: radius: 4
Standard form:
x2 � y2 � 16
�x � 0�2 � �y � 0�2 � 42
�0, 0�;
53. Center: solution point:
Standard form: �x � 1�2 � �y � 2�2 � 5
�0 � 1�2 � �0 � 2�2 � r 2 ⇒ 5 � r2
�x � ��1��2 � �y � 2�2 � r2
�0, 0���1, 2�;51. Center: radius: 4
Standard form:
�x � 2�2 � �y � 1�2 � 16
�x � 2�2 � �y � ��1��2 � 42
�2, �1�;
55. Endpoints of a diameter:
Center:
Standard form: �x � 3�2 � �y � 4�2 � 25
�0 � 3�2 � �0 � 4�2 � r 2 ⇒ 25 � r 2
�x � 3�2 � �y � 4�2 � r2
�0 � 62
, 0 � 8
2 � � �3, 4�
�0, 0�, �6, 8� 57.
Center: radius: 5�0, 0�,
1−1−2
−2−3−4
−6
−3−4
1234
6
2 3 4 6x
y
(0, 0)
x2 � y2 � 25
59.
Center: radius: 3
−3 −2 1
1
2 4 5x
−1
−2
−3
−4
−5
−6
−7
y
(1, −3)
�1, �3�,
�x � 1�2 � �y � 3�2 � 9 61.
Center: radius:
–1 1 2 3
1
3
x
y
( )12
, 12
32�1
2, 12�,�x �
12�2
� �y �12�2
�94 63.
250,000
200,000
150,000
100,000
50,000
1 2 3 4 5 6 7 8t
Year
Dep
reci
ated
val
ue
y
y � 225,000 � 20,000t, 0 ≤ t ≤ 8
65. (a)
(c)
0 1800
8000
y
x
(b)
(d) When yards, the area is a maximum of square yards.
(e) A regulation NFL playing field is 120 yards long and yards wide. The actual area is 6400 square yards.531
3
751119x � y � 862
3
A � xy � x�5203 � x�
y �5203 � x
2y �1040
3 � 2x
2x � 2y �1040
3
Section 1.2 Linear Equations in One Variable 43
71. The viewing window is incorrect. Change the viewingwindow. Examples will vary. For example,will not appear in the standard window setting.
y � x2 � 20
73.
Terms: 9x5, 4x3, �7
9x5 � 4x3 � 7 75. �18x � �2x � 3�2x � �2x � 2�2x
77. 6�t2 � t2�6 � �t�1�3 � 3��t�
Section 1.2 Linear Equations in One Variable
■ You should know how to solve linear equations.
■ An identity is an equation whose solution consists of every real number in its domain.
■ To solve an equation you can:
(a) Add or subtract the same quantity from both sides.
(b) Multiply or divide both sides by the same nonzero quantity.
(c) Remove all symbols of grouping and all fractions.
(d) Combine like terms.
(e) Interchange the two sides.
■ Check the answer!
■ A “solution” that does not satisfy the original equation is called an extraneous solution.
■ Be able to find intercepts algebraically.
a � 0ax � b � 0,
Vocabulary Check
1. equation 2. solve 3. identities; conditional
4. 5. extraneousax � b � 0
67.
(a) and (b)
Year (20 ↔ 1920)
Lif
e ex
pect
ancy
20 40 60 80 100
100
80
60
40
20
t
y
y � �0.0025t 2 � 0.574t � 44.25, 20 ≤ t ≤ 100
(c) For the year 1948, let years.
(d) For the year 2005, let years.
For the year 2010, let years.
(e) No. The graph reaches a maximum of yearswhen or during the year 2014. After this time,the model has life expectancy decreasing, which is notrealistic.
t � 114.8,y � 77.2
t � 110: y � 77.1
t � 105: y � 77.0
t � 48: y � 66.0
69. False. A graph is symmetric with respect to the axis if,whenever is on the graph, is also on thegraph.
�x, �y��x, y�x-
44 Chapter 1 Equations, Inequalities, and Mathematical Modeling
1.
(a)
is not a solution.
(c)
is a solution.x � 4
17 � 17
5�4� � 3 �?
3�4� � 5
x � 0
�3 � 5
5�0� � 3 �?
3�0� � 5
5x � 3 � 3x � 5
(b)
is not a solution.
(d)
is not a solution.x � 10
47 � 35
5�10� � 3 �?
3�10� � 5
x � �5
�28 � �10
5��5� � 3 �?
3��5� � 5
3.
(a)
is a solution.
(c)
is not a solution.x � 4
51 � 30
48 � 8 � 5 �?
32 � 2
3�4�2 � 2�4� � 5 �?
2�4�2 � 2
x � �3
16 � 16
27 � 6 � 5 �?
18 � 2
3��3�2 � 2��3� � 5 �?
2��3�2 � 2
3x2 � 2x � 5 � 2x2 � 2 (b)
is a solution.
(d)
is not a solution.x � �5
60 � 48
75 � 10 � 5 �?
50 � 2
3��5�2 � 2��5� � 5 �?
2��5�2 � 2
x � 1
0 � 0
3 � 2 � 5 �?
2 � 2
3�1�2 � 2�1� � 5 �?
2�1�2 � 2
5.
(a)
is a solution.x � �12
3 � 3
�5 � 8 �?
3
52��1�2� �
4��1�2� �
?3
52x
�4x
� 3
(c) is undefined.
is not a solution.x � 0
52�0� �
40
(b)
is not a solution.
(d)
is not a solution.x �14
�6 � 3
10 � 16 �?
3
52�1�4� �
41�4
�?
3
x � 4
�38
� 3
58
� 1 �?
3
52�4� �
44
�?
3
7.
(a)
is not a solution.x � 3
�7 � 4
�3�3� � 2 �?
4
�3x � 2 � 4
(b)
is not a solution.x � 2
�4 � 4
�3�2� � 2 �?
4 (c)
is not a solution.x � 9
�25 � 4
�3�9� � 2 �?
4 (d)
is not a solution.x � �6
��20 � 4
�3��6� � 2 �?
4
Section 1.2 Linear Equations in One Variable 45
11. is an identity by the DistributiveProperty. It is true for all real values of x.2�x � 1� � 2x � 2 13. is conditional. There are
real values of x for which the equation is not true.�6�x � 3� � 5 � �2x � 10
15.
This is an identity by simplification. It is true for all realvalues of x.
4�x � 1� � 2x � 4x � 4 � 2x � 2x � 4 � 2�x � 2� 17.
Thus, is an identity bysimplification. It is true for all real values of x.
x2 � 8x � 5 � �x � 4�2 � 11
�x � 4�2 � 11 � x2 � 8x � 16 � 11 � x2 � 8x � 5
19. is conditional. There are real values
of x for which the equation is not true.
3 �1
x � 1�
4xx � 1
21. Original equation
Subtract 32 from both sides.
Simplify.
Divide both sides by 4.
Simplify. x �514
4x4
�514
4x � 51
4x � 32 � 32 � 83 � 32
4x � 32 � 83
9.
(a)
is a solution.
(c)
is a solution.x �72
0 � 0
1472 �
772 �
702 �
?0
6�72�2
� 11�72� � 35 �
?0
x � �53
0 � 0
503 �
553 �
1053 �
?0
6��53�2
� 11��53� � 35 �
?0
6x2 � 11x � 35 � 0
(b)
is not a solution.
(d)
is not a solution.x �53
�1103 � 0
503 �
553 �
1053 �
?0
6�53�2
� 11�53� � 35 �
?0
x � �27
�153749 � 0
2449 �
15449 �
171549 �
?0
6��27�2
� 11��27� � 35 �
?0
23.
x � 4
x � 11 � 11 � 15 � 11
x � 11 � 15 25.
x � �9
�2x�2
�18�2
�2x � 18
7 � 7 � 2x � 25 � 7
7 � 2x � 25 27.
x � 5
5x5
�255
5x � 25
5x � 5 � 5 � 20 � 5
5x � 5 � 20
8x � 3x � 5 � 3x � 3x � 20
8x � 5 � 3x � 20
29.
x � 9
�x � �9
�x � 3 � 3 � �6 � 3
�x � 3 � �6
2x � 3x � 3 � 3x � 3x � 6
2x � 3 � 3x � 6
2x � 10 � 7 � 3x � 6
2�x � 5� � 7 � 3�x � 2� 31.
No solution
�9 � 8
�5x � 5x � 9 � 8 � 5x � 5x
�5x � 9 � 8 � 5x
x � 6x � 9 � 8 � 5x
x � 3�2x � 3� � 8 � 5x 33.
x � �4
5x � 2 � 4x � 2
4�5x4 � � 4�1
2� � 4�x� � 4�12�
4�5x4
�12� � 4�x �
12�
5x4
�12
� x �12
46 Chapter 1 Equations, Inequalities, and Mathematical Modeling
37.
x � 9
�0.50x � �4.5
�0.50x � 7.5 � 3
0.25x � 7.5 � 0.75x � 3
0.25x � 0.75�10 � x� � 3
39. or
The second way is easier since you are not working withfractions until the end of the solution.
x �73 x �
73
3x � 7 x �43 � 1
3x � 3 � 4 x � 1 �43
3�x � 1� � 4 3�x � 1� � 4 41. or
The first way is easier. The fraction is eliminated in thefirst step.
x � 13
x � 3�133 �
13x �133 x � 13
13x � 5 �23 x � 2 � 15
13x �23 � 53�1
3��x � 2� � 3�5�
13�x � 2� � 5 13�x � 2� � 5
43.
Contradiction; no solution
8 � �4
x � 8 � x � 4
x � 8 � 2x � 4 � x
x � 8 � 2�x � 2� � x 45.
x � 10
�31x � �310
400 � 16x � 15x � 18 � 72
4�100 � 4x� � 3�5x � 6� � 72
12�100 � 4x3 � � 12�5x � 6
4 � � 12�6�
100 � 4x
3�
5x � 64
� 6
35.
z � �65
5z � �6
6z � 30 � z � 24 � 0
6�z � 5� � �z � 24� � 0
4�32��z � 5� � 4�1
4��z � 24� � 4�0�
432�z � 5� �
14�z � 24� � 4�0�
32�z � 5� �14�z � 24� � 0
47.
x � 4
5x � 20
15x � 12 � 10x � 8
3�5x � 4� � 2�5x � 4�
5x � 45x � 4
�23
49.
x � 3
6x � 18
10x � 13 � 4x � 5
10x � 13
x�
4x � 5x
10 �13x
� 4 �5x
51.
z � 0
3z � 6 � 2z � 4 � 2
3�z � 2� � �2 �2
z � 2��z � 2�
3 � 2 �2
z � 253.
Contradiction; no solution
3 � 0
1 � 2 � 0
x � 4x � 4
� 2 � 0
x
x � 4�
4x � 4
� 2 � 0
Section 1.2 Linear Equations in One Variable 47
57.
Multiply both sides by
x � 5
2x � 10
1�x � 3� � 1�x � 3� � 10
�x � 3��x � 3�. 1
x � 3�
1x � 3
�10
�x � 3��x � 3�
1
x � 3�
1x � 3
�10
x2 � 9
59.
Multiply both sides by
A check reveals that is an extraneous solution since it makes the denominator zero, so there is no solution.x � 3
x � 3
3x � 9
3 � 4x � 12 � x
3 � 4�x � 3� � x
x�x � 3�.3x�x � 3� �
4x
�1
x � 3
3
x2 � 3x�
4x
�1
x � 3
55. Multiply both sides by
A check reveals that is an extraneous solution—it makes the denominator zero. There is no real solution.x � 4
4 � x
12 � 3x
2 � 3x � 10
2 � x � 2 � 2x � 8
2 � 1�x � 2� � 2�x � 4�
�x � 4��x � 2�. 2
�x � 4��x � 2� �1
x � 4�
2x � 2
61.
x � 0
�2x � 0
4x � 9 � 6x � 9
x2 � 4x � 4 � 5 � x2 � 6x � 9
�x � 2�2 � 5 � �x � 3�2 63.
The equation is an identity; every real number is a solution.
4 � 4
x2 � 4x � 4 � x2 � 4x � 4
�x � 2�2 � x2 � 4�x � 1�
65.
The x-intercept is at 3. The solution to and the x-intercept of are the same.They are both The x-intercept is �3, 0�.x � 3.
y � 2�x � 1� � 40 � 2�x � 1� � 4
x � 3
3 � x
6 � 2x
0 � 2x � 6
0 � 2x � 2 � 4
−6 12
−8
4
0 � 2�x � 1� � 4y � 2�x � 1� � 4 67.
The x-intercept is at 10. The solution toand the x-intercept of are the same. They are both The x-intercept is �10, 0�.x � 10.
y � 20 � �3x � 10�0 � 20 � �3x � 10�
x � 10
3x � 30
0 � 30 � 3x
0 � 20 � 3x � 10
−20 40
35
−5
0 � 20 � �3x � 10�y � 20 � �3x � 10�
48 Chapter 1 Equations, Inequalities, and Mathematical Modeling
73.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, �3�.��12, 0�
y � �3�2�0� � 1� ⇒ y � �3
0 � �3�2x � 1� ⇒ 0 � 2x � 1 ⇒ x � �12
y � �3�2x � 1� 75.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, 103 �.�5, 0�
2�0� � 3y � 10 ⇒ 3y � 10 ⇒ y �103
2x � 3�0� � 10 ⇒ 2x � 10 ⇒ x � 5
2x � 3y � 10
77. Multiply both sides by 5.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, 83�.��20, 0�
2�0� � 40 � 15y � 0 ⇒ 40 � 15y � 0 ⇒ y �4015
�83
2x � 40 � 15�0� � 0 ⇒ 2x � 40 � 0 ⇒ x � �20
2x5
� 8 � 3y � 0 ⇒ 2x � 40 � 15y � 0
79.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, �0.3�.�1.6, 0�
4y � 0.75�0� � 1.2 � 0 ⇒ 4y � 1.2 � 0 ⇒ y ��1.2
4� �0.3
4�0� � 0.75x � 1.2 � 0 ⇒ � 0.75x � 1.2 � 0 ⇒ x �1.2
0.75� 1.6
4y � 0.75x � 1.2 � 0
71.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, 12�.�125 , 0�
y � 12 � 5�0� ⇒ y � 12
0 � 12 � 5x ⇒ 5x � 12 ⇒ x �125
y � 12 � 5x69.
The x-intercept is at The solution to and the x-intercept of are the same.They are both The x-intercept is �7
5, 0�.x �75.
y � �38 � 5�9 � x�0 � �38 � 5�9 � x�7
5.
x �75
5x � 7
0 � 7 � 5x
0 � �38 � 45 � 5x
−6 10
−2
10
0 � �38 � 5�9 � x�y � �38 � 5�9 � x�
81.
x �1
3 � a, a � 3
x�3 � a� � 1
3x � ax � 1
4x � 4 � ax � x � 5
4�x � 1� � ax � x � 5 83.
x �5
4 � a, a � �4
x�4 � a� � 5
4x � ax � 5
6x � ax � 2x � 5
85.
Multiply both sides by 2.
x �18
36 � a, a � �36
x�36 � a� � 18
36x � ax � 18
18x �12
ax � 9
19x �12
ax � x � 9 87.
x ��17
�2a � 10�
�1710 � 2a
, a � 5
x��2a � 10� � �17
�2ax � 10x � �17
�2ax � 10x � 18 � 1
�2ax � 6x � 18 � �4x � 1
�2ax � 6�x � 3� � �4x � 1
Section 1.2 Linear Equations in One Variable 49
91. Multiply both sides by
x ��5.405��7.398�
2� 19.993
2x � �5.405��7.398�
2x � �4.405��7.398� � 7.398
2x � �4.405��7.398� � 7.398
7.398x. 2
7.398�
4.405x
�1x
93.
h � 10 feet
h �471 � 50�
10��
471 � 50(3.14)10(3.14)
� 10
471 � 50� � 10�h
471 � 50� � 10�h
471 � 2��25� � 2��5h�
95. (a) Female:
inches
(c)
The lengths of the male and female femurs areapproximately equal when the lengths are 100 inches.
x � 61.2
26.440.432
� x
26.44 � 0.432x
For y � 16: 16 � 0.432x � 10.44
y � 0.432x � 10.44 (b) Male:
Yes, it is likely that both bones came from the sameperson because the estimated height of a male with a 19-inch thigh bone is 69.4 inches.
(d)
inches
It is unlikely that a female would be over 8 feet tall,so if a femur of this length was found, it most likelybelonged to a very tall male.
x � 100.59
1.71 � 0.017x
0.432x � 10.44 � 0.449x � 12.15
69.4 � x
31.15 � 0.449x
For y � 19: 19 � 0.449x � 12.15
y � 0.449x � 12.15
Height x Female Femur Male FemurLength Length
60 15.48 14.79
70 19.80 19.28
80 24.12 23.77
90 28.44 28.26
100 32.76 32.75
110 37.08 37.24
89.
� 138.889
x �62.50.45
�0.45x � �62.5
0.275x � 362.5 � 0.725x � 300
0.275x � 0.725�500 � x� � 300
97.
