chapter 1 equations, inequalities, and mathematical modelingsection 1.1 graphs of equations 69 4....
TRANSCRIPT
C H A P T E R 1Equations, Inequalities, and Mathematical Modeling
Section 1.1 Graphs of Equations . . . . . . . . . . . . . . . . . . . . 68
Section 1.2 Linear Equations in One Variable . . . . . . . . . . . . . 76
Section 1.3 Modeling with Linear Equations . . . . . . . . . . . . . . 90
Section 1.4 Quadratic Equations and Applications . . . . . . . . . . 104
Section 1.5 Complex Numbers . . . . . . . . . . . . . . . . . . . . 121
Section 1.6 Other Types of Equations . . . . . . . . . . . . . . . . . 128
Section 1.7 Linear Inequalities in One Variable . . . . . . . . . . . . 147
Section 1.8 Other Types of Inequalities . . . . . . . . . . . . . . . . 157
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
C H A P T E R 1Equations, Inequalities, and Mathematical Modeling
Section 1.1 Graphs of Equations
68
■ 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.
2 � 2
�0, 2�: 2 �? �0 � 4
y � �x � 4
(b)
Yes, the point is on the graph.
3 � �9
�5, 3�: 3 �? �5 � 4
2.
(a)
Yes, the point is on the graph.
0 � 0
4 � 6 � 2 �?
0
�2, 0�: �2�2 � 3�2� � 2 �?
0
y � x2 � 3x � 2
(b)
No, the point is not on the graph.
12 � 8
4 � 6 � 2 �?
8
��2, 8�: ��2�2 � 3��2� � 2 �?
8
3.
(a)
No, the point is not on the graph.
5 � 4 � 1
�1, 5�: 5 �? 4 � �1 � 2�
y � 4 � �x � 2�(b)
Yes, the point is on the graph.
0 � 4 � 4
�6, 0�: 0 �? 4 � �6 � 2�
Vocabulary Check
1. solution or solution point 2. graph 3. intercepts
4. axis 5. circle; 6. point-plotting�h, k�; ry-
Section 1.1 Graphs of Equations 69
4.
(a)
Yes, the point is on the graph.
�163 � �16
3
83 �243 �
?�16
3
83 � 8 �?
�163
13 � 8 � 2 � 4 �?
�163
�2, �163 �: 13�2�3 � 2�2�2 �
?�16
3
y �13x3 � 2x2
(b)
No, the point is not on the graph.
�27 � 9
�9 � 18 �?
9
13��27� � 2�9� �?
9
��3, 9�: 13��3�3 � 2��3�2 �?
9
5.
x541−1−2−3
4
7
2
1
−1
2
3
5
y
y � �2x � 5
7.
1 2
3
4
5
4 5x
−2 −1−1
−2
yy � x2 � 3x
x 0 1 2
y 7 5 3 1 0
�52, 0��2, 1��1, 3��0, 5���1, 7��x, y�
52�1
x 0 1 2 3
y 4 0 0
�3, 0��2, �2��1, �2��0, 0���1, 4��x, y�
�2�2
�1
6. y �34x � 1
0 1 2
0 12�
14�1�
52
43�2
8. 5 � x2
�2 �1 0 1 2
1 4 5 4 1
–4 –3 –2 –1 2 3 4
–4
–3
–2
1
2
3
4
x
y
–4 –3 –1 1 3 4
–2
–1
1
2
3
6
4
x
y 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
70 Chapter 1 Equations, Inequalities, and Mathematical Modeling
10.
�0, 9�
y � 9
y � 32
y-intercept: y � �0 � 3�2
(�3, 0)
x � �3
0 � x � 3
x-intercept: 0 � �x � 3�2
y � �x � 3�2 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
12.
�0, 1�, �0, �1�
y � ±1
y-intercepts: y2 � 0 � 1
��1, 0�
x � �1
x-intercept: 0 � x � 1
y2 � x � 1 13. y-axis symmetry
–4 –3 –1 1 3 4
1
2
−2
3
4
x
y
14.
1 2 3 6 7 8
−4
−3
−2
−1
1
2
3
4
x
y 15. Origin symmetry
–4 –3 –2 1 2 3 4
–4
–3
–2
1
2
3
4
x
y
16.
–4 –3 –2 –1 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 18.
x-axis symmetry
x � y2 � 0
x � ��y�2 � 0
x � y2 � 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 20.
y-axis symmetry
y � x4 � x2 � 3
y � ��x�4 � ��x�2 � 3
y � x4 � x2 � 3
Section 1.1 Graphs of Equations 71
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
22.
y-axis symmetry
y � �9 � x2
y � �9 � ��x�2
y � �9 � x2
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
24.
Origin symmetry
xy � 4
��x���y� � 4
xy � 4 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
26.
x-intercept:
y-intercept:
No symmetry
�0, �3�
�32, 0�
(0, −3)
32 , 0( )
y
x−1−2−3 1 2 3
−1
−2
−3
1
2
y � 2x � 3 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
28.
x-intercept:
y-intercept:
No symmetry
�0, 0�
��2, 0�, �0, 0�x
321−1−3−4−5
−6
−5
−4
−3
−2
2
−1
(−2, 0) (0, 0)
yy � �x2 � 2x 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
72 Chapter 1 Equations, Inequalities, and Mathematical Modeling
30.
x-intercept:
y-intercept:
No symmetry
�0, �1�
�1, 0�
–3 –2 –1 2 3
–4
–3
–2
1
2
x(1, 0)
(0, −1)
yy � x3 � 1 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
32.
Domain:
y-intercept:
No symmetry
�0, 1�
���, 1�
–4 –3 –2 –1 1 2–1
2
3
4
5
x(1, 0)
(0, 1)
yy � �1 � x
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
34.
x-intercepts:
y-intercept:
y-axis symmetry
�0, 1�
�±1, 0�
x2 31−2−3
2
−1
−2
−3
1
3
(−1, 0) (1, 0)
(0, 1)
yy � 1 � �x� 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
36.
x-intercept:
y-intercept:
x-axis symmetry
�0, ±�5 ���5, 0�
x21−1−2−3−4−6
4
−1
−4
−3
1
3
(−5, 0)
0, 5( )
0, − 5( )
yx � y2 � 5 37.
Intercepts: �6, 0�, �0, 3�
−10
−10
10
10
y � 3 �12x
Section 1.1 Graphs of Equations 73
38.
Intercepts: �0, �1�, �32, 0�
−10
−10
10
10
y �23x � 1 39.
Intercepts: �3, 0�, �1, 0�, �0, 3�
−10 10
−10
10
y � x2 � 4x � 3 40.
Intercepts: ��2, 0�, �1, 0�, �0, �2�
−10
−10
10
10
y � x2 � x � 2
41.
Intercept: �0, 0�
−10
−10
10
10
y �2x
x � 142.
Intercept: �0, 4�
−10
−10
10
10
y �4
x2 � 143.
Intercept: �0, 0�
−10
−10
10
10
y � 3�x
44.
Intercepts: ��1, 0�, �0, 1�
−10
−10
10
10
y � 3�x � 1 45.
Intercepts: �0, 0�, ��6, 0�
−10
−10
10
10
y � x�x � 6 46.
Intercepts: �0, 0�, �6, 0�
−10
−10
10
10
y � �6 � x��x
47.
Intercepts: ��3, 0�, �0, 3�
−10
−10
10
10
y � �x � 3� 48.
Intercepts: �±2, 0�, �0, 2�
−10
−10
10
10
y � 2 � �x� 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�; 54.
�x � 3�2 � �y � 2�2 � 25
�x � 3�2 � �y � ��2��2 � 52
� �42 � ��3�2 � �25 � 5
r � ��3 � ��1��2 � ��2 � 1�2
50.
x2 � y2 � 25
�x � 0�2 � �y � 0�2 � 52 51. Center: radius: 4
Standard form:
�x � 2�2 � �y � 1�2 � 16
�x � 2�2 � �y � ��1��2 � 42
�2, �1�; 52.