(a)
The intercept is �0, 36.8�.y-
y
t20 4 6 8 10 12
32
40
44
48
52
56
60
Year (0 ↔ 1990)
Con
sum
ptio
n of
gas
(in
billi
ons
of g
allo
ns)
y � 1.64t � 36.8, �1 ≤ t ≤ 12 (b) Let
intercept:
(c)
This corresponds with the year 2017. Explanations will vary.
t �28.21.64
� 17.2
28.2 � 1.64t
65 � 1.64t � 36.8
�0, 36.8�y-
y � 1.64�0� � 36.8 � 36.8t � 0:
50 Chapter 1 Equations, Inequalities, and Mathematical Modeling
99.
m � 23,437.5 miles
75000.32
� m
7500 � 0.32m
10,000 � 0.32m � 2500 101. False.
This is a quadratic equation. The equation cannot be written in the form ax � b � 0.
x�3 � x� � 10 ⇒ 3x � x2 � 10
103. Equivalent equations are derived from the substitution principle and simplification techniques.They have the same solution(s).
and are equivalent equations.2x � 52x � 3 � 8
105. (a)
(b) Since the sign changes from negative at 1 to positive at 2, the root is somewhere between 1 and 2.
(c)
(d) Since the sign changes from negative at to positive at the root is somewhere between and
To improve accuracy, evaluate the expression at subintervals within this interval and determine where the sign changes.
1.8 < x < 1.9
1.9.1.81.9,1.8
1 < x < 2
x 0 1 2 3 4
73.80.6�2.6�5.8�93.2x � 5.8
�1
x 2
0.60.28�0.04�0.36�0.68�13.2x � 5.8
1.91.81.71.61.5
107.x2 � 5x � 362x2 � 17x � 9
��x � 9��x � 4��2x � 1��x � 9� �
x � 42x � 1
, x � �9
109.
Intercepts: �0, �5�, �53, 0�
x−3 −2 −1 1 2 3 4 5
−1
1
2
−2
−3
−4
−5
y
y � 3x � 5 111.
Intercepts: �0, 0�, ��5, 0�
x1
1
4
5
6
7
2−7 −6 −4 −3 −2 −1−1
−2
y
y � �x2 � 5x � �x�x � 5�
Section 1.3 Modeling with Linear Equations 51
Section 1.3 Modeling with Linear Equations
■ You should be able to set up mathematical models to solve problems.
■ You should be able to translate key words and phrases.
(a) Equality: (b) Addition:
Equals, equal to, is, are, was, will be, represents Sum, plus, greater, increased by, more than,exceeds, total of
(c) Subtraction: (d) Multiplication:
Difference, minus, less than, decreased by, Product, multiplied by, twice, times, percent ofsubtracted from, reduced by, the remainder
(e) Division: (f) Consecutive:
Quotient, divided by, ratio, per Next, subsequent
■ You should know the following formulas:
(a) Perimeter: (b) Area:
1. Square: 1. Square:
2. Rectangle: 2. Rectangle:
3. Circle: 3. Circle:
4. Triangle: 4. Triangle:
(c) Volume
1. Cube:
2. Rectangular solid:
3. Cylinder:
4. Sphere:
(d) Simple Interest: (e) Compound Interest:
(f) Distance: (g) Temperature:
■ You should be able to solve word problems. Study the examples in the text carefully.
F �9
5 C � 32d � rt
A � P�1 �r
n�nt
I � Prt
V � �4
3��r 3
V � �r 2 h
V � lwh
V � s3
A � �1
2�bhP � a � b � c
A � �r2C � 2�r
A � lwP � 2l � 2w
A � s2P � 4s
1.
The sum of a number and 4 A number increased by 4
x � 4 3.
The ratio of a number and 5The quotient of a number and 5A number divided by 5
u
55.
The difference of a number and 4 is divided by 5. A number decreased by 4 is divided by 5.
y � 4
5
Vocabulary Check
1. mathematical modeling 2. verbal model; 3. 4.algebraic equation
5. 6. 7. 8. I � PrtA � P�1 �r
12�12t
V � �r2hV � s3
P � 2l � 2wA � �r2
52 Chapter 1 Equations, Inequalities, and Mathematical Modeling
7.
The product of and the sum of a number and 2.Negative 3 is multiplied by a number increased by 2.
�3
�3�b � 2� 9.
The difference of a number and 5 is multiplied by 12times the number. 12 is multiplied by a number and thatproduct is multiplied by the number decreased by 5.
12x�x � 5�
11. Verbal Model: (Sum) (first number) (second number)
Labels: Sum first number second number
Expression: S � n � �n � 1� � 2n � 1
� n � 1� n,� S,
��
13. Verbal Model: Product (first odd integer)(second odd integer)
Labels: Product first odd integer second odd integer
Expression: P � �2n � 1��2n � 1� � 4n2 � 1
� 2n � 1 � 2 � 2n � 1� 2n � 1,� P,
�
15. Verbal Model: (Distance) (rate) (time)
Labels: Distance rate time
Expression: d � 50t
� t� 50 mph,� d,
��
17. Verbal Model: (Amount of acid) (amount of solution)
Labels: Amount of acid (in gallons) amount of solution (in gallons)
Expression: A � 0.20x
� x� A,
� 20% �
19. Verbal Model:
Labels: Perimeter widthlength (width)
Expression: P � 2x � 2�2x� � 6x
� 2x� 2� x,� P,
Perimeter � 2�width� � 2�length�
21. Verbal Model: (Total cost) (unit cost)(number of units) (fixed cost)
Labels: Total cost fixed cost unit cost number of units
Expression: C � 25x � 1200
� x� $25,� $1200,� C,
��
23. Verbal Model: Thirty percent of the list price L
Expression: 0.30L
25. Verbal Model: percent of 500 that is represented by thenumber N
Equation: , p is in decimal formN � p�500�
27.
A � 4x � 8x � 12x
Area � Area of top rectangle � Area of bottom rectangle
4
42x
x
x
8
Section 1.3 Modeling with Linear Equations 53
31. Verbal Model: Difference (one number) (another number)
Labels: Difference one number another number
Equation:
Answer: The two numbers are 37 and 185.
5x � 185
x � 37
148 � 4x
148 � 5x � x
� x� 5x,� 148,
��
33. Verbal Model: Product (smaller number) (larger number)
Labels: Smaller number larger number
Equation:
Answer: Smaller number larger number � n � 1 � �4� n � �5,
n � �5
n2 � n � n2 � 5
n�n � 1� � n2 � 5
� n � 1� n,
� �smaller number)2 � 5��
29. Verbal Model: Sum (first number) (second number)
Labels: Sum first number second number
Equations:
Answer: First number second number � n � 1 � 263� n � 262,
n � 262
524 � 2n
525 � 2n � 1
525 � n � �n � 1�
� n � 1� n,� 525,
��
35.
� 13.5
� 0.30�45�
� �30%��45�
x � percent � number 37.
p � 0.27 � 27%
4321600 � p
432 � p�1600�
432 � percent � 1600 39.
x � 2400
12
0.005� x
12 � 0.005x
12 �12
% � number
41. Verbal Model: Loan payments
Labels: Loan payments
Annual income
Equation:
The family’s annual income is $22,316.98.
I � 22,316.98
13,077.75 � 0.586I
� I
� 13,077.75
� 58.6% � Annual income
54 Chapter 1 Equations, Inequalities, and Mathematical Modeling
45. Verbal Model:
Labels: 2002 price for a pound of tomatoes: $1.66
1990 price for a pound of tomatoes: $0.86
Percentage increase:
Equation:
Answer: Percentage increase � 93%
0.930 � p
0.80 � 0.86p
1.66 � 0.86p � 0.86
p
� �1990 price for a pound of tomatoes�
� �percentage increase��1990 price for a pound of tomatoes� �2002 price for a pound of tomatoes�
43. Verbal Model:
Labels: 2002 price for a gallon of unleaded gasoline: $1.36
1990 price for a gallon of unleaded gasoline: $1.16
Percentage increase:
Equation:
Answer: Percentage increase � 17.2%
0.172 � p
0.20 � 1.16p
1.36 � 1.16p � 1.16
p
� �1990 price for a gallon of unleaded gasoline�
� �percentage increase��1990 price for a gallon of unleaded gasoline� �2002 price for a gallon of unleaded gasoline�
47. Verbal Model: (Sale price) (list price) (discount)
Labels: Sale price list price discount
Equation:
Answer: The list price of the pool is $1450.
1450 � L
1210.75 � 0.835L
1210.75 � L � 0.165L
� 0.165L� L,� $1210.75,
��
49. (a)
w
l
(b)
� 5w
� 2�1.5w� � 2w
P � 2l � 2w
l � 1.5w (c)
Width:
Length:
Dimensions: 7.5 meters � 5 meters
l � 1.5w � 7.5 meters
w � 5 meters
5 � w
25 � 5w
51. Verbal Model: Average
Labels: Average
Equation:
Answer: You must score 97 (or better) on test #4 to earn an A for the course.
x � 97
360 � 263 � x
90 �87 � 92 � 84 � x
4
test #4 � xtest #3 � 84,test #2 � 92,test #1 � 87,� 90,
��test #1� � �test #2� � �test #3� � �test #4�
4
Section 1.3 Modeling with Linear Equations 55
55.
Multiply both sides by
The salesman averaged 58 miles per hour for 3 hours and 52 miles per hour for 2 hours and 45 minutes.
t2 �317 � 174
52� 2.75 hours
t1 �17458
� 3 hours
x � 174 miles
�6x � �1044
52x � 18,386 � 58x � 17,342
52x � 58�317 � x� � 5.75�58��52�
58�52�. x
58�
317 � x52
� 5.75
d1
r1�
d2
r2� T
t1 � t2 � T
�time on first part� � �time on second part) � �Total time�
57. (a) Time for the first family:
Time for the other family: hours
(b)
(c) d � r t � 42�16042
�16050 � � 25.6 miles
t �dr
�100
42 � 50�
10092
� 1.08 hours
t2 �dr2
�16050
� 3.2
t1 �dr1
�16042
� 3.81 hours
53.
Total time �total distance
rate�
300 kilometers100 kilometers�hour
� 3 hours
Rate �distance
time�
50 kilometers12 hour
� 100 kilometers�hour
59. Verbal Model:
Labels: Let wind speed, then the rate to thecity the rate from the city the distance to thecity kilometers, the distance traveled so far in the returntrip kilometers.
Equation:
Wind speed: kilometers per hour6623
6623
� x
180,000 � 2700x
900,000 � 1500x � 720,000 � 1200x
1500�600 � x� � 1200�600 � x�
1500600 � x
�1200
600 � x
� 1500 � 300 � 1200
� 1500� 600 � x,
� 600 � x,x �
time �distance
rate61. Verbal Model:
Equation:
The radio wave travels from Mission Control to the moonin 1.28 seconds.
t � 1.28 seconds
t �3.84 � 108 meters
3.0 � 108 meters per second
time �distance
rate
56 Chapter 1 Equations, Inequalities, and Mathematical Modeling
63. Verbal Model:
Label: Let h height of the building in feet.
Equation:
The Chrysler building is 1044 feet tall.
h � 1044 feet
13
h � 348
h
87�
41�3
h feet
87 feet�
4 feet1�3 foot
�
�height of building��length of building's shadow�
��height of stake�
�length of stake's shadow�
65. Verbal Model:
Label: Let length of person’s shadow.
Equation:
x � 4.36 feet
44x � 192
50x � 192 � 6x
50x � 6�32 � x�
50
32 � x�
6x
x �
50 ft
6 ft
32 ft x
�height of silo��length of silo’s shadow�
��height of person�
�height of person’s shadow�
67. Verbal Model: interest in (total interest)
Labels: Let x amount in the 3% fund. Then amount in the fund.
Equation:
25,000 � x � $16,666.67 at 412%
x � $8333.33 at 3%
�125 � �0.015x
1000 � 0.03x � 1125 � 0.045x
1000 � 0.03x � 0.045�25,000 � x�41
2%25,000 � x ��
�412% fund��Interest in 3% fund� � �
69. Verbal Model: (profit on minivans) (profit on SUVs) (total profit)
Labels: Let x amount invested in minivans. Then, amount invested in SUVs.
Equation:
$450,000 is invested in minivans and is invested in SUVs.600,000 � x � $150,000
x � 450,000
�0.04x � �18,000
0.24x � 168,000 � 0.28x � 150,000
0.24x � 0.28�600,000 � x� � 0.25�600,000�
600,000 � x ��
��
71. Verbal Model:
Label: Let x amount of 100% concentrate
Equation:
Approximately 32.1 gallons of the 100% concentrate will be needed.
x � 32.1 gallons
41.25 � 0.60x � 22
41.25 � 22 � 0.40x � 1.00x
0.75�55� � 0.40�55 � x� � 1.00x
�
�Final concentration��Amount� � �Solution 1 concentration��Amount� � �Solution 2 concentration��Amount�
Section 1.3 Modeling with Linear Equations 57
73. Verbal Model: (price per pound of peanuts)(number of pounds of peanuts) (price per pound of walnuts)(number ofpounds of walnuts) (price per pound of nut mixture)(number of pounds of nut mixture)
Labels: Let x number of pounds of $2.49 peanuts. Then number of pounds of $3.89 walnuts.
Equation:
Use 50 pounds of peanuts and 50 pounds of walnuts.
100 � x � 50 pounds of $3.89 walnuts
x � 50 pounds of $2.49 peanuts
x ��70
�1.40
�1.40x � �70
2.49x � 389 � 3.89x � 319
2.49x � 3.89�100 � x� � 3.19�100�
100 � x ��
��
75. Verbal Model:
Labels:
Equation:
At most the company can manufacture 8064 units.
x �75,000
9.3� 8064.52 units
$85,000 � $10,000 � $9.30x
Total cost � $85,000, variable costs � $9.30x, fixed costs � $10,000
Total cost � �Fixed cost� � �Variable cost per unit� � �Number of units�
77.
2Ab
� h
2A � bh
A �12
bh 79.
S
1 � R� C
S � C�1 � R�
S � C � RC 81.
3V
4�a2 � b
�3�4�V
�a2 � b
34
V � �a2b
V �43
�a2b 83.
2�h � v0t�
t2� a
2�h � v0t� � at2
h � v0t �1
2at2
h � v0t �1
2at2
85.
CC2
C2 � C� C1
C2 � C
CC2�
1C1
1C
�1
C2�
1C1
1C
�1
C1�
1C2
C �1
1C1
�1
C2
89.
x � 6 feet from 50-pound child.
125x � 750
50x � 750 � 75x
50x � 75�10 � x�
W1x � W2�L � x�87.
L � a � d
d� n
L � a � d � nd
L � a � nd � d
L � a � �n � 1�d
58 Chapter 1 Equations, Inequalities, and Mathematical Modeling
Section 1.4 Quadratic Equations and Applications
■ You should be able to solve a quadratic equation by factoring, if possible.
■ You should be able to solve a quadratic equation of the form by extracting square roots.
■ You should be able to solve a quadratic equation by completing the square.
■ You should know and be able to use the Quadratic Formula: For
■ You should be able to determine the types of solutions of a quadratic equation by checking the discriminant
(a) If there are two distinct real solutions. The graph has two x-intercepts.
(b) If there is one repeated real solution. The graph has one x-intercept.
(c) If there is no real solution. The graph has no x-intercepts.
■ You should be able to use your calculator to solve quadratic equations.
■ You should be able to solve applications involving quadratic equations. Study the examples in the text carefully.
b2 � 4ac < 0,
b2 � 4ac � 0,
b2 � 4ac > 0,
b2 � 4ac.
x ��b ± �b2 � 4ac
2a.
ax2 � bx � c � 0, a � 0,
u2 � d
93.
When
59�64.4 � 32� � 18�C.
F � 64.4�,
C �59�F � 32� 95.
When
95�50� � 32 � 122�F.
C � 50�,
F �95C � 32
97. False, it should be written as
z3 � 8z2 � 9
.
99.
(a) If then a and b have the same sign andis negative.
(b) If then a and b have opposite signs andis positive.x � �b�a
ab < 0,
x � �b�aab > 0,
ax � b � 0 ⇒ x � �ba
101. �5x4��25x2��1 �5x4
25x2�
x2
5, x � 0
91.
r � 3�4.47�
� 1.12 inches
17.88
4�� r3
17.88 � 4�r3
5.96 �43
�r3
V �43
�r3
103. 3
x � 5�
2
5 � x�
3
x � 5�
��1��2�x � 5
�3 � 2
x � 5�
1
x � 5
105.10
7�3�
107�3
��3�3
�10�3
21 107.
� �11 � �6
� ���6 � �11 �
�5�6 � 5�11
�5
5
�6 � �11�
5
�6 � �11��6 � �11
�6 � �11
Section 1.4 Quadratic Equations and Applications 59
1.
General form: 2x2 � 8x � 3 � 0
2x2 � 3 � 8x 3.
General form: x2 � 6x � 6 � 0
x2 � 6x � 9 � 3
�x � 3�2 � 3 5.
General form: 3x2 � 90x � 10 � 0
3x2 � 10 � 90x
15�3x2 � 10� � 18x
7.
x � 0 or x � �12
3x � 0 or 2x � 1 � 0
3x�2x � 1� � 0
6x2 � 3x � 0 9.
x � 4 or x � �2
x � 4 � 0 or x � 2 � 0
�x � 4��x � 2� � 0
x2 � 2x � 8 � 0 11.
x � �5
x � 5 � 0
�x � 5�2 � 0
x2 � 10x � 25 � 0
13.
x � 3 or x � �12
3 � x � 0 or 1 � 2x � 0
�3 � x��1 � 2x� � 0
3 � 5x � 2x2 � 0 15.
x � �6 or x � 2
x � 6 � 0 or x � 2 � 0
�x � 6��x � 2� � 0
x2 � 4x � 12 � 0
x2 � 4x � 12 17.
x � �203 or x � �4
3x � 20 � 0 or x � 4 � 0
�3x � 20��x � 4� � 0
3x2 � 32x � 80 � 0
4�34x2 � 8x � 20� � 4�0�
34x2 � 8x � 20 � 0
Vocabulary Check
1. quadratic equation 2. factoring, extracting square roots, completing the square,and the Quadratic Formula
3. discriminant 4. position equation; velocity of theobject; initial height of the object
5. Pythagorean Theorem
�16t 2 � v0t � s0;
19.
x � �a
x � a � 0
�x � a�2 � 0
x2 � 2ax � a2 � 0 21.
x � ±7
x2 � 49 23.
x � ±�11
x2 � 11
25.
x � ±3�3
x2 � 27
3x2 � 81 27.
x � 16 or x � 8
x � 12 ± 4
x � 12 � ±4
�x � 12�2 � 16 29.
x � �2 ± �14
x � 2 � ±�14
�x � 2�2 � 14
31.
x �1 ± 3�2
2
2x � 1 ± 3�2
2x � 1 � ±�18
�2x � 1�2 � 18 33.
The only solution to the equation is x � 2.
x � 2
�7 � 3 or 2x � 4
x � 7 � x � 3 or x � 7 � �x � 3
x � 7 � ± �x � 3�
�x � 7�2 � �x � 3�2
60 Chapter 1 Equations, Inequalities, and Mathematical Modeling
39.
x � 1 ±�63
x � 1 ± �23
x � 1 � ±�23
�x � 1�2 �23
x2 � 2x � 12 � �13
� 12
x2 � 2x � �13
9x2 � 18x � �3 41.
x � 2 ± 2�3
x � 2 � ±�12
�x � 2�2 � 12
x2 � 4x � 22 � 8 � 22
x2 � 4x � 8
x2 � 4x � 8 � 0
�x2 � 4x � 8 � 0
8 � 4x � x2 � 0
45.
�1
�x � 1�2 � 4
1
x2 � 2x � 5�
1x2 � 2x � 12 � 12 � 5
43.
x ��5 ± �89
4
x � �54
±�89
4
x �54
� ±�89
4
�x �54�
2
�8916
x2 �52
x � �54�
2
� 4 � �54�
2
x2 �52
x � 4
2x2 � 5x � 8
2x2 � 5x � 8 � 0
35.
x � 4 or x � �8
x � �2 ± 6
x � 2 � ±6
�x � 2�2 � 36
x2 � 4x � 22 � 32 � 22
x2 � 4x � 32
x2 � 4x � 32 � 0 37.
x � �6 ± �11
x � 6 � ±�11
�x � 6�2 � 11
x2 � 12x � 62 � �25 � 62
x2 � 12x � �25
x2 � 12x � 25 � 0
47.
�4
�x � 2�2 � 7
4
x2 � 4x � 3�
4x2 � 4x � 4 � 4 � 3
49.
�1
���x � 3�2 � 9�
1�9 � �x � 3�2
�1
��1��x � 3�2 � 9
1
�6x � x2�
1��1�x2 � 6x � 32 � 32�
51. (a)
(b) The x-intercepts are
(c)
(d) The x-intercepts of the graph are solutions to theequation 0 � �x � 3�2 � 4.
x � �1 or x � �5
�3 ± 2 � x
±�4 � x � 3
4 � �x � 3�2
0 � �x � 3�2 � 4
��1, 0� and ��5, 0�.
−10 5
−6
4
y � �x � 3�2 � 4
Section 1.4 Quadratic Equations and Applications 61
53. (a)
(b) The x-intercepts are �1, 0� and �3, 0�.
−3
−4
6
2
y � 1 � �x � 2�2 (c)
(d) The x-intercepts of the graph are solutions to the equation 0 � 1 � �x � 2�2.
x � 3 or x � 1
x � 2 ± 1
x � 2 � ±1
�x � 2�2 � 1
0 � 1 � �x � 2�2
55. (a)
(b) The x-intercepts are and
(d) The x-intercepts of the graph are solutions to theequation 0 � �4x2 � 4x � 3.
�32, 0�.��1
2, 0�
−5 7
−3
5
y � �4x2 � 4x � 3 (c)
x �32 or x � �
12
x �12 ± 1
x �12 � ±�1
�x �12�2
� 1
x2 � x � �12�2 �
34 � �1
2�2
x2 � x �34
4�x2 � x� � 3
4x2 � 4x � 3
0 � �4x2 � 4x � 3
57. (a)
(b) The x-intercepts are and �1, 0�.��4, 0�
−8 4
−7
1
y � x2 � 3x � 4 (c)
(d) The x-intercepts of the graph are solutions to the equation 0 � x2 � 3x � 4.
x � �4 or x � 1
x � 4 � 0 or x � 1 � 0
0 � �x � 4��x � 1�
0 � x2 � 3x � 4
59.
No real solution
b2 � 4ac � ��5�2 � 4�2��5� � �15 < 0
2x2 � 5x � 5 � 0 61.
Two real solutions
b2 � 4ac � ��1�2 � 4�2���1� � 9 > 0
2x2 � x � 1 � 0
63.
No real solution
b2 � 4ac � ��5�2 � 4�13��25� � �
253 < 0
13x2 � 5x � 25 � 0 65.
Two real solutions
b2 � 4ac � �1.2�2 � 4�0.2���8� � 7.84 > 0
0.2x2 � 1.2x � 8 � 0
67.
��1 ± 3
4�
12
, �1
��1 ± �12 � 4�2���1�
2�2�
x ��b ± �b2 � 4ac
2a
2x2 � x � 1 � 0 69.
��8 ± 16
32�
14
, �34
��8 ± �82 � 4�16���3�
2�16�
x ��b ± �b2 � 4ac
2a
16x2 � 8x � 3 � 0 71.
��2 ± 2�3
�2� 1 ± �3
��2 ± �22 � 4��1��2�
2��1�
x ��b ± �b2 � 4ac
2a
�x2 � 2x � 2 � 0
2 � 2x � x2 � 0
62 Chapter 1 Equations, Inequalities, and Mathematical Modeling
77.
��12 ± 6�7
�18�
23
±�73
��12 ± �122 � 4��9��3�
2��9�
x ��b ± �b2 � 4ac
2a
�9x2 � 12x � 3 � 0
12x � 9x2 � �3
79.
� �43
��24 ± 0
18
��24 ± �242 � 4�9��16�
2�9�
x ��b ± �b2 � 4ac
2a
9x2 � 24x � 16 � 0 81.
��4 ± 8�2
8� �
12
± �2
��4 ± �42 � 4�4���7�
2�4�
x ��b ± �b2 � 4ac
2a
4x2 � 4x � 7 � 0
4x2 � 4x � 7 83.
��28 ± 0
�98�
27
��28 ± �282 � 4��49���4�
2��49�
x ��b ± �b2 � 4ac
2a
�49x2 � 28x � 4 � 0
28x � 49x2 � 4
85.
��8 ± 2�6
�4� 2 ±
�62
��8 ± �82 � 4��2���5�
2��2�
t ��b ± �b2 � 4ac
2a
�2t2 � 8t � 5 � 0
8t � 5 � 2t2 87.
�12 ± 2�11
2� 6 ± �11
����12� ± ���12�2 � 4�1��25�
2�1�
y ��b ± �b2 � 4ac
2a
y2 � 12y � 25 � 0
�y � 5�2 � 2y
73.
��14 ± 2�5
2� �7 ± �5
��14 ± �142 � 4�1��44)
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 14x � 44 � 0 75.
��8 ± 4�5
2� �4 ± 2�5
��8 ± �82 � 4�1���4�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 8x � 4 � 0
89.
� �38
±�265
8
��3 ± �265
8
��3 ± �32 � 4�4���16�
2�4�
x ��b ± �b2 � 4ac
2a
4x2 � 3x � 16 � 0
4x2 � 3x � 16
12
x2 �38
x � 2
91.
x � 0.976, �0.643
x �1.7 ± ���1.7�2 � 4�5.1���3.2�
2�5.1�
5.1x2 � 1.7x � 3.2 � 0 93.
x � �14.071, 1.355
x ����0.852� ± ���0.852�2 � 4��0.067��1.277�
2��0.067�
�0.067x2 � 0.852x � 1.277 � 0
95.
x � 1.687, �0.488
x �506 ± ���506�2 � 4�422���347�
2�422�
422x2 � 506x � 347 � 0 97.
x � �2.200, �0.290
x ��31.55 ± ��31.55�2 � 4�12.67��8.09�
2�12.67�
12.67x2 � 31.55x � 8.09 � 0
Section 1.4 Quadratic Equations and Applications 63
99. Complete the square.
x � 1 ± �2
x � 1 � ± �2
�x � 1�2 � 2
x2 � 2x � 12 � 1 � 12
x2 � 2x � 1
x2 � 2x � 1 � 0 101.
x � 6 or x � �12
x � 3 � 9 or x � 3 � �9
x � 3 � ±9
�x � 3�2 � 81 Extract square roots.
103. Complete the square.
x �12 ± �3
x �12 � ±�12
4
�x �12�2
�124
x2 � x � �12�2
�114 � �1
2�2
x2 � x �114
x2 � x �114 � 0 105. Extract square roots.
x � �12
2x � �1
For x � ��x � 1�:
0 � 1 No solution
For x � ��x � 1�:
x � ± �x � 1�
x2 � �x � 1�2
�x � 1�2 � x2
107. Quadratic Formula
�34
±�97
4
�3 ± �97
4
x ����3� ± ���3�2 � 4�2���11�
2�2�
0 � 2x2 � 3x � 11
3x � 4 � 2x2 � 7 109. (a)
(b)
(c)
Since w must be greater than zero, we have feet and the length is w � 14 � 48 feet.
w � 34
w � �48 or w � 34
�w � 48��w � 34� � 0
w2 � 14w � 1632 � 0
w�w � 14� � 1632
w
w + 14
111.
Since x must be positive, we have inches. The dimensions of the box are 6 inches � 6 inches � 2 inches.x � 6
x � �14 or x � 6
0 � �x � 14��x � 6�
0 � x2 � 8x � 84
84 � x2 � 4x�2�
S � x2 � 4xh
113.
Thus,
—CONTINUED—
a � 1, b � �150, and c � 2500.
4�x2 � 150x � 2500� � 0
4x2 � 600x � 10,000 � 0
20,000 � 600x � 4x2 � 10,000100 ft
200 ft
100 2− x
200 2− x
x
x
�200 � 2x��100 � 2x� �1
2�100��200�
64 Chapter 1 Equations, Inequalities, and Mathematical Modeling
113. —CONTINUED—
not possible since the lot is only 100 feet wide.
The person must go around the lot 19.098 feet
24 inches�
19.098 feet
2 feet� 9.5 times.
x �150 � 111.8034
2� 19.098 feet
x �150 � 111.8034
2� 130.902 feet,
x �150 ± ���150�2 � 4�1��2500�
2�1��
150 ± 111.8034
2
115.
(a)
(b) Model:
Labels: Rate 500 miles per hour
Distance
Equation:
The bomb will travel approximately 6.2 miles horizontally.
d � 500�0.0124� � 6.2 miles
� d
Time �20�53600
� 0.0124 hour
�
�Rate� � �time� � �distance�
� 44.72 seconds
� 20�5 seconds
t � �2000
t2 � 2000
�16t2 � 32,000 � 0
s � �16t2 � 32,000 117.
(a)
(b) When
When
When
During the interval the baseball’s speeddecreased due to gravity.
(c) The ball hits the ground when
By the Quadratic Formula, or Assuming that the ball is not caught and drops to theground, it will be in the air for approximately 9.209seconds.
t � 9.209.t � �0.042
�16t 2 � 146 23t � 6 1
4 � 0
s � 0.
3 ≤ t ≤ 5,
s�5� � 339.58 feett � 5:
s�4� � 336.92 feett � 4:
s�3� � 302.25 feett � 3:
s � �16t 2 � 14623t � 61
4
s0 � 614 feet
v0 � 100 mph ��100��5280�
3600� 1462
3 ft�sec
s � �16t 2 � v0t � s0
119.
(a)
The average admission price reached or surpassed $5.00 in 1999.
(b)
By the Quadratic Formula, or 42.80. Since we choose which corresponds to 1999.
(c) For 2008, let Answers will vary.P�18� � $6.87.t � 18:
t � 8.68 � 97 ≤ t ≤ 13,t � 8.68
�0.0081t 2 � 0.417t � 3.01 � 0
�0.0081t 2 � 0.417t � 1.99 � 5.00
P � �0.0081t 2 � 0.417t � 1.99, 7 ≤ t ≤ 13
7 8 9 10 11 12 13
$4.51 $4.81 $5.09 $5.35 $5.60 $5.83 $6.04P
t
Section 1.4 Quadratic Equations and Applications 65
121. Pythagorean Theorem
Each leg in the right triangle is approximately 3.54 centimeters.
�5�2
2� 3.54 centimeters
�5
�2
x ��25
2
x2 �25
2x
x
5
2x2 � 25
x2 � x2 � 52
123.
Using the positive value for we have one plane moving northbound at miles per hour and one plane moving eastbound at miles per hour.r � 550
r � 50 � 600r,
r ��900 ± �9002 � 4�18���5,931,100�
2�18��
�900 ± 60�118,847
36
18r2 � 900r � 5,931,100 � 0
9�r � 50�2 � 9r2 � 24402
dN2 � dE
2 � 24402
dE � �3 hours��r mph�
2440 miles
3r
3(r + 50)
dN � �3 hours��r � 50 mph�
125.
x � 50,000 units
0 � �x � 50,000�2
0 � x2 � 100,000x � 2,500,000,000
0 � 0.0002x2 � 20x � 500,000
x�20 � 0.0002x� � 500,000 127.
Using the positive value for x, we have
x ��20 � �7150
0.25� 258 units.
x ��20 ± �202 � 4�0.125���13,500�
2�0.125�
0.125x2 � 20x � 13,500 � 0
0.125x2 � 20x � 500 � 14,000
129.
Choosing the positive value for x, we have
x ��0.04 � �7.0416
0.004� 653 units.
��0.04 ± �7.0416
0.004
x ��0.04 ± ��0.04�2 � 4�0.002���880�
2�0.002�
0.002x2 � 0.04x � 880 � 0
800 � 0.04x � 0.002x2 � 1680
66 Chapter 1 Equations, Inequalities, and Mathematical Modeling
131.
(a)
The total money in circulation reached or surpassed $600 billion in 2001.
(b)
By the Quadratic Formula, or Since we choose which corresponds to 2001.
(c) For 2008, let billionAnswers will vary.
M�18� � $991.98t � 18:
t � 11.15 ≤ t ≤ 13,�13.1.t � 11.1
1.835t 2 � 3.58t � 267.0 � 0
1.835t 2 � 3.58t � 333.0 � 600
M � 1.835t 2 � 3.58t � 333.0, 5 ≤ t ≤ 13
5 6 7 8 9 10 11 12 13
(in billions) $396.78 $420.54 $447.98 $479.08 $513.86 $552.30 $594.42 $640.20 $689.66M
t
133.
The other two distances are 354.5 miles and 433.5 miles.
x �1576 ± �15762 � 4�2��307,344�
2�2� � 354.5 or 433.5
0 � 2x2 � 1576x � 307,344
313,600 � x2 � 620,944 � 1576x � x2
5602 � x2 � (788 � x�2
Distance from New Orleans to Austin � 1348 � 560 � x � 788 � x
Distance from Oklahoma City to Austin � x
Distance from Oklahoma City to New Orleans � 560
�Distance from OklahomaCity to New Orleans �
2
� �Distance from OklahomaCity to Austin �
2
� � Distance from New Orleans to Austin�
2
135. False
so, the quadratic equation has no real solutions.
b2 � 4ac � ��1�2 � 4��3���10� < 0,
137. The student should have subtracted 15x from both sidesso that the equation is equal to zero. By factoring out anx, there are two solutions, and x � 6.x � 0
139.
(a) Let
or
or x � �5 x � �103
x � 4 � �1 x � 4 �23
u � �1 u �23
u � 1 � 0 3u � 2 � 0
�3u � 2��u � 1� � 0
3u2 � u � 2 � 0
u � x � 4
3�x � 4�2 � �x � 4� � 2 � 0
(b)
or
(c) The method of part (a) reduces the number of algebraicsteps.
x � �5x � �103
x � 5 � 03x � 10 � 0
�3x � 10��x � 5� � 0
3x2 � 25x � 50 � 0
3x2 � 24x � 48 � x � 4 � 2 � 0
3�x2 � 8x � 16� � �x � 4� � 2 � 0
141.