�x � 7�2 � �y � 4�2 � 49
�x � ��7��2 � �y � ��4��2 � 72
74 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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� 56.
Midpoint of diameter (center of circle):
x2 � y2 � 17
�x � 0�2 � �y � 0�2 � ��17�2
��4 � 4
2,
�1 � 1
2 � � �0, 0�
�1
2�68 � �1
2��2��17 � �17
�1
2�64 � 4
�1
2���8�2 � ��2�2
r �1
2���4 � 4�2 � ��1 � 1�2
57.
Center: radius: 5
1−1−2
−2−3−4
−6
−3−4
1234
6
2 3 4 6x
y
(0, 0)
�0, 0�,
x2 � y2 � 25 58.
Center: radius: 4
x
−2
−3
1
2
3
5
−5
531−1−2−3−5 2
y
(0, 0)
�0, 0�,
x2 � y2 � 16 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
60.
Center: radius: 1
–2 –1 1 2
–1
1
3
x
y
(0, 1)
�0, 1�,
x2 � �y � 1�2 � 1 61.
Center: radius:
–1 1 2 3
1
3
x
y
( )12
, 12
32�1
2, 12�,�x �
12�2
� �y �12�2
�94 62.
Center: radius:
y
x−1 1 2 3 4
−1
−2
−3
−4
1
(2, −3)
43�2, �3�,
�x � 2�2 � �y � 3�2 �169
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 64.
t1 65432
1600
3200
4800
6400
8000
Year
Dep
reci
ated
val
ue
y
y � 8100 � 929t, 0 ≤ t ≤ 6
Section 1.1 Graphs of Equations 75
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
66. (a)
(c)
0 1800
9000
y
x
(b) meters so:
(d) and
A square will give the maximum area of 8100 square meters.
(e) The dimensions of a Major League Soccer field can varybetween 110 and 120 yards in length and between 70 and 80yards in width.
y � 90x � 90
A � lw � x�180 � x�
w � y � 180 � x
2x � 2y � 360
P � 360
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
68. (a)5 10 20 30 40 50 60 70 80 90 100
430.43 107.33 26.56 11.60 6.36 3.94 2.62 1.83 1.31 0.96 0.71y
x
(b) y
x20 40 60 80 100
50100150200250300350400450
Res
ista
nce
(in
ohm
s)
Diameter of wire (in mils)
(c) When
(d) As the diameter of the wire increases, the resistancedecreases.
y �10,77085.52 � 0.37 � 1.10327.
x � 85.5,
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- 70. True. The graph can have no intercepts, one, two or
many. For example, a circle centered at the origin has two intercepts. A circle of radius 1, centered at has no intercepts.y-
�7, 7�,y-
76 Chapter 1 Equations, Inequalities, and Mathematical Modeling
71. The viewing window is incorrect. Change the viewingwindow. Examples will vary. For example,will not appear in the standard window setting.
y � x2 � 2072.
(a)
To be symmetric with respect to the axis, can beany non-zero real number; must be zero.
(b)
To be symmetric with respect to the origin, must bezero; can be any non-zero real number.b
a
y � �ax2 � bx3
�y � ax2 � bx3
�y � a��x�2 � b��x�3
bay-
y � a��x�2 � b��x�3 � ax2 � bx3
y � ax2 � bx3
73.
Terms: 9x5, 4x3, �7
9x5 � 4x3 � 7 74. ��7 � 7 � 7 � 7� � ��7�4 � �74
75. �18x � �2x � 3�2x � �2x � 2�2x 76.
� 5��20 � 3� � 5�2�5 � 3�
�55��20 � 3�
20 � 9�
55��20 � 3�11
55
�20 � 3�
55�20 � 3
��20 � 3�20 � 3
77. 6�t2 � t2�6 � �t�1�3 � 3��t� 78. 3��y � � y1�2�1�3� y1�6 � 6�y
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
Section 1.2 Linear Equations in One Variable 77
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
2.