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
x2 � 3x � 18 � 0
�x � 3��x � 6� � 0
�x � ��3���x � 6� � 0
�3 and 6 143. 8 and 14
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
x2 � 22x � 112 � 0
�x � 8��x � 14� � 0
Section 1.5 Complex Numbers 67
149. by the Additive Inverse Property.7x 4 � ��7x 4� � 0
159. Answers will vary.
151.
� x2 � 3x � 18
�x � 3��x � 6� � x2 � 6x � 3x � 18
153.
� x3 � 3x2 � 2x � 8
� x3 � x2 � 2x � 4x2 � 4x � 8
�x � 4��x2 � x � 2� � x�x2 � x � 2� � 4�x2 � x � 2� 155.
� x2�x � 3��x2 � 3x � 9�
� x2�x3 � 33�
x5 � 27x2 � x2�x3 � 27�
157.
� �x � 5��x2 � 2�
x3 � 5x2 � 2x � 10 � x2�x � 5� � 2�x � 5�
Section 1.5 Complex Numbers
■ Standard form: .
If then is a real number.
If and then is a pure imaginary number.
■ Equality of Complex Numbers: if and only if and
■ Operations on complex numbers
(a) Addition:
(b) Subtraction:
(c) Multiplication:
(d) Division:
■ The complex conjugate of
■ The additive inverse of
■ ��a � �a i for a > 0.
a � bi is �a � bi.
�a � bi��a � bi� � a2 � b2
a � bi is a � bi:
a � bi
c � di�
a � bi
c � di�
c � di
c � di�
ac � bd
c2 � d 2�
bc � ad
c2 � d 2i
�a � bi��c � di� � �ac � bd� � �ad � bc�i
�a � bi� � �c � di� � �a � c� � �b � d�i
�a � bi� � �c � di� � �a � c� � �b � d�i
b � da � ca � bi � c � di
a � bib � 0,a � 0
a � bib � 0,
a � bi
145. and
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
x2 � 2x � 1 � 0
x2 � 2x � 1 � 2 � 0
�x � 1�2 � ��2 �2� 0
��x � 1� � �2� ��x � 1� � �2� � 0
�x � �1 � �2�� �x � �1 � �2 �� � 0
1 � �21 � �2 147. by the Associative Property ofMultiplication.�10x�y � 10�xy�
68 Chapter 1 Equations, Inequalities, and Mathematical Modeling
1.
b � 6
a � �10
a � bi � �10 � 6i 3.
b � 3 � 8 ⇒ b � 5
a � 1 � 5 ⇒ a � 6
�a � 1� � �b � 3�i � 5 � 8i 5. 4 � ��9 � 4 � 3i
7.
� 2 � 3�3 i
2 � ��27 � 2 � �27 i 9. ��75 � �75 i � 5�3 i 11. 8 � 8 � 0i � 8
13.
� �1 � 6i
�6i � i2 � �6i � 1 15.
� 0.3i
��0.09 � �0.09 i 17. �5 � i� � �6 � 2i� � 11� i
19.
� 4
�8 � i� � �4 � i� � 8 � i � 4 � i
21.
� 3 � 3�2 i
��2 � ��8 � � �5 � ��50 � � �2 � 2�2 i � 5 � 5�2 i
23.
� �14 � 20i
13i � �14 � 7i� � 13i � 14 � 7i 25.
� 16 �
76i
� �96 �
156 i �
106 �
226 i
��32 �
52i� � �5
3 �113 i� � �
32 �
52i �
53 �
113 i
27.
� 3 � i � 2 � 5 � i
�1 � i��3 � 2i� � 3 � 2i � 3i � 2i2 29.
� 12 � 30i
6i�5 � 2i� � 30i � 12i2 � 30i � 12
31.
� 14 � 10 � 24
��14 � �10 i���14 � �10 i� � 14 � 10i2
Vocabulary Check
1. (a) iii (b) i (c) ii 2.
3. principal square 4. complex conjugates
��1; �1
33.
� �9 � 40i
� 16 � 40i � 25
�4 � 5i�2 � 16 � 40i � 25i 2
35.
� �10
� 4 � 12i � 9 � 4 � 12i � 9
�2 � 3i�2 � �2 � 3i�2 � 4 � 12i � 9i2 � 4 � 12i � 9i2 37. The complex conjugate of is
�6 � 3i��6 � 3i� � 36 � �3i�2 � 36 � 9 � 45
6 � 3i.6 � 3i
39. The complex conjugate of is
� 1 � 5 � 6
��1 � �5i���1 � �5i� � ��1�2 � ��5i�2
�1 � �5i.�1 � �5i 41. The complex conjugate of is
�2�5i���2�5i� � �20i2 � 20
�2�5i.��20 � 2�5i
43. The complex conjugate of is
��8���8� � 8
�8.�8 45.5i
�5i
��i�i
��5i
1� �5i
Section 1.5 Complex Numbers 69
47.
�2�4 � 5i�16 � 25
�8 � 10i
41�
841
�1041
i
2
4 � 5i�
24 � 5i
�4 � 5i4 � 5i
49.
�9 � 6i � i 2
9 � 1�
8 � 6i10
�45
�35
i
3 � i3 � i
�3 � i3 � i
�3 � i3 � i
51.
��6i � 5i 2
1� �5 � 6i
6 � 5i
i�
6 � 5ii
��i�i
53.
� �1201681
�27
1681i
��27i � 120i2
81 � 1600�
�120 � 27i1681
3i
�4 � 5i�2 �3i
16 � 40i � 25i2�
3i�9 � 40i
��9 � 40i�9 � 40i
55.
� �1
2�
5
2i
��1 � 5i
2
�2 � 2i � 3 � 3i
1 � 1
2
1 � i�
3
1 � i�
2�1 � i� � 3�1 � i��1 � i��1 � i�
57.
�62 � 297i
949�
62
949�
297
949i
��100 � 297i � 162
949
��100 � 72i � 225i � 162i2
625 � 324
��4 � 9i
25 � 18i�
25 � 18i
25 � 18i
�4i2 � 9i
9 � 18i � 16
�3i � 8i2 � 6i � 4i2
9 � 24i � 6i � 16i2
i
3 � 2i�
2i
3 � 8i�
i�3 � 8i� � 2i�3 � 2i��3 � 2i��3 � 8i�
59.
� �2�3
��6 � ��2 � ��6i���2i� � �12i 2 � �2�3 ���1� 61. ���10 �2� ��10 i �2
� 10i2 � �10
63.
� �21 � 5�2 � � �7�5 � 3�10 �i � �21 � �50 � � �7�5 � 3�10 �i � 21 � 3�10 i � 7�5 i � �50 i2
�3 � ��5 ��7 � ��10 � � �3 � �5 i��7 � �10 i�
65.
� 1 ± i
�2 ± 2i
2
�2 ± ��4
2
x ����2� ± ���2�2 � 4�1��2�
2�1�
x2 � 2x � 2 � 0; a � 1, b � �2, c � 2 67.
��16 ± 4i
8� �2 ±
1
2i
��16 ± ��16
8
x ��16 ± ��16�2 � 4�4��17�
2�4�
4x2 � 16x � 17 � 0; a � 4, b � 16, c � 17
70 Chapter 1 Equations, Inequalities, and Mathematical Modeling
69.
x � �12
8� �
3
2 or x � �
20
8� �
5
2
��16 ± �16
8�
�16 ± 4
8
x ��16 ± ��16�2 � 4�4��15�
2�4�
4x2 � 16x � 15 � 0; a � 4, b � 16, c � 15 71. Multiply both sides by 2.
�12 ± 6�2i
6� 2 ± �2i
�12 ± ��72
6
x ����12� ± ���12�2 � 4�3��18�
2�3�
3x2 � 12x � 18 � 0
32
x2 � 6x � 9 � 0
73. Multiply both sides by 5.
�5 ± 5�15
7�
57
±5�15
7
�10 ± �1500
14�
10 ± 10�1514
x ����10� ± ���10�2 � 4�7���50�
2�7�
7x2 � 10x � 50 � 0
1.4x2 � 2x � 10 � 0 75.
� �1 � 6i
� 6i � 1
� �6��1�i � ��1�
�6i 3 � i 2 � �6i2i � i2
77.
� �5i
� �5��1���1�i
�5i5 � �5i2i2i 79.
� �375�3 i
� 125�3�3 � ��1�i
� 53��3 �3i 3
���75 �3 � �5�3 i�3 81.1
i3�
1
�i�
1
�i�
i
i�
i
�i2�
i
1� i
83. (a)
(b)
�11,240 � 4630i
877�
11,240877
�4630877
i z � �340 � 230i29 � 6i ��29 � 6i
29 � 6i�
�29 � 6i
340 � 230i�
20 � 10i � 9 � 16i�9 � 16i��20 � 10i��
19 � 16i
�1
20 � 10i 1z
�1z1
�1z2
z1 � 9 � 16i, z2 � 20 � 10i
85. (a)
(b)
(c)
(d) ��2i�4 � ��2�4i 4 � 16�1� � 16
�2i�4 � 24i 4 � 16�1� � 16
��2�4 � 16
24 � 16 87. False, if then
That is, if the complex number is real, the number equalsits conjugate.
a � bi � a � bi � a.b � 0
89. False
� 1 � ��1� � 1 � i � i � 1
� �1�11 � �1�37��1� � �1�18��1� � �1�27�i� � �1�15�i�
i44 � i150 � i 74 � i109 � i61 � �i4�11 � �i4�37�i2� � �i4�18�i2� � �i4�27�i� � �i4�15�i�
Section 1.6 Other Types of Equations 71
Section 1.6 Other Types of Equations
■ You should be able to solve certain types of nonlinear or nonquadratic equations by rewriting them in a form in whichyou can factor, extract square roots, complete the square, or use the Quadratic Formula.
■ For equations involving radicals or rational exponents, raise both sides to the same power.
■ For equations that are of the quadratic type, use either factoring, the Quadratic Formula,or completing the square.
■ For equations with fractions, multiply both sides by the least common denominator to clear the fractions.
■ For equations involving absolute value, remember that the expression inside the absolute value can be positive or negative.
■ Always check for extraneous solutions.
au2 � bu � c � 0, a � 0,
Vocabulary Check
1. polynomial 2. extraneous 3. quadratic type
91.
The complex conjugate of this product is
The product of the complex conjugates is:
Thus, the complex conjugate of the product of two complex numbers is the product of their complex conjugates.
� �a1a2 � b1b2� � �a1b2 � a2b1�i
�a1 � b1i��a2 � b2i� � a1a2 � a1b2i � a2b1i � b1b2i2
�a1a2 � b1b2� � �a1b2 � a2b1�i.
� �a1a2 � b1b2� � �a1b2 � a2b1�i
�a1 � b1i��a2 � b2i� � a1a2 � a1b2i � a2b1i � b1b2i2
93. �4 � 3x� � �8 � 6x � x2� � �x2 � 3x � 12 95.
� 3x2 �232 x � 2
�3x �12��x � 4� � 3x2 � 12x �
12x � 2
97.
x � �31
�x � 31
�x � 12 � 19 99.
x �272
2x � 27
2x � 27 � 0
20x � 24 � 18x � 3 � 0
4�5x � 6� � 3�6x � 1� � 0 101.
a �1
2�3V
�b�
�3V�b
2�b
� 3V
4�b� a
3V
4�b� a2
3V � 4�a2b
V �4
3�a2b
103. Let liters withdrawn and replaced.
x � 1 liter
0.50x � 0.50
2.50 � 0.50x � 1.00x � 3.00
0.50�5 � x� � 1.00x � 0.60�5�
x � #
72 Chapter 1 Equations, Inequalities, and Mathematical Modeling
5.
x2 � 6x � 36 � 0 ⇒ x � 3 ± 3�3i
x � 6 � 0 ⇒ x � �6
�x � 6��x2 � 6x � 36� � 0
x3 � 63 � 0
x3 � 216 � 0
7.
x � 3 � 0 ⇒ x � �3
5x � 0 ⇒ x � 0
5x�x � 3�2 � 0
5x�x2 � 6x � 9� � 0
5x3 � 30x2 � 45x � 0 9.
x � 1 � 0 ⇒ x � 1
x � 1 � 0 ⇒ x � �1
x � 3 � 0 ⇒ x � 3
�x � 3��x � 1��x � 1� � 0
�x � 3��x2 � 1� � 0
x2�x � 3� � �x � 3� � 0
x3 � 3x2 � x � 3 � 0
11.
x2 � x � 1 � 0 ⇒ x �1
2±�3
2i
x � 1 � 0 ⇒ x � �1
x � 1 � 0 ⇒ x � 1
�x � 1��x � 1��x2 � x � 1� � 0
�x � 1��x3 � 1� � 0
x3�x � 1� � �x � 1� � 0
x4 � x3 � x � 1 � 0
1.
2x2 � 9 � 0 ⇒ x � ±3�2
2
2x2 � 0 ⇒ x � 0
2x2�2x2 � 9� � 0
4x4 � 18x2 � 0 3.
2x � 3 � 0 ⇒ x � 3
2x � 3 � 0 ⇒ x � �3
x2 � 9 � 0 ⇒ x � ±3i
�x2 � 9��x � 3��x � 3� � 0
x4 � 81 � 0
13.
x � 1 � 0 ⇒ x � 1
x � 1 � 0 ⇒ x � �1
x � �3 � 0 ⇒ x � �3
x � �3 � 0 ⇒ x � ��3
�x � �3 ��x � �3 ��x � 1��x � 1� � 0
�x2 � 3��x2 � 1� � 0
x4 � 4x2 � 3 � 0
15.
1x � 4 � 0 ⇒ x � 4
1x � 4 � 0 ⇒ x � �4
2x � 1 � 0 ⇒ x �12
2x � 1 � 0 ⇒ x � �12
�2x � 1��2x � 1��x � 4��x � 4� � 0
�4x2 � 1��x2 � 16� � 0
4x4 � 65x2 � 16 � 0
Section 1.6 Other Types of Equations 73
17.
x2 � x � 1 � 0 ⇒ x � �1
2 ±
�3
2i
x � 1 � 0 ⇒ x � 1
x2 � 2x � 4 � 0 ⇒ x � 1 ± �3i
x � 2 � 0 ⇒ x � �2
�x � 2��x2 � 2x � 4��x � 1��x2 � x � 1� � 0
�x3 � 8��x3 � 1� � 0
x6 � 7x3 � 8 � 0
19.
1 � 5x � 0 ⇒ x � �1
5
1 � 3x � 0 ⇒ x � �1
3
�1 � 3x��1 � 5x� � 0
1 � 8x � 15x2 � 0
1
x2�
8
x� 15 � 0 21.
��x � �5 is not a solution.� �x �
12 ⇒ x �
14
�2�x � 1���x � 5� � 0
2��x �2� 9�x � 5 � 0
2x � 9�x � 5 � 0
2x � 9�x � 5
23.
x1�3 � 1 � 0 ⇒ x1�3 � 1 ⇒ x � �1�3 � 1
2x1�3 � 5 � 0 ⇒ x1�3 � �52 ⇒ x � �� 5
2�3� �
1258
�2x1�3 � 5��x1�3 � 1� � 0
2�x1�3�2 � 3x1�3 � 5 � 0
2x2�3 � 3x1�3 � 5 � 0
3x1�3 � 2x2�3 � 5
25.
(a)
(b) x-intercepts: ��1, 0�, �0, 0�, �3, 0�
−9 9
−7
5
y � x3 � 2x2 � 3x
(c)
(d) The intercepts of the graph are the same as the solutions to the equation.x-
x � 3 � 0 ⇒ x � 3
x � 1 � 0 ⇒ x � �1
x � 0
0 � x�x � 1��x � 3�
0 � x3 � 2x2 � 3x
27.
(a)
(b) x-intercepts: �±1, 0�, �±3, 0�
−5 5
20
−20
y � x4 � 10x2 � 9
(c)
(d) The x-intercepts of the graph are the same as the solutions to the equation.
x � 3 � 0 ⇒ x � 3
x � 3 � 0 ⇒ x � �3
x � 1 � 0 ⇒ x � 1
x � 1 � 0 ⇒ x � �1
0 � �x � 1��x � 1��x � 3��x � 3�
0 � �x2 � 1��x2 � 9�
0 � x4 � 10x2 � 9
74 Chapter 1 Equations, Inequalities, and Mathematical Modeling
29.
x � 50
2x � 100
�2x � 10
�2x � 10 � 0 31.
x � 26
x � 10 � 16
�x � 10 � 4
�x � 10 � 4 � 0 33.
x � �16
2x � �32
2x � 5 � �27
33�2x � 5 � �3
3�2x � 5 � 3 � 0
35.
x � 2 � 0 ⇒ x � 2
x � 5 � 0 ⇒ x � �5
�x � 5��x � 2� � 0
x2 � 3x � 10 � 0
16 � 8x � x2 � 26 � 11x
4 � x � �26 � 11x
��26 � 11x � 4 � x 37.
x � 0
�2x � 0
x � 1 � 3x � 1
�x � 1 � �3x � 1
39.
9 � x
4 � x � 5
2 � �x � 5
4 � 2�x � 5
x � 1 � 2�x � 5 � x � 5
��x �2� �1 � �x � 5 � 2
�x � 1 � �x � 5
�x � �x � 5 � 1 41.
1014 � x
101 � 4x
81 � 4x � 20
81 � 4�x � 5�
9 � 2�x � 5
�90 � �20�x � 5
x � 5 � 100 � 20�x � 5 � x � 5
��x � 5 �2� �10 � �x � 5 �2
�x � 5 � 10 � �x � 5
�x � 5 � �x � 5 � 10
43.
x � 14 � 0 ⇒ x � 14
x � 2 � 0 ⇒ x � 2, extraneous
�x � 2��x � 14� � 0
x2 � 16x � 28 � 0
x2 � 8x � 16 � 8x � 12
x2 � 8x � 16 � 4�2x � 3�
�x � 4�2 � �2�2x � 3�2
x � 4 � 2�2x � 3
�x � 4 � �2�2x � 3
x � 2 � 2x � 2 � 2�2x � 3
x � 2 � 2x � 3 � 2�2x � 3 � 1
��x � 2�2� ��2x � 3 � 1�2
�x � 2 � �2x � 3 � 1
�x � 2 � �2x � 3 � �1 45.
x � 5 � 4 � 9
x � 5 � 3�64
�x � 5�3 � 82
�x � 5�3�2 � 8
Section 1.6 Other Types of Equations 75
47.
x � �3 ± 16�2
x � 3 � ±�512
x � 3 � ±�83
�x � 3�2 � 83
�x � 3�2�3 � 8 49.
x � ±�14
x2 � 14
x2 � 5 � 9
x2 � 5 � 3�272
�x2 � 5�3 � 272
�x2 � 5�3�2 � 27
51.
5x � 2 � 0 ⇒ x �25, extraneous
�x � 1�1�2 � 0 ⇒ x � 1 � 0 ⇒ x � 1
�x � 1�1�2�5x � 2� � 0
�x � 1�1�2�3x � 2�x � 1�� � 0
3x�x � 1�1�2 � 2�x � 1�3�2 � 0
53.
(a)
(b) x-intercepts:
(c)
(d) The x-intercepts of the graph are the same as the solutions to the equation.
x � 6 � 0 ⇒ x � 6
x � 5 � 0 ⇒ x � 5
�x � 5��x � 6� � 0
x2 � 11x � 30 � 0
x2 � 11x � 30
x � �11x � 30
0 � �11x � 30 � x
�5, 0�, �6, 0�
−1 11
−6
2
y � �11x � 30 � x 55.
(a)
(b) x-intercepts:
(c)
(d) The x-intercepts of the graph are the same as thesolutions to the equation.
x � 4 � 0 ⇒ x � 4
x � 0
x�x � 4� � 0
x2 � 4x � 0
x2 � 16x � 64 � 20x � 64
x2 � 16x � 64 � 4�5x � 16�
�x � 8�2 � �2�5x � 16�2
x � 8 � 2�5x � 16
2x � 16 � 4�5x � 16
7x � 36 � 5x � 20 � 4�5x � 16
7x � 36 � 4 � 4�5x � 16 � 5x � 16
��7x � 36 �2� �2 � �5x � 16 �2
�7x � 36 � 2 � �5x � 16
��7x � 36 � ��5x � 16 � 2
0 � �7x � 36 � �5x � 16 � 2
�0, 0�, �4, 0�−0.5
0.5
5−3
y � �7x � 36 � �5x � 16 � 2
76 Chapter 1 Equations, Inequalities, and Mathematical Modeling
57.
x � 2 � 0 ⇒ x � 2
2x � 3 � 0 ⇒ x � �3
2
�2x � 3��x � 2� � 0
2x2 � x � 6 � 0
2x2 � 6 � x
�2x��x� � �2x��3
x� � �2x��1
2�
x �3
x�
1
259.
x ��3 ± ��3�2 � 4�3���1�
2�3��
�3 ± �21
6
a � 3, b � 3, c � �1
0 � 3x2 � 3x � 1
1 � 3x2 � 3x
x � 1 � x � 3x�x � 1�
x�x � 1�1
x� x�x � 1�
1
x � 1� x�x � 1��3�
1
x�
1
x � 1� 3
61.
x � 4 � 0 ⇒ x � 4
x � 5 � 0 ⇒ x � �5
0 � �x � 5��x � 4�
0 � x2 � x � 20
20 � x � x2
20 � x
x� x 63.
�2 ± �124
6�
2 ± 2�31
6�
1 ± �31
3
x ����2� ± ���2�2 � 4�3���10�
2�3�
a � 3, b � �2, c � �10
3x2 � 2x � 10 � 0
x � x � 2 � 3x2 � 12
�x � 2��x � 2�x
x2 � 4� �x � 2��x � 2�
1
x � 2� 3�x � 2��x � 2�
x
x2 � 4�
1
x � 2� 3
65.
��2x � 1� � 5 ⇒ x � �2
2x � 1 � 5 ⇒ x � 3
2x � 1 � 5
67.
First equation:
Only are solutions to the original equation. and are extraneous.x � 1x � ��3 x � �3 and x � �3
x � ±�3
x2 � 3 � 0
x � x2 � x � 3
x � x2 � x � 3
Second equation:
x � 3 � 0 ⇒ x � �3
x � 1 � 0 ⇒ x � 1
�x � 1��x � 3� � 0
x2 � 2x � 3 � 0
�x � x2 � x � 3
Section 1.6 Other Types of Equations 77
69.
First equation:
Only are solutions to the original equation. and are extraneous.x ��1 � �17
2x � �2x � 3 and x �
�1 � �17
2
x � 2 � 0 ⇒ x � �2
x � 3 � 0 ⇒ x � 3
�x � 3��x � 2� � 0
x2 � x � 6 � 0
x � 1 � x2 � 5
x � 1 � x2 � 5
Second equation:
x ��1 ± �17
2
x2 � x � 4 � 0
�x � 1 � x2 � 5
��x � 1� � x2 � 5
71.
(a)
(b) x-intercept:
(c)
(d) The x-intercept of the graph is the same as the solution to the equation.
x � 1 � 0 ⇒ x � �1
0 � �x � 1�2
0 � x2 � 2x � 1
0 � �x2 � 2x � 1
0 � x � 1 � 4x � x2 � x
0 � �x � 1� � 4x � x�x � 1�
0 �1
x�
4
x � 1� 1
��1, 0�
4−4
−24
24
y �1
x�
4
x � 1� 1 73.
(a)
(b) x-intercepts:
(c)
OR
OR
(d) The x-intercepts of the graph are the same asthe solutions to the equation.
x � �3
�x � 3
�x � 1 � 2 x � 1
��x � 1� � 2 x � 1 � 2
2 � x � 1 0 � x � 1 � 2
�1, 0�, ��3, 0�
−10 8
−4
8
y � x � 1 � 2
75.
Using the positive value for we have x � ±�1.5 � �29.13
6.4 ±1.038.x2,
x2 �1.5 ± �1.52 � 4�3.2���2.1�
2�3.2�
3.2x4 � 1.5x2 � 2.1 � 0
77.
Use the Quadratic Formula with and
Considering only the positive value for we have:
x 16.756.
�x 4.0934
�x ,
�x �6 ± �36 � 4�1.8���5.6�
2�1.8�
6 ± 8.7361
3.6
c � �5.6.b � �6,a � 1.8,
1.8��x �2� 6�x � 5.6 � 0
1.8x � 6�x � 5.6 � 0 Given equation 79. and 5
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
x2 � 3x � 10 � 0
�x � 2��x � 5� � 0
�x � ��2���x � 5� � 0
�2
78 Chapter 1 Equations, Inequalities, and Mathematical Modeling
81. and
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
21x2 � 31x � 42 � 0
�3x � 7��7x � 6� � 0
x �67 ⇒ 7x � 6 ⇒ 7x � 6 is a factor.
x � �73 ⇒ 3x � �7 ⇒ 3x � 7 is a factor.
67�
73 83. and 4
One possible equation is:
Any non-zero multiple of this equation would also havethese solutions.
x3 � 4x2 � 3x � 12 � 0
�x2 � 3��x � 4� � 0
�x � �3��x � �3��x � 4� � 0
�x � �3��x � ���3���x � 4� � 0
�3, ��3,
85. and
One possible equation is:
Any non-zero multiple of this equation would also have these solutions.
x4 � 1 � 0
�x2 � 1��x2 � 1� � 0
�x � 1��x � 1��x � i��x � i� � 0
�x � ��1���x � 1��x � i��x � ��i�� � 0
�i�1, 1, i,
87. Let the number of students in the original group. Then, the original cost per student.
When six more students join the group, the cost per student becomes
Model:
Multiply both sides by to clear fractions.
Using the positive value for x we conclude that the original number was students.x � 34
x �90 ± ���90�2 � 4��15��20,400�
2��15��
90 ± 1110
�30
�15x2 � 90x � 20,400 � 0
2x �3400 � 15x��x � 6� � 3400x
�1700
x� 7.5��x � 6� � 1700
�Cost per student� � �Number of students� � Total cost
1700x
� 7.50.
1700x
�x �
89. Model:
Labels: Let average speed of the plane. Then wehave a travel time of If the averagespeed is increased by 40 mph, then
Now, we equate these two equations and solve for x.
t �145
x � 40�
1
5.
t �12
60�
145
x � 40
t � 145�x.x �
Time �Distance
RateEquation:
Using the positive value for x found by the Quadratic Formula, we have mph and
mph. The airspeed required to obtain the decrease in travel time is 191.5 miles per hour.x � 40 � 191.5
x 151.5
0 � x2 � 40x � 29,000
725x � 29,000 � 725x � x2 � 40x
145�5��x � 40� � 145�5�x � x�x � 40�
145
x�
145
x � 40�
1
5
Section 1.6 Other Types of Equations 79
91.
r 0.04 � 4%
��1.220996�1�60 � 1��12� � r
�1.220996�1�60 � 1 �r
12
1.220996 � �1 �r
12�60
3052.49 � 2500�1 �r
12��12��5�
A � P�1 �r
n�nt
93.
(a)
The number of medical doctors reached 816,000 lateduring the year 1999.
(b)
The model predicts the number of medical doctorswill reach 900,000 during the year 2005.
The actual number of medical doctors in 2005 isabout 800,000.
t 15.27
t � �436.03111.6 �
2
111.6�t � 436.03
463.97 � 111.6�t � 900
t 9.95
t � �352.03111.6 �
2
111.6�t � 352.03
463.97 � 111.6�t � 816
M � 463.97 � 111.6�t, 4 ≤ t ≤ 12
95.
(a)
(b) when pounds per square inch.x 15T � 212�
T � 75.82 � 2.11x � 43.51�x, 5 ≤ x ≤ 40
(c)
By the Quadratic Formula, we have
Since x is restricted to let pounds per square inch.
(d)
05 40
300
(14.806119, 212)
x � 14.8065 ≤ x ≤ 40,
�x 3.84787 ⇒ x 14.806
�x 16.77298 ⇒ x 281.333
0 � �2.11x � 43.51�x � 136.18
212 � 75.82 � 2.11x � 43.51�xAbsolute TemperaturePressure,
5 162.56
10 192.31
15 212.68
20 228.20
25 240.62
30 250.83
35 259.38
40 266.60
Tx
97.
Rounding x to the nearest whole unit yields units.x 500
� 10.01x � 500.25
� 10.01x � 5.0025
0.01x � 1 � 6.0025
�0.01x � 1 � 2.45
1x � 37.55 � 40 � �0.01x � 1
80 Chapter 1 Equations, Inequalities, and Mathematical Modeling
99. Model:
Labels:
Equation:
The distance between bases is approximately 90 feet.
x � ±�8128.125 ±90
x2 �16,256.25
2
2x2 � 16,256.25
x2 � x2 � �127.5�2
distance from home to 2nd � 127.5distance from 1st to 2nd � x,Distance from home to 1st � x,
�Distance fromhome to 1st �
2
� �distance from1st to 2nd �
2
� �distance fromhome to 2nd �
2
101.
(a)
When
(b)
when h is between 11 and 12 inches.S � 350
S � 350, h 11.4.
0150
450
S � 8��64 � h2
(c)
(d) Solving graphically or numerically yields anapproximate solution. An exact solution is obtained algebraically.
h 11.4
h2 129.935
64 � h2 193.935
122,500 � 64�2�64 � h2�
�350�2 � �8��64 � h2 �2
350 � 8��64 � h2
8 9 10 11 12 13
284.3 302.6 321.9 341.8 362.5 383.6S
h
103. Model:
Labels: Work done rate of first person time worked by first person
rate of second person time worked by second person
Equation:
(Choose the positive value for r.)
It would take approximately 23 hours and 26 hours individually.
r 23
r ����21� ± ���21�2 � 4�1���36�
2�1��
21 ± �585
2
0 � r2 � 21r � 36
12r � 36 � 12r � r2 � 3r
r�r � 3�12
r� r�r � 3�
12
r � 3� r�r � 3�
12
r�
12
r � 3� 1
� 12�1
r � 3,
� 12,�1
r,� 1,
� Portion doneby first person� � � portion done
by second person� � �workdone�
105.
v2�s
R� g
v2�s � gR
v2 �gR�s
v ��gR�s
107. False—See Example 7 on page 137.
Section 1.6 Other Types of Equations 81
109. The distance between is 13.
Both and are a distance of 13from �1, 2�.
�6,�10���4, �10�
x � 6 � 0 ⇒ x � 6
x � 4 � 0 ⇒ x � �4
�x � 4��x � 6� � 0
x2 � 2x � 24 � 0
x2 � 2x � 1 � 144 � 169
�x � 1�2 � ��12�2 � 132
��x � 1�2 � ��10 � 2�2 � 13
�1, 2� and �x, �10� 111. The distance between is 17.
Both and are a distance of 17from �0, 0�.
�8, �15��8, 15�
� ±15
y2 � ±�225
y2 � 225
64 � y2 � 289
�8�2 � �y�2 � 172
��8 � 0�2 � �y � 0�2 � 17
�0, 0� and �8, y�
113.
OR
Thus, or From the original equationwe know that
Some possibilities are:
or
or
or
or
or a � 14a � 4b � 14,
a � 13a � 5b � 13,
a � 12a � 6b � 12,
a � 11a � 7b � 11,
a � 10a � 8b � 10,
a � 9b � 9,
b ≥ 9.a � b.a � 18 � b
a � 18 � b
�a � b � 18
9 � a � b � 9
9 � a � b � 9
9 � 9 � a � b
a � b
�a � �b
9 � a � �b � 9
9 � a � ��b � 9�
115.
This formula gives the relationship between a and b. From the original equation we know that and
Choose a b value, where and then solve for a, keeping in mind that
Some possibilities are:
b � 25, a � �5
b � 24, a � 14
b � 23, a � 11
b � 22, a � 16
b � 21, a � 19
b � 20, a � 20
a ≤ 20.b ≥ 20b ≥ 20.
a ≤ 20
a � �b2 � 40b � 380
�a � b2 � 40b � 380
20 � a � b2 � 40b � 400
�20 � a � b � 20
20 � �20 � a � b
117.83x
�32x
�166x
�96x
�256x 119.
��3z2 � 2z � 4
z�z � 2�
�2z � 3z2 � 6z � 2z � 4
z�z � 2�
�2z � 3z�z � 2� � 2�z � 2�
z�z � 2�
2
z � 2� �3 �
2z� �
2z � 2
� 3 �2z
121.
x � 11
x � 11 � 0
�x � 11�2 � 0
x2 � 22x � 121 � 0
82 Chapter 1 Equations, Inequalities, and Mathematical Modeling
Section 1.7 Linear Inequalities in One Variable
■ You should know the properties of inequalities.
(a) Transitive: and implies
(b) Addition: and implies
(c) Adding or Subtracting a Constant: if
(d) Multiplying or Dividing a Constant: For
1. If then and
2. If then and
■ You should be able to solve absolute value inequalities.
(a) if and only if
(b) if and only if x < �a or x > a.�x� > a
�a < x < a.�x� < a
a
c>
b
c.ac > bcc < 0,
a
c<
b
c.ac < bcc > 0,
a < b,
a < b.a ± c < b ± c
a � c < b � d.c < da < b
a < c.b < ca < b
1. Interval:
(a) Inequality:
(b) The interval is bounded.
�1 ≤ x ≤ 5
��1, 5� 3. Interval:
(a) Inequality:
(b) The interval is unbounded.
x > 11
�11, �� 5. Interval:
(a) Inequality:
(b) The interval is unbounded.
x < �2
���, �2�
7.
Matches (b).
x < 3 9.
Matches (d).
�3 < x ≤ 4
Vocabulary Check
1. solution set 2. graph 3. negative
4. equivalent 5. double 6. union
11.
Matches (e).
�x� < 3 ⇒ �3 < x < 3
13.
(a)
Yes, is asolution.
x � 3
3 > 0
5�3� � 12 >?