(a)
is not a solution.
(c)
is not a solution.x � 8
�17 � 23
7 � 24 �?
40 � 17
7 � 3�8� �?
5�8� � 17
x � 8
x � �3
16 � �32
7 � 9 �?
�15 � 17
7 � 3��3� �?
5��3� � 17
x � �3
7 � 3x � 5x � 17
(b)
is not a solution.
(d)
is a solution.x � 3
�2 � �2
7 � 9 �?
15 � 17
7 � 3�3� �?
5�3� � 17
x � 3
x � 0
7 � �17
7 � 0 �?
0 � 17
7 � 3�0� �?
5�0� � 17
x � 0
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
4.
(a)
is not a solution.
(c)
is not a solution.x � 0
�3 � �11
0 � 0 � 3 �?
0 � 0 � 11
5�0�3 � 2�0� � 3 �?
4�0�3 � 2�0� � 11
x � 0
x � 2
41 � 25
40 � 4 � 3 �?
32 � 4 � 11
5 � 8 � 4 � 3 �?
4�8� � 4 � 11
5�2�3 � 2�2� � 3 �?
4�2�3 � 2�2� � 11
x � 2
5x3 � 2x � 3 � 4x3 � 2x � 11
(b)
is a solution.
(d)
is not a solution.x � 10
5017 � 4009
5000 � 20 � 3 �?
4000 � 20 � 11
5�1000� � 20 � 3 �?
4�1000� � 20 � 11
5�10�3 � 2�10� � 3 �?
4�10�3 � 2�10� � 11
x � 10
x � �2
�47 � �47
�40 � 4 � 3 �?
�32 � 4 � 11
5��8� � 4 � 3 �?
4��8� � 4 � 11
5��2�3 � 2��2� � 3 �?
4��2�3 � 2��2� � 11
x � �2
78 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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
6.
(a)
is a solution.x � �1
4 � 4
3 � 1 �?
4
3 �1
1�?
4
3 �1
�1 � 2�?
4
x � �1
3 �1
x � 2� 4
(b)
Division by zero is undefined.
is not a solution.x � �2
3 �1
0�?
4
3 �1
�2 � 2�?
4
x � �2 (c)
is not a solution.x � 0
312 � 4
3 �1
0 � 2�?
4
x � 0 (d)
is not a solution.x � 5
317 � 4
3 �1
5 � 2�?
4
x � 5
8.
(a)
is not a solution.x � 2
3��6 � 3
3�2 � 8 �?
3
x � 2
3�x � 8 � 3
(b)
is not a solution.x � �5
3��13 � 3
3��5 � 8 �?
3
x � �5 (c)
is a solution.x � 35
3 � 3
3�27 �?
3
3�35 � 8 �?
3
x � 35 (d)
is not a solution.x � 8
0 � 3
3�8 � 8 �?
3
x � 8
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
Section 1.2 Linear Equations in One Variable 79
10.
(a)
is a solution.
(c)
is not a solution.x � �13
�1439 � 0
109 �
639 �
909 �
?0
10�19� � 7 � 10 �
?0
10��13 �2
� 21��13 � � 10 �
?0
x � �13
x �25
0 � 0
85 �425 �
505 �
?0
10� 425� �
425 � 10 �
?0
10�25 �2
� 21�25 � � 10 �
?0
x �25
10x2 � 21x � 10 � 0
(b)
is a solution.
(d)
is not a solution.x � �2
�12 � 0
40 � 42 � 10 �?
0
10�4� � 42 � 10 �?
0
10��2�2 � 21��2� � 10 �?
0
x � �2
x � �52
0 � 0
1252 �
1052 �
202 �
?0
10�254 � �
1052 � 10 �
?0
10��52 �2
� 21��52 � � 10 �
?0
x � �52
11. is an identity by the DistributiveProperty. It is true for all real values of x.2�x � 1� � 2x � 2 12. is conditional. There are real
values of x for which the equation is not true (for example, x � 0).