0
x � 3
5x � 12 > 0
(b)
No, is not asolution.
x � �3
�27 >� 0
5��3� � 12 >?
0
x � �3 (c)
Yes, is a solution.x �52
12 > 0
5�52� � 12 >
?0
x �52 (d)
No, is not a solution.
x �32
�92 >� 0
5�32� � 12 >
?0
x �32
15.
(a)
Yes, is asolution.
x � 4
0 <12
< 2
0 <? 4 � 2
4<?
2
x � 4
0 <x � 2
4< 2
(b)
No, is not asolution.
x � 10
0 < 2 <� 2
0 <? 10 � 2
4<?
2
x � 10 (c)
No, is not a solution.
x � 0
0 <� �12
< 2
0 <? 0 � 2
4<?
2
x � 0 (d)
Yes, is a solution.x �72
0 <38
< 2
0 <? �7�2� � 2
4<?
2
x �7
2
Section 1.7 Linear Inequalities in One Variable 83
17.
(a)
Yes, is asolution.
x � 13
3 ≥ 3
�13 � 10� ≥?
3
x � 13
�x � 10� ≥ 3
(b)
Yes, is a solution.
x � �1
11 ≥ 3
��1 � 10� ≥?
3
x � �1 (c)
Yes, is a solution.
x � 14
4 ≥ 3
�14 � 10� ≥?
3
x � 14 (d)
No, is not a solution.
x � 9
1 ≥� 3
�9 � 10� ≥?
3
x � 9
19.
1 2 3 4 5
x
x < 3
14�4x� < 14�12�
4x < 12 21.
2 310−1
x
−2
32
x < 32
�12��2x� < ��1
2���3�
�2x > �3 23.
10 11 12 13 14
x
x ≥ 12
x � 5 ≥ 7
25.
0 1 2 3 4
x
x > 2
�2x < �4
2x � 7 < 3 � 4x 27.
−2 −1 10 2
x
27
x ≥ 27
7x ≥ 2
2x � 1 ≥ 1 � 5x 29.
43 5 6 7
x
x < 5
4 � 2x < 9 � 3x
4 � 2x < 3�3 � x�
31.
432 5 6
x
x ≥ 4
�14x ≤ �1
34x � 6 ≤ x � 7 33.
210 3 4
x
x ≥ 2
4x �12 ≥ 3x �
52
12�8x � 1� ≥ 3x �52 35.
−2−3−4−5−6
x
x ≥ �4
3.6x ≥ �14.4
3.6x � 11 ≥ �3.4
37.
30 1 2−1
x
�1 < x < 3
�2 < 2x < 6
�1 < 2x � 3 < 9 39.
−2−4−6
−
0 2 64 8
x29
215
1�92 < x < 15
2
1�9 < 2x < 15
�12 < 2x � 3 < 12
1�4 <2x � 3
3< 4 41.
0 1−1
−
x
−43
41
� 34
< x < �14
�14
> x > �34
34
> x � 1 >14
43.
121110 13 14
x
13.510.5
10.5 ≤ x ≤ 13.5
4.2 ≤ 0.4x ≤ 5.4
3.2 ≤ 0.4x � 1 ≤ 4.4 45.
−2−4−6 0 2 4 6
x
�6 < x < 6
�x� < 6 47.
−2−3 −1 0 1 2 3
x
x < �2 x > 2
x2
< �1 or x2
> 1
�x2� > 1
84 Chapter 1 Equations, Inequalities, and Mathematical Modeling
49.
No solution. The absolute value of a number cannot beless than a negative number.
�x � 5� < � 1 51.
201510 25 30
x
14 26
14 ≤ x ≤ 26
�6 ≤ x � 20 ≤ 6
�x � 20� ≤ 6
53.
or
−2 −1 10 2 3 4
x
32
−
x ≤ �32 x ≥ 3
�4x ≥ 6 �4x ≤ �12
3 � 4x ≥ 9 3 � 4x ≤ �9
�3 � 4x� ≥ 9 55.
or
0−5−10−15 5 10 15
x
11
x ≥ 11 x ≤ �5
x � 3 ≥ 8 x � 3 ≤ �8
x � 3
2≥ 4
x � 3
2≤ �4
�x � 32 � ≥ 4
57.
x
3 654
4 < x < 5
5 > x > 4
�10 < �2x < �8
�1 < 9 � 2x < 1
�9 � 2x� < 1
�9 � 2x� � 2 < �1 59.
or
−12
− −
−16 −4−8
x
112
292
x ≥ �112 x ≤ �
292
x � 10 ≥ 92 x � 10 ≤ �
92
�x � 10� ≥ 92
2�x � 10� ≥ 9
61.
−10 10
−10
10
6x > 2
6x > 12 63.
−10 10
−10
10
5 � 2x ≤ 2
5�2x ≥ �4
5 � 2x ≥ 1 65.
10
−10
−10 24
�6 ≤ x ≤ 22
�14 ≤ x � 8 ≤ 14
�x � 8� ≤ 14
67.
or
−15 1
−10
10
x ≥ �12 x ≤ �
272
x � 7 ≥ 132 x � 7 ≤ �
132
�x � 7� ≥ 132
2�x � 7� ≥ 13 69.
(a)
(b)
2 � 3x ≤ 32
� 32x ≤ 3
2x � 3 ≤ 0
2 � 3y ≤ 0
2 � 3x ≥ 2
� 32x ≥ 4
2x � 3 ≥ 1
2 � 3y ≥ 1−4 8
−5
3y � 2x � 3
Section 1.7 Linear Inequalities in One Variable 85
71.
(a)
(b)
x ≤ 4
�12x ≥ �2
�12x � 2 ≥ 0
y ≥ 0
4 ≥ x ≥ �2
�2 ≤ �12x ≤ 1
0 ≤ �12x � 2 ≤ 3
0 ≤ y ≤ 3
−6 6
−2
6y � �12x � 2 73.
(a)
(b)
or
or x ≥ 7 x ≤ �1
x � 3 ≥ 4 x � 3 ≤ �4
�x � 3� ≥ 4
y ≥ 4
1 ≤ x ≤ 5
�2 ≤ x � 3 ≤ 2
�x � 3� ≤ 2
y ≤ 2
−5 10
−2
8y � �x � 3�
75.
�5, ��
x ≥ 5
x � 5 ≥ 0 77.
��3, ��
x ≥ �3
x � 3 ≥ 0 79.
���, 72� x ≤ 7
2
�2x ≥ �7
7 � 2x ≥ 0 81.
All real numbers within8 units of 10.
�x � 10� < 8
83. The midpoint of the interval is 0. The interval represents all realnumbers x no more than 3 unitsfrom 0.
�x� ≤ 3
�x � 0� ≤ 3
��3, 3� 85. The graph shows all real numbersat least 3 units from 7.
�x � 7� ≥ 3
87. All real numbers within 10 units of 12
�x � 12� < 10
89. All real numbers more than 4 units from
�x � 3� > 4
�x � ��3�� > 4
�3 91. Let
Type A account charges:
Type B account charges:
If you write more than six checks a month, then thecharges for the type A account are less than the chargesfor the type B account.
6 < x
1.50 < 0.25x
6.00 � 0.25x < 4.50 � 0.50x
4.50 � 0.50x
6.00 � 0.25x
x � the number of checks written in a month.
93.
r > 3.125%
r > 0.03125
2r > 0.0625
1 � 2r > 1.0625
1000�1 � r�2�� > 1062.50 95.
120.95x ≥ 36 units
120.95x > 35.7995
120.95x > 750
115.95x > 95x � 750
115.95R > C
86 Chapter 1 Equations, Inequalities, and Mathematical Modeling
97. Let
Revenue:
Cost:
Profit:
In whole dozens, 134 ≤ x ≤ 234.
13313 ≤ x ≤ 2331
3
200 ≤ 1.50x ≤ 350
50 ≤ 1.50x � 150 ≤ 200
50 ≤ P ≤ 200
� 1.50x � 150
� 2.95x � �150 � 1.45x�
P � R � C
C � 150 � 1.45x
R � 2.95x
x � daily sales level (in dozens) of doughnuts. 99. (a)
(b) From the graph we see that when Algebraically we have:
IQ scores are not a good predictor of GPAs. Otherfactors include study habits, class attendance, andattitude.
x ≥ 129
8.638 ≤ 0.067x
3 ≤ 0.067x � 5.638
x ≥ 129.y ≥ 3
075 150
5
y � 0.067x � 5.638
101.
(a)
Rounding to the nearest year, The average salary was at least $32,000 but not morethan $42,000 between 1991 and 2000.
(b)
According to the model, the average salary willexceed $48,000 in 2006.
t > 16
1.05t > 17
1.05t � 31 > 48
1 ≤ t ≤ 10.
0.95 ≤ t ≤ 10.48
1 ≤ 1.05t ≤ 11
32 ≤ 1.05t � 31 ≤ 42
S � 1.05t � 31.0, 0 ≤ t ≤ 12 103.
Since we have
106.864 ≤ area ≤ 109.464.
�10.3375�2 ≤ area ≤ �10.4625�2
A � s2,
10.3375 ≤ s ≤ 10.4625
�0.0625 ≤ s � 10.4 ≤ 0.0625
� 116 ≤ s � 10.4 ≤ 1
16
�s � 10.4� ≤ 116
105.
You might have been undercharged or overcharged by $0.19.
110�$1.89� � $0.19
14.9 ≤ x ≤ 15.1 gallons
� 110 ≤ x � 15 ≤ 1
10
�x � 15� ≤ 110 107.
Two-thirds of the workers could perform the task in thetime interval between 13.7 minutes and 17.5 minutes.
13.7 < t < 17.5
�1.9 < t � 15.6 < 1.91615141312 17 18 19
t
17.513.7 �1 <t � 15.6
1.9 < 1
�t � 15.6
1.9 � < 1
109.
The minimum relative humidity is20 and the maximum is 80.
20 ≤ h ≤ 80
�30 ≤ h � 50 ≤ 30
�h � 50� ≤ 30 111. False. If c is negative, thenac ≥ bc.
113.
or
Matches (b).
x ≥ a � 2
x � a ≥ 2
x ≤ a � 2
x � a ≤ �2
�x � a� ≥ 2
Section 1.8 Other Types of Inequalities 87
117. and
Midpoint: �3 � ��5�2
, 6 � ��8�
2 � � ��1, �1�
d � ���5 � 3�2 � ��8 � 6�2 � ���8�2 � ��14�2 � �260 � 2�65
��5, �8��3, 6�
119.
x � 10
6x � 60
6x � 24 � 36
�12 � 6x � 12 � 36
�6�2 � x� � 12 � 36 121.
or
x � �12 x �
17
2x � �1 7x � 1
2x � 1 � 0 7x � 1 � 0
�7x � 1��2x � 1� � 0
14x2 � 5x � 1 � 0
123. ��3, 10� 125. Answers will vary.
Section 1.8 Other Types of Inequalities
■ You should be able to solve inequalities.
(a) Find the critical number.
1. Values that make the expression zero
2. Values that make the expression undefined
(b) Test one value in each test interval on the real number line resulting from the critical numbers.
(c) Determine the solution intervals.
Vocabulary Check
1. critical; test intervals 2. zeroes; undefined values 3. P � R � C
115. and
Midpoint: ��4 � 12
, 2 � 12
2 � � ��32
, 7�d � ��1 � ��4��2 � �12 � 2�2 � �52 � 102 � �125 � 5�5
�1, 12���4, 2�
1.
(a)
No, is not asolution.
x � 3
6 <� 0
�3�2 � 3 <?
0
x � 3
x2 � 3 < 0
(b)
Yes, is a solution.x � 0
�3 < 0
�0�2 � 3 <?
0
x � 0 (c)
Yes, is a solution.x �32
�34 < 0
�32�2
� 3 <?
0
x �32 (d)
No, is not asolution.
x � �5
22 <� 0
��5�2 � 3 <?
0
x � �5
88 Chapter 1 Equations, Inequalities, and Mathematical Modeling
3.
(a)
Yes, is asolution.
x � 5
7 ≥ 3
5 � 25 � 4
≥?
3
x � 5
x � 2x � 4
≥ 3
(b)
is undefined.
No, is not asolution.
x � 4
60
4 � 24 � 4
≥?
3
x � 4 (c)
No, is not asolution.
x � �92
517
≥� 3
�
92 � 2
�92 � 4
≥?
3
x � �9
2(d)
Yes, is a solution.x �92
13 ≥ 3
92 � 292 � 4
≥?
3
x �9
2
5.
Critical numbers: x � �32
, x � 2
x � 2 � 0 ⇒ x � 2
2x � 3 � 0 ⇒ x � �32
2x2 � x � 6 � �2x � 3��x � 2� 7.
Critical numbers: x �72
, x � 5
⇒ x � 5 x � 5 � 0
⇒ x �72
2x � 7 � 0
�2x � 7x � 5
2 �3
x � 5�
2�x � 5� � 3x � 5
9.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set:
−1−2−3 0 1 2 3
x
��3, 3�
16 � 9 � 7x � 4�3, ��
0 � 9 � �9x � 0��3, 3�
16 � 9 � 7x � �4���, �3�
x2 � 9
�x � 3��x � 3� ≤ 0?
���, �3�, ��3, 3�, �3, ��
x � ±3
�x � 3��x � 3� ≤ 0
x2 � 9 ≤ 0
x2 ≤ 9 11.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set:
−8 −4−6
−7
0 2
3
4
x
−2
��7, 3�
�12��2� � 24x � 5�3, ��
�7���3� � �21x � 0��7, 3�
��3���13� � 39x � �10���, �7�
�x � 7��x � 3�
�x � 7��x � 3� < 0?
���, �7�, ��7, 3�, �3, ��
x � �7, x � 3
�x � 7��x � 3� < 0
x2 � 4x � 21 < 0
x2 � 4x � 4 < 25
�x � 2�2 < 25
Section 1.8 Other Types of Inequalities 89
13.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set:
x
210−1−2−3−4−5−6
���, �5� � �1, ��
�7��1� � 7x � 2�1, ��
�5���1� � �5x � 0��5, 1�
��1���7� � 7x � �6���, �5�
�x � 5��x � 1�
�x � 5��x � 1� ≥ 0?
���, �5�, ��5, 1�, �1, ��
x � �5, x � 1
�x � 5��x � 1� ≥ 0
x2 � 4x � 5 ≥ 0
x2 � 4x � 4 ≥ 9 15.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set:
0−1−2−3 1 2
x
��3, 2�
�6��1� � 6x � 3�2, ��
�3���2� � �6x � 0��3, 2�
��1���6� � 6x � �4���, �3�
�x � 3��x � 2�
�x � 3��x � 2� < 0?
���, �3�, ��3, 2�, �2, ��
x � �3, x � 2
�x � 3��x � 2� < 0
x2 � x � 6 < 0
x2 � x < 6
17.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set:
10−1−2−3
x
��3, 1�
�5��1� � 5x � 2�1, ��
�3���1� � �3x � 0��3, 1�
��1���5� � 5x � �4���, �3�
�x � 3��x � 1�
�x � 3��x � 1� < 0?
���, �3�, ��3, 1�, �1, ��
x � �3, x � 1
�x � 3��x � 1� < 0
x2 � 2x � 3 < 0
19.
Complete the square.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Negative
Positive
Solution set: ��� < �4 � �21� � ��4 � �21, �� 20
x
−8 −6 −4 −2−10
−4 + 21−4 − 214 � 16 � 5 � 15x � 2��4 � �21, ��0 � 0 � 5 � �5x � 0��4 � �21, �4 � �21 �100 � 80 � 5 � 15x � �10���, �4 � �21 �
x2 � 8x � 5
x2 � 8x � 5 ≥ 0?
��4 � �21, ����4 � �21, �4 � �21�,���, �4 � �21�,x � �4 ± �21
x � �4 ± �21
x � 4 � ±�21
�x � 4�2 � 21
x2 � 8x � 16 � 5 � 16
x2 � 8x � 5 � 0
x2 � 8x � 5 ≥ 0
90 Chapter 1 Equations, Inequalities, and Mathematical Modeling
21.
Critical numbers:
Test intervals:
Test: Is ?
Interval x-Value Value of Conclusion
Negative
Positive
Negative
Positive
Solution set: ��1, 1� � �3, ��
�5��3��1� � 15x � 4�3, ��
�3��1���1� � �3x � 2�1, 3�0 1 2 43
x
−1
�1���1���3� � 3x � 0��1, 1�
��1���3���5� � �15x � �2���, �1�
�x � 1��x � 1��x � 3�
�x � 1��x � 1��x � 3� > 0
���, �1�, ��1, 1�, �1, 3�, �3, ��
x � ±1, x � 3
�x � 1��x � 1��x � 3� > 0
�x2 � 1��x � 3� > 0
x2�x � 3� � 1�x � 3� > 0
x3 � 3x2 � x � 3 > 0
23.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Negative
Positive
Negative
Positive
Solution set: ��3, 2� � �3, ��
�2��7��1� � 14x � 4�3, ��
�0.5��5.5���0.5� � �1.375x � 2.5�2, 3�
x
−4 −3 −2 −1 0 1 2 3 4 5
��2��3���3� � 18x � 0��3, 2�
��6���1���7� � �42x � �4���, �3�
�x � 2��x � 3��x � 3�
�x � 2��x � 3��x � 3� ≥ 0?
���, �3�, ��3, 2�, �2, 3�, �3, ��
x � 2, x � ±3
�x � 2��x � 3��x � 3� ≥ 0
�x � 2��x2 � 9� ≥ 0
x2�x � 2� � 9�x � 2� ≥ 0
x3 � 2x2 � 9x � 18 ≥ 0
x3 � 2x2 � 9x � 2 ≥ �20
25.
Critical number:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Positive
Positive
Solution set:
210−1
x
−2
12
x �12
�1�2 � 1x � 1�12, ��
��1�2 � 1x � 0���, 12��2x � 1�2
�2x � 1�2 ≤ 0?
���, 12�, �12, ��
x �12
�2x � 1�2 ≤ 0
4x2 � 4x � 1 ≤ 0 27.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is: ���, 0� � �0, 32�
2x2�2x � 3� < 0?
���, 0�, �0, 32�, �32, ��
x � 0, x �32
2x2�2x � 3� < 0
4x3 � 6x2 < 0
Section 1.8 Other Types of Inequalities 91
29.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is: ��2, 0� � �2, ��
x�x � 2��x � 2� ≥ 0?
���, �2�, ��2, 0�, �0, 2�, �2, ��
x � 0, x � ±2
x�x � 2��x � 2� ≥ 0
x3 � 4x ≥ 0 31.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is: ��2, ��
�x � 1�2�x � 3�3 ≥ 0?
���, �2�, ��2, 1�, �1, �)
x � 1, x � �2
�x � 1�2�x � 2�3 ≥ 0
33.
(a)
(b) y ≥ 3 when 0 ≤ x ≤ 2.
y ≤ 0 when x ≤ �1 or x ≥ 3.
−5 7
−2
6
y � �x2 � 2x � 3 35.
(a)
(b) y ≤ 6 when x ≤ 4.
y ≥ 0 when �2 ≤ x ≤ 0, 2 ≤ x < �.
−12 12
−8
8
y �18x3 �
12x
37.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is:
−1−2 210
x
���, �1� � �0, 1�
1 � x2
x> 0?
���, �1�, ��1, 0�, �0, 1�, �1, ��
x � 0, x � ±1
1 � x2
x> 0
1
x� x > 0 39.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is:
−2 −1 10 2 3 4 5
x
���, �1� � �4, ��
4 � x
x � 1< 0?
���, �1�, ��1, 4�, �4, ��
x � �1, x � 4
4 � x
x � 1< 0
x � 6 � 2�x � 1�x � 1
< 0
x � 6
x � 1� 2 < 0
41.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is: �5, 15�
15 � x
x � 5> 0?
���, 5�, �5, 15�, �15, ��
x � 5, x � 153 6 9
5
12 15 18
x 15 � x
x � 5> 0
3x � 5 � 4�x � 5�x � 5
> 0
3x � 5
x � 5� 4 > 0
3x � 5
x � 5> 4
92 Chapter 1 Equations, Inequalities, and Mathematical Modeling
43.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is: ��5, �32� � ��1, ��
7�x � 1��x � 5��2x � 3�
> 0?
��3
2, �1�, ��1, ��
���, �5�, ��5, �3
2�,
2
−1−2−3−4−5
−
0
3
x
x � �1, x � �5, x � �3
2
7x � 7
�x � 5��2x � 3�> 0
4�2x � 3� � �x � 5��x � 5��2x � 3�
> 0
4
x � 5�
1
2x � 3> 0
4
x � 5>
1
2x � 3
45.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is: ��34, 3� � �6, ��
30 � 5x
�x � 3��4x � 3�≤ 0?
−2−4
−33
4 6 820
4x
���, �3
4�, ��3
4, 3�, �3, 6�, �6, ��
x � 3, x � �3
4, x � 6
30 � 5x
�x � 3��4x � 3�≤ 0
4x � 3 � 9�x � 3��x � 3��4x � 3�
≤ 0
1
x � 3�
9
4x � 3≤ 0
1
x � 3≤
9
4x � 3
47.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is:
−1−2−3 0 1 2 3
x
��3, �2� � �0, 3�
x�x � 2��x � 3��x � 3�
≤ 0?
���, �3�, ��3, �2�, ��2, 0�, �0, 3�, �3, ��
x � 0, x � �2, x � ±3
x�x � 2��x � 3��x � 3�
≤ 0
x2 � 2x
x2 � 9≤ 0
Section 1.8 Other Types of Inequalities 93
51.
(a)
(b) y ≥ 6 when 2 < x ≤ 4.
y ≤ 0 when 0 ≤ x < 2.
−6 12
8
−4
y �3x
x � 253.
(a) when or
This can also be expressed as
(b) for all real numbers
This can also be expressed as �� < x < �.
x.y ≤ 2
�x� ≥ 2.
x ≥ 2.x ≤ �2y ≥ 1
−6 6
−2
6
y �2x2
x2 � 4
49.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is:
x
−1
−
43210
23
���, �1� � ��23, 1� � �3, ��
��3x � 2��x � 3��x � 1��x � 1� < 0?
���, �1�, ��1, �23�, ��2
3, 1�, �1, 3�, �3, ��
x � �23
, x � 3, x � ±1
��3x � 2��x � 3�
�x � 1��x � 1� < 0
�3x2 � 7x � 6�x � 1��x � 1� < 0
5x � 5 � 2x2 � 2x � x2 � 1
�x � 1��x � 1� < 0
5�x � 1� � 2x�x � 1� � �x � 1��x � 1�
�x � 1��x � 1� < 0
5
x � 1�
2xx � 1
� 1 < 0
5
x � 1�
2xx � 1
< 1
55.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the domain set is: ��2, 2�
4 � x2 ≥ 0?
���, �2�, ��2, 2�, �2, ��
x � ±2
�2 � x��2 � x� ≥ 0
4 � x2 ≥ 0 57.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the domain set is: ���, 3� � �4, ��
�x � 3��x � 4� ≥ 0?
���, 3�, �3, 4�, �4, ��
x � 3, x � 4
�x � 3��x � 4� ≥ 0
x2 � 7x � 12 ≥ 0
94 Chapter 1 Equations, Inequalities, and Mathematical Modeling
63.
The zeros are
Critical numbers: ,
Test intervals:
By testing an x-value in each test interval in the inequality,we see that the solution set is: ��0.13, 25.13�
�25.13, ��
���, �0.13�, ��0.13, 25.13�,
x 25.13 x �0.13
x ��12.5 ± ��12.5�2 � 4��0.5��1.6�
2��0.5�.
�0.5x2 � 12.5x � 1.6 > 0 65.
Critical numbers:
Test intervals:
By testing an x-value in each test interval in the inequality,we see that the solution set is: �2.26, 2.39�
���, 2.26�, �2.26, 2.39�, �2.39, ��
x 2.39, x 2.26
�7.82x � 18.68
2.3x � 5.2> 0
1 � 3.4�2.3x � 5.2�
2.3x � 5.2> 0
1
2.3x � 5.2� 3.4 > 0
1
2.3x � 5.2> 3.4
67.
(a)
It will be back on the ground in 10 seconds.
(b)
4 < t < 6 seconds
�t � 4��t � 6� < 0
t2 � 10t � 24 < 0
�16�t2 � 10t � 24� > 0
�16t2 � 160t � 384 > 0
�16t2 � 160t > 384
t � 0, t � 10
�16t�t � 10� � 0
�16t2 � 160t � 0
s � �16t2 � v0t � s0 � �16t2 � 160t 69.
By the Quadratic Formula we have:
Critical numbers:
Test: Is
Solution set:
13.8 meters ≤ L ≤ 36.2 meters
25 � 5�5 ≤ L ≤ 25 � 5�5
�L2 � 50L � 500 ≥ 0?
L � 25 ± 5�5
�L2 � 50L � 500 ≥ 0
L�50 � L� ≥ 500
LW ≥ 500
2L � 2W � 100 ⇒ W � 50 � L
59.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequali-ty, we see that the domain set is: ��5, 0� � �7, ��
x�x � 5��x � 7� ≥ 0?
���, �5�, ��5, 0�, �0, 7�, �7, ��
x � 0, x � �5, x � 7
x
�x � 5��x � 7� ≥ 0
x
x2 � 2x � 35≥ 0 61.
Critical numbers:
Test intervals:
By testing an x-value in each test interval in the inequality,we see that the solution set is: ��3.51, 3.51�
���, �3.51�, ��3.51, 3.51�, �3.51, ��
x ±3.51
0.4�x2 � 12.35� < 0
0.4x2 � 4.94 < 0
0.4x2 � 5.26 < 10.2
Section 1.8 Other Types of Inequalities 95
71. and
Critical numbers: (These were obtained by using the Quadratic Formula.)
Test intervals:
By testing values in each test interval in the inequality, we see that the solution set is orThe price per unit is
For For Therefore, for 40,000 ≤ x ≤ 50,000, $50.00 ≤ p ≤ $55.00.x � 50,000, p � $50.x � 40,000, p � $55.
p �Rx
� 75 � 0.0005x.
40,000 ≤ x ≤ 50,000.�40,000, 50,000�x-
�50,000, ���0, 40,000�, �40,000, 50,000�,
x � 40,000, x � 50,000
�0.0005x2 � 45x � 1,000,000 ≥ 0
�0.0005x2 � 45x � 250,000 ≥ 750,000
P ≥ 750,000
� �0.0005x2 � 45x � 250,000
� �75x � 0.0005x2� � �30x � 250,000�
P � R � C
C � 30x � 250,000R � x�75 � 0.0005x�
73.
(a)
(b) will be greater than 75% whenwhich corresponds to 2011.
(c) when t 30.41.C � 75
t 31,C
0 230
80
C � 0.0031t3 � 0.216t 2 � 5.54t � 19.1, 0 ≤ t ≤ 23
(d)
will be between 85% and 100% when is between 37and 42. These values correspond to the years 2017 to2022.
(e) when or
(f ) The model is a third-degree polynomial and ast →�, C →�.
37 ≤ t ≤ 42.36.82 ≤ t ≤ 41.8985 ≤ C ≤ 100
tC
24 70.5
26 71.6
28 72.9
30 74.6
32 76.8
34 79.6
Ct
36 83.2
37 85.4
38 87.8
39 90.5
40 93.5
41 96.8
42 100.4
43 104.4
Ct
96 Chapter 1 Equations, Inequalities, and Mathematical Modeling
75.
Since we have
Since the only critical number is Theinequality is satisfied when R1 ≥ 2 ohms.
R1 � 2.R1 > 0,
R1 � 2
2 � R1
≥ 0.
2R1
2 � R1
� 1 ≥ 0
2R1
2 � R1
≥ 1
R ≥ 1,
2R1
2 � R1
� R
2R1 � R�2 � R1�
2R1 � 2R � RR1
1
R�
1
R1
�1
277. True
The test intervals are and�4, ��.
���, �3�, ��3, 1�, �1, 4�,
x3 � 2x2 � 11x � 12 � �x � 3��x � 1��x � 4�
79.
To have at least one real solution, Thisoccurs when This can be written as���, �4� � �4, ��.
b ≤ �4 or b ≥ 4.b2 � 16 ≥ 0.
x2 � bx � 4 � 0 81.
To have at least one real solution,
Critical numbers:
Test intervals:
Test: Is
Solution set: ���, �2�30� � �2�30, ��b2 � 120 ≥ 0?
���, �2�30 �, ��2�30, 2�30 �, �2�30, ��
b � ±�120 � ±2�30
�b � �120 ��b � �120 � ≥ 0
b2 � 120 ≥ 0
b2 � 4�3��10� ≥ 0.
3x2 � bx � 10 � 0
83. (a) If a > 0 and c ≤ 0, then b can be any real number. Ifand then for to be greater than
or equal to zero, b is restricted to or
(b) The center of the interval for b in Exercises 79–82 is 0.
b > 2�ac.b < �2�ac
b2 � 4acc > 0,a > 085. 4x2 � 20x � 25 � �2x � 5�2
87.
� �x � 2��x � 2��x � 3�
x2�x � 3) � 4�x � 3� � �x2 � 4��x � 3� 89.
� 2x2 � x
� �2x � 1��x�
Area � �length��width�
Review Exercises for Chapter 1
Review Exercises for Chapter 1 97
1.
–3 –2 –1 1 2 3
–5
–4
–3
–2
–1
1
x
y
y � 3x � 5
x 0 1 2
y 1�2�5�8�11
�1�2
5.
Line with x-intercept and y-intercept �0, 3�
��32, 0�
y � 2x � 3
–5 –4 –3 –1 1 2 3
–2
1
3
4
5
6
x
y y � 2x � 3 � 0
3.
–3 –2 –1 1 2 54
–3
–2
4
5
x
y
y � x2 � 3x
x 0 1 2 3 4
y 4 0 0 4�2�2
�1
7.
Domain: ���, 5�
–2 –1 1 2 3 4 5 6
–2
1
3
4
5
6
x
yy � �5 � x
x 5 4 1
y 0 1 2 3
�4
x 0 1
y 1 3 5�1
�1�2
9.
is a parabola. y � �2x2
–3 –2 –1 1 2 3
–5
–4
–3
–2
1
x
y y � 2x2 � 0
x 0
y 0 �8�2
±2±1
11.
x-intercepts:
y-intercept:
The intercepts are and The intercept is �0, 5�.y-�5, 0�.�1, 0�x-
y � �0 � 3�2 � 4 ⇒ y � 9 � 4 ⇒ y � 5
⇒ x � 5 or x � 1
0 � �x � 3�2 � 4 ⇒ �x � 3�2 � 4 ⇒ x � 3 � ±2 ⇒ x � 3 ± 2
y � �x � 3�2 � 4
13.
Intercepts:
No y-axis symmetry
No x-axis symmetry
No origin symmetry�y � �4��x� � 1 ⇒ y � �4x � 1 ⇒
�y � �4x � 1 ⇒ y � 4x � 1 ⇒
y � �4��x� � 1 ⇒ y � 4x � 1 ⇒
�14, 0�, �0, 1�
x1−1−2−3−4
4
2−1
3
1
4
−2
−3
−4
yy � �4x � 1
98 Chapter 1 Equations, Inequalities, and Mathematical Modeling
15.
Intercepts:
y-axis symmetry
No x-axis symmetry
No origin symmetry
x1
6
−13 42−1−3−4
3
−2
1
2
4
y
�y � 5 � ��x�2 ⇒ y � �5 � x2 ⇒
�y � 5 � x2 ⇒ y � �5 � x2 ⇒
y � 5 � ��x�2 ⇒ y � 5 � x2 ⇒
�±�5, 0�, �0, 5�
y � 5 � x2 17.
Intercepts:
No y-axis symmetry
No x-axis symmetry
No origin symmetry
x1 2 4
7
−13−2−3−4
1
2
4
5
6
y
�y � ��x�3 � 3 ⇒ y � x3 � 3 ⇒
�y � x3 � 3 ⇒ y � �x3 � 3 ⇒
y � ��x�3 � 3 ⇒ y � �x3 � 3 ⇒
�� 3�3, 0�, �0, 3�
y � x3 � 3
19.
Domain:
Intercepts:
No y-axis symmetry
No x-axis symmetry
No origin symmetry�y � ��x � 5 ⇒ y � ���x � 5 ⇒
�y � �x � 5 ⇒ y � ��x � 5 ⇒
y � ��x � 5 ⇒
��5, 0�, �0, �5���5, ��
x1
7
−6−1
−−5
1
3
4
5
6
−1−24 2−3
yy � �x � 5
21.
Center:
Radius: 3
�0, 0�
–4 –2 –1 1 2 4
–4
–2
–1
1
2
4
x(0, 0)
yx2 � y2 � 9 23.
Center:
Radius: 4
��2, 0�
�x � ��2��2 � �y � 0�2 � 42
–8 –4 –2 4
–6
–2
2
6
x(−2, 0)
y�x � 2�2 � y2 � 16
25.
Center:
Radius: 6
x2
8
8−2
−8
2
4
−4
−2−4−6 4
( (12, −1
y
�12, �1�
�x �12�
2� �y � ��1��2 � 62
�x �12�
2
� �y � 1�2 � 36 27. Endpoints of a diameter: and
Center:
Radius:
Standard form:
�x � 2�2 � �y � 3�2 � 13
�x � 2�2 � �y � ��3��2 � ��13�2
r � ��2 � 0�2 � ��3 � 0�2 � �4 � 9 � �13
�0 � 42
, 0 � ��6�
2 � � �2, �3�
�4, �6��0, 0�
Review Exercises for Chapter 1 99
29.
(a)
(c) When pounds.F �504 � 12.5x � 10,
0 ≤ x ≤ 20F �54x,
(b)
x
F
4 8 12 2416 20
5
10
15
20
25
Length (in inches)
Forc
e (i
n po
unds
)
30x 0 4 8 12 16 20
F 0 5 10 15 20 25
31.
All real numbers are solutions.