3�x � 2� � 5x � 4
13. is conditional. There arereal values of x for which the equation is not true.�6�x � 3� � 5 � �2x � 10 14. is an identity by simplification.
It is true for all real values of x.
3�x � 2� � 5 � 3x � 6 � 5 � 3x � 1
3�x � 2� � 5 � 3x � 1
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� 16. is an identity by simplification. It is true for all real values of x.
� 21 � 3x � 3�7 � x�
�7�x � 3� � 4x � �7x � 21 � 4x
�7�x � 3� � 4x � 3�7 � x�
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 18. is an identity by simplification. It is true for all real values of x.x2 � 2�3x � 2� � x2 � 6x � 4
19. is conditional. There are real values
of x for which the equation is not true.
3 �1
x � 1�
4xx � 1
20. is conditional. There are real values of x for
which the equation is not true (for example, x � 1).
5
x�
3
x� 24
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 22. Original equation
Remove parentheses.
Simplify.
Add 2 to both sides.
Simplify.
Divide both sides by 3.
Simplify. x � 3
3x
3�
9
3
3x � 9
3x � 2 � 2 � 7 � 2
3x � 2 � 7
3x � 12 � 10 � 7
3�x � 4� � 10 � 7
80 Chapter 1 Equations, Inequalities, and Mathematical Modeling
23.
x � 4
x � 11 � 11 � 15 � 11
x � 11 � 15 24.
�12 � x
7 � 19 � 19 � x � 19
7 � 19 � x
7 � x � x � 19 � x
7 � x � 19 25.
x � �9
�2x�2
�18�2
�2x � 18
7 � 7 � 2x � 25 � 7
7 � 2x � 25
26.
x � 3
7x7
�217
7x � 21
7x � 2 � 2 � 23 � 2
7x � 2 � 23 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 28.
x � �5
4x � �20
7x � 3 � 3 � 3x � 3x � 17 � 3 � 3x
7x � 3 � 3x � 17
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� 30.
x � �58
8x � �5
3x � 9 � 5x � 9 � 4 � 5x � 5x � 9
3x � 9 � 4 � 5x
3x � 9 � 5 � 5x � 1
3�x � 3� � 5�1 � x� � 1 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
32.
The solution is the set of all realnumbers.
9x � 10 � 9x � 10
9x � 10 � 5x � 4x � 10
9x � 10 � 5x � 2�2x � 5� 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
34.
x � �5
�6x � 30
2x � 5x � 30 � 3x
10�x5
�x2� � 10�3 �
3x10�
x5
�x2
� 3 �3x10
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�
4�32
�z � 5� �14
�z � 24� � 4�0�
32
�z � 5� �14
�z � 24� � 0 36.
x � 6
7x � 42
7x � 2 � 40
6x � �x � 2� � 40
4�3x
2 � � 4�1
4��x � 2� � 4�10�
4�3x
2�
1
4�x � 2� � 4�10�
3x
2�
1
4�x � 2� � 10
Section 1.2 Linear Equations in One Variable 81
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 38.
x � 50
0.20x � 10
0.60x � 40 � 0.40x � 50
0.60x � 0.40�100 � x� � 50
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 40. or
The second way is easier because you are not working with fractions until the end of the solution.
x �34 x �
34
4x � 3 x �154 � 3
4x � 12 � 15 x � 3 �154
4�x � 3� � 15 4�x � 3� � 15
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 42. or
The first way is easier because the fraction is eliminated in the first step.
z � 12
43 � 34z �
43 � 9 z � 12
34z � 9 z � 4 � 8
34z � 3 � 6 �43�3
4�z � 4� � �43�6
34�z � 4� � 6 34�z � 4� � 6
43.
Contradiction; no solution
8 � �4
x � 8 � x � 4
x � 8 � 2x � 4 � x
x � 8 � 2�x � 2� � x 44.