0 � 0 Identity
2 � 4x � x2 � 2 � 4x � x2
6 � �x2 � 4x � 4� � 2 � 4x � x2
6 � �x � 2�2 � 2 � 4x � x2 33.
All real numbers are solutions.
0 � 0 Identity
�x3 � x2 � 7x � 3 � �x3 � x2 � 7x � 3
�x3 � 7x � x2 � 3 � �x3 � x2 � 7x � 7 � 4
�x3 � x�7 � x� � 3 � x��x2 � x� � 7�x � 1� � 4
35.
x � 20
3x � 2x � 10 � 10
3x � 2�x � 5� � 10 37.
x � �1
2
10x � �5
4x � 9 � �6x � 4
4x � 12 � 3 � 8 � 6x � 4
4�x � 3� � 3 � 2�4 � 3x� � 4 39.
x � �5
�4x � 20
x � 15 � 5x � 5
5�x5
� 3� � �x � 1�5
x5
� 3 �2x2
� 1
41.
x � 9
8x � 72
18x � 72 � 10x
18�x � 4� � 10x
18x
�10
x � 443.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, �1�.�13, 0�
y � 3�0� � 1 ⇒ y � �1
0 � 3x � 1 ⇒ x �13
y � 3x � 1
45.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, �8�.�4, 0�
y � 2�0 � 4� ⇒ y � �8
0 � 2�x � 4� ⇒ x � 4
y � 2�x � 4� 47.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, 23�.�43, 0�
y � �12
�0� �23
⇒ y �23
0 � �12
x �23
⇒ x �2�31�2
�43
y � �12
x �23
49.
x-intercept:
y-intercept:
The x-intercept is and the y-intercept is �0, � 519�.�2, 0�
3.8y � 0.5�0� � 1 � 0 ⇒ y ��13.8
� �5
19
3.8�0� � 0.5x � 1 � 0 ⇒ x �1
0.5� 2
3.8y � 0.5x � 1 � 0 51.
The height is 10 inches.
10 � h
188.40 � 18.84h
244.92 � 56.52 � 18.84h
244.92 � 2�3.14��3�2 � 2�3.14��3�h
100 Chapter 1 Equations, Inequalities, and Mathematical Modeling
53. Verbal Model:
Labels: Let September’s profit. Then
Equation:
Answer: September profit: $325,000, October profit: $364,000
x � 0.12x � 364,000
x � 325,000
2.12x � 689,000
x � �x � 0.12x� � 689,000
x � 0.12x � October’s profit.x �
September’s profit � October’s profit � 689,000
55. Let height of the streetlight.
By similar triangles we have:
The streetlight is 24 feet tall.
x � 24
5x � 120
x
20�
65
x
15 ft5 ft
6 ft
x � 57. Let the number of original investors.
Each person’s share is
If three more people invest, each person’s share is
Since this is $2500 less than the original cost, we have:
There are currently nine investors.
x � �12, extraneous or x � 9
�2500�x � 12��x � 9� � 0
�2500�x2 � 3x � 108� � 0
�2500x2 � 7500x � 270,000 � 0
90,000x � 270,000 � 2500x2 � 7500x � 90,000x
90,000�x � 3� � 2500x�x � 3� � 90,000x
90,000
x� 2500 �
90,000x � 3
90,000x � 3
.
90,000x
.
x �
59. Let the number of liters of pure antifreeze.
x �2
0.70�
20
7� 2
6
7 liters
0.70x � 2
3 � 0.30x � 1.00x � 5
0.30�10 � x� � 1.00x � 0.50�10�
30% of �10 � x� � 100% of x � 50% of 10
x � 61.
3V
�r 2� h
3V � �r 2 h
V �1
3�r 2h
63.
t �23 hour or 40 minutes
15t � 10
55t � 40t � 10
rate time distance
1st car 40 mph t 40t
2nd car 55 mph t 55t
65.
x � 3 � 0 ⇒ x � 3
2x � 5 � 0 ⇒ x � �52
0 � �2x � 5��x � 3�
0 � 2x2 � x � 15
15 � x � 2x2 � 0
Review Exercises for Chapter 1 101
67.
±�2 � x
2 � x2
6 � 3x2 69.
x � �4 ± 3�2
x � 4 � ±�18
�x � 4�2 � 18 71.
x � 6 ± �6
x � 6 � ±�6
�x � 6�2 � 6
x2 � 12x � 36 � �30 � 36
x2 � 12x � �30
x2 � 12x � 30 � 0
73.
��5 ± �241
4
x ��5 ± �52 � 4�2���27�
2�2�
2x2 � 5x � 27 � 0
�2x2 � 5x � 27 � 0 75.
(a) when feet and feet.
(b)
(c) The bending moment is greatest when feet.x � 10
00
22
55,000
x � 20x � 0500x�20 � x� � 0
M � 500x�20 � x�
77. 6 � ��4 � 6 � 2i 79. i2 � 3i � �1 � 3i
81. �7 � 5i� � ��4 � 2i� � �7 � 4� � �5i � 2i� � 3 � 7i 83. 5i�13 � 8i� � 65i � 40i2 � 40 � 65i
85.
� �4 � 46i
�10 � 8i��2 � 3i� � 20 � 30i � 16i � 24i2 87.
�2317
�1017
i
�23 � 10i
17
�24 � 10i � i2
16 � 1
6 � i4 � i
�6 � i4 � i
�4 � i4 � i
89.
�2113
�1
13i
� � 813
� 1� � �1213
i � i�
�8
13�
1213
i � 1 � i
�8 � 12i4 � 9
�2 � 2i1 � 1
4
2 � 3i�
21 � i
�4
2 � 3i�
2 � 3i2 � 3i
�2
1 � i�
1 � i1 � i
91.
� ±�1
3 i � ±
�3
3i
x � ±��1
3
x2 � �1
3
3x2 � �1
3x2 � 1 � 0
93.
x � 1 ± 3i
x � 1 � ±��9
�x � 1�2 � �9
x2 � 2x � 1 � �10 � 1
x2 � 2x � 10 � 0 95.
x � 0 or 5 � x �12
5
x3 � 0 or 5x � 12 � 0
x3�5x � 12� � 0
5x4 � 12x3 � 0
102 Chapter 1 Equations, Inequalities, and Mathematical Modeling
97.
or
x � ±�3 x � ±�2
x2 � 3 x2 � 2
x2 � 3 � 0 x2 � 2 � 0
�x2 � 2��x2 � 3� � 0
x4 � 5x2 � 6 � 0 99.
x � 5
x � 4 � 9
��x � 4 �2� �3�2
�x � 4 � 3
101.
No solution
x � 3, extraneous or x � 11, extraneous
�x � 3��x � 11� � 0
x2 � 14x � 33 � 0
x2 � 2x � 1 � 16�x � 2�
�x � 1�2 � ��4�x � 2 �2
x � 1 � �4�x � 2
2x � 3 � 4 � 4�x � 2 � x � 2
��2x � 3 �2� �2 � �x � 2 �2
�2x � 3 � �x � 2 � 2 103.
x � 126 or x � �124
x � 1 ± 125
x � 1 � ±�253
�x � 1�2 � 253
�x � 1�2�3 � 25
�x � 1�2�3 � 25 � 0
105.
or
x � �4
5x2 � 20x � 1 � 0 �x � 4�1�2 � 0
�x � 4�1�2�5x2 � 20x � 1� � 0
�x � 4�1�2�1 � 5x�x � 4�� � 0
�x � 4�1�2 � 5x�x � 4�3�2 � 0
x � �2 ±�95
5
x ��20 ± 2�95
10
x ��20 ± �400 � 20
10
x ��20 ± �400 � 20
10
107.
±�10 � x
10 � x2
5x � 10 � x2 � 2x � 3x
5�x � 2� � 1�x��x � 2� � 3x
5x
� 1 �3
x � 2109.
x �43
9x � 12
15x � 5x � 10 � x � 2
3�5x� � 5�x � 2� � x � 2
3
x � 2�
1x
�15x
111.
x � �51 or 5 � x � 15
x � 5 � �10 or x � 5 � 10
x � 5 � 10
Review Exercises for Chapter 1 103
113.
or
The only solutions to the original equation are or �x � �3 and x � �1 are extraneous.�x � 1.
x � 3
x � �3 or x � 1 x � 3 or x � �1
�x � 3��x � 1� � 0 �x � 3��x � 1� � 0
x2 � 2x � 3 � 0 x2 � 2x � 3 � 0
x2 � 3 � �2x x2 � 3 � 2x
x2 � 3 � 2x 115.
143,203 units
x � 143,202.5
0.001x � 143.2025
0.001x � 2 � 145.2025
�0.001x � 2 � 12.05
�12.05 � ��0.001x � 2
29.95 � 42 � �0.001x � 2
117. Interval:
Inequality:
The interval is bounded.
�7 < x ≤ 2
��7, 2� 119. Interval:
Inequality:
The interval is unbounded.
x ≤ �10
���, �10� 121.
���, 12�
x ≤ 12
2x ≤ 24
9x � 8 ≤ 7x � 16
123.
�3215, ��
x ≥ 3215
�152 x ≤ �16
20 � 8x ≤ 4 �12x
4�5 � 2x� ≤ 12�8 � x� 125.
��23, 17�
�23 < x ≤ 17
�2 < 3x ≤ 51
�19 < 3x � 17 ≤ 34 127.
��4, 4�
�4 ≤ x ≤ 4
x ≤ 4
129.
or
or
���, �1� � �7, ��
x > 7 x < �1
x � 3 > 4 x � 3 < �4
x � 3 > 4 131. If the side is 19.3 cm, then with the possible error of 0.5 cm we have:
353.44 cm2 ≤ area ≤ 392.04 cm2
18.8 ≤ side ≤ 19.8
133.
Critical numbers:
Test intervals:
Test: Is ?
By testing an x-value in each test interval in the inequality, we see that the solution set is: ��3, 9�
�x � 3��x � 9� < 0
���, �3�, ��3, 9�, �9, ��
x � �3, x � 9
�x � 3��x � 9� < 0
x2 � 6x � 27 < 0 135.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality, we see that the solution set is: ��4
3, 12�
�3x � 4��2x � 1� < 0?
���, �43, �, ��4
3, 12�, �12, ��
x � �43, x �
12
�3x � 4��2x � 1� < 0
6x2 � 5x � 4 < 0
6x2 � 5x < 4
137.
Critical numbers:
Test intervals:
Test: Is
By testing an value in each test interval in the inequality,we see that the solution set is: ��4, 0� � �4, ��
x-
x�x � 4��x � 4� ≥ 0?
�4, ���0, 4�,��4, 0�,���, �4�,
x � 0, x � ±4
x�x � 4��x � 4� ≥ 0
x3 � 16x ≥ 0 139.
Critical numbers:
Test intervals:
By testing an value in each test interval in the inequality,we see that the solution set is: ���, �5� � ��2, ��
x-
��2, ����5, �2�,���, �5�,
x � �2, x � �5
�x � 2x � 5
< 0
x � 8 � 2�x � 5�
x � 5< 0
x � 8x � 5
� 2 < 0
104 Chapter 1 Equations, Inequalities, and Mathematical Modeling
143.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is: ��4, �3� � �0, ��
�x � 4��x � 3�x
≥ 0?
���, �4�, ��4, �3�, ��3, 0�, �0, ��
x � �4, x � �3, x � 0
�x � 4��x � 3�x
≥ 0
x2 � 7x � 12
x≥ 0
145.
r > 4.9%
r > 0.0488
1 � r > 1.0488
�1 � r�2 > 1.1
5000�1 � r�2 > 5500 147. False
���18���2� � �36 � 6
��18��2 � ��18i���2i� � �36i2 � �6
149. Rational equations, equations involving radicals, and absolute value equations, may have “solutions” that are extraneous. So checking solutions, in the original equations, is crucial to eliminate these extraneous values.
Problem Solving for Chapter 1
1.
Time (in seconds)
Dis
tanc
e (i
n fe
et)
y
x
141.
Critical numbers:
Test intervals:
Test: Is
By testing an x-value in each test interval in the inequality,we see that the solution set is: ��5, �1� � �1, ��
��x � 5��x � 1��x � 1� ≤ 0?
���, �5�, ��5, �1�, ��1, 1�, �1, ��
x � �5, x � ±1
��x � 5�
�x � 1��x � 1�≤ 0
2x � 2 � 3x � 3
�x � 1��x � 1)≤ 0
2�x � 1� � 3�x � 1�
�x � 1��x � 1�≤ 0
2
x � 1≤
3
x � 1
3. (a)
thus:
—CONTINUED—
A � �a�20 � a�
a � b � 20 ⇒ b � 20 � a,
A � �ab (b)a 4 7 10 13 16
A 64 91 100 91 64�����
Problem Solving for Chapter 1 105
3. —CONTINUED—
(c)
or a � 7.877a � 12.123
� 10 ±10����� � 3�
�20� ± 20���� � 3�
2�
�20� ± �400�2 � 1200�
2�
a �20� ± ���20��2 � 4��300�
2�
�a2 � 20�a � 300 � 0
300 � 20�a � �a2
300 � �a�20 � a� (d)
(e) The a-intercepts occur at and Both yield anarea of 0. When and you have a verticalline of length 40. Likewise when and youhave a horizontal line of length 40. They represent theminimum and maximum values of a.
(f) The maximum value of A is This occurswhen and the ellipse is actually a circle.a � b � 10
100� � 314.159.
a � 20, b � 0a � 0, b � 20
a � 20.a � 0
00 20
350
A � na�20 � a�
5.
(a)
(b)
seconds
(c) The speed at which the water drains decreases as theamount of the water in the bathtub decreases.
t �225�12.5
��3� 146.2
0 � ��12.5 ���3225
t�2
t �225
��3�5 � �12.5 � � 60.6 seconds
�12.5 � 5 ���3225
t
12.5 � �5 ���3225
t�2
h � �5 �8��31800
t�2
� �5 ���3225
t�2
l � 60�, w � 30�, h0 � 25�, d � 2�
h � ��h0 �2�d 2�3
lw t�2
7. (a) 5, 12, and 13; 8, 15, and 17
7, 24, and 25
(b) which is divisible by 3, 4, and 5
which is divisible by 3, 4, and 5
which is also divisible by 3, 4,and 5
(c) Conjecture: If where a, b, and c arepositive integers, then is divisible by 60.abc
a2 � b2 � c2
7 � 24 � 25 � 4200
8 � 15 � 17 � 2040
5 � 12 � 13 � 780
9.
x1 � x2 �ca
x1 � x2 � �ba
x2 �ba
x �ca
� 0
ax2 � bx � c � 0
106 Chapter 1 Equations, Inequalities, and Mathematical Modeling
13. (a)
The terms are:
The sequence is bounded so is in the Mandelbrot Set.
(b)
The terms are:
The sequence is unbounded so is not in the Mandelbrot Set.
(c)
The terms are:
The sequence is bounded so is in the Mandelbrot Set.c � �2
�2, 2, 2, 2, 2, . . .
c � �2
c � 1 � i
1 � i, 1 � 3i, �7 � 7i, 1 � 97i, �9407 � 193i, . . .
c � 1 � i
c � i
i, �1 � i, �i, �1 � i, �i, �1 � i, �i,�1 � i, �i, . . .
c � i
15.
From the graph we see that on the intervals ���, �2� � ��1, 2� � �2, ��.x4 � x3 � 6x2 � 4x � 8 > 0
−5
−5 5
10
� �x � 2�2�x � 1��x � 2�
y � x4 � x3 � 6x2 � 4x � 8
11. (a)
�1 � i
2�
12
�12
i
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Practice Test for Chapter 1 107
Practice Test for Chapter 1
1. Graph 3x � 5y � 15. 2. Graph y � �9 � x.
3. Solve 5x � 4 � 7x � 8. 4. Solve x
3� 5 �
x
5� 1.
5. Solve 3x � 1
6x � 7�
2
5. 6. Solve �x � 3�2 � 4 � �x � 1�2.
7. Solve A �12�a � b�h for a. 8. 301 is what percent of 4300?
9. Cindy has in quarters and nickels. How many of each coin does she have if there are 53 coins in all?$6.05
10. Ed has $15,000 invested in two funds paying and 11% simple interest, respectively. How much is invested in each if the yearly interest is $1582.50?
912%
11. Solve by factoring.28 � 5x � 3x2 � 0
12. Solve by taking the square root of both sides.�x � 2�2 � 24
13. Solve by completing the square.x2 � 4x � 9 � 0
14. Solve by the Quadratic Formula.x2 � 5x � 1 � 0
15. Solve by the Quadratic Formula.3x2 � 2x � 4 � 0
16. The perimeter of a rectangle is 1100 feet. Find the dimensions so that the enclosed area will be 60,000 square feet.
17. Find two consecutive even positive integers whose product is 624.
18. Solve by factoring.x3 � 10x2 � 24x � 0 19. Solve 3�6 � x � 4.
20. Solve �x2 � 8�2�5 � 4. 21. Solve x4 � x2 � 12 � 0.
22. Solve 4 � 3x > 16. 23. Solve �x � 3
2 � < 5.
24. Solve x � 1
x � 3< 2. 25. Solve �3x � 4� ≥ 9.