Contradiction; no solution
13 � 10
2x � 13 � 2x � 10
8x � 16 � 6x � 3 � 2x � 10
8�x � 2� � 3�2x � 1� � 2�x � 5�
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 46.
1
2� y
49 � 98y
49 � 2y � 100y
17 � y � 32 � y � 100y
�y�17 � y
y� �y�
32 � y
y� 100�y�
17 � y
y�
32 � y
y� 100
82 Chapter 1 Equations, Inequalities, and Mathematical Modeling
47.
x � 4
5x � 20
15x � 12 � 10x � 8
3�5x � 4� � 2�5x � 4�
5x � 45x � 4
�23
48.
x � 0
15x � 0
20x � 6 � 5x � 6
2�10x � 3� � 1�5x � 6�
10x � 3
5x � 6�
1
2
49.
x � 3
6x � 18
10x � 13 � 4x � 5
10x � 13
x�
4x � 5x
10 �13x
� 4 �5x
50.
9
7� x
9 � 7x
9
x� 7
15
x�
6
x� 7
15
x� 4 �
6
x� 3 51.
z � 0
3z � 6 � 2z � 4 � 2
3�z � 2� � �2 �2
z � 2��z � 2�
3 � 2 �2
z � 2
52. Multiply both sides by
x �53
3x � 5
3x � 5 � 0
1�x � 5� � 2x � 0
x�x � 5�. 1x
�2
x � 5� 0 53.
Contradiction; no solution
3 � 0
1 � 2 � 0
x � 4x � 4
� 2 � 0
x
x � 4�
4x � 4
� 2 � 0
54. Multiply both sides by
x �116
6x � 11
14x � 7 � 16x2 � 8x � �16x2 � 4
7�2x � 1� � 8x�2x � 1� � �4�2x � 1��2x � 1�
�2x � 1��2x � 1�. 7
2x � 1�
8x2x � 1
� �4
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
Section 1.2 Linear Equations in One Variable 83
56. Multiply both sides by
x � �133
3x � �13
18x � 2 � 15x � 15
12x � 4 � 6x � 6 � 15x � 15
4�3x � 1� � 6�x � 1� � 15�x � 1�
�x � 1��3x � 1� 4x � 1
� �x � 1��3x � 1� 63x � 1
� �x � 1��3x � 1� 153x � 1
�x � 1��3x � 1�. 4
x � 1�
63x � 1
�15
3x � 1
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
58.
Multiply both sides by
x �7
4
4x � 7
4x � 3 � 4
x � 3 � 3x � 6 � 4
�x � 3� � 3�x � 2� � 4
�x � 3��x � 2�.1
x � 2�
3
x � 3�
4
�x � 3��x � 2�
1
x � 2�
3
x � 3�
4
x2 � x � 6
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
60. Multiply both sides by
x � �3
4x � 18 � 3x � 15
6x � 18 � 2x � 3x � 15
6�x � 3� � 2x � 3�x � 5�
x�x � 3�. 6
x�
2
x � 3�
3�x � 5�x�x � 3�
Division by zero is undefined. Thus, is not asolution, and the original equation has no solution.
x � �3
�2 �2
0�
�6
�3�0�
Check: 6
�3�
2
�3 � 3�
3��3 � 5��3��3 � 3�
84 Chapter 1 Equations, Inequalities, and Mathematical Modeling
61.
x � 0
�2x � 0
4x � 9 � 6x � 9
x2 � 4x � 4 � 5 � x2 � 6x � 9
�x � 2�2 � 5 � �x � 3�2 62.
x �15
5x � 1
x2 � 2x � 1 � 2x � 4 � x2 � x � 2
�x � 1�2 � 2�x � 2� � �x � 1��x � 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� 64.
This is a contradiction. Thus, the equation has no solution.
1 � 4
4x2 � 4x � 1 � 4x2 � 4x � 4
�2x � 1�2 � 4�x2 � 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 66.
Intercept:
The solution to is the same as the x-intercept of They are both x � �
32.y �
43x � 2.
0 �43x � 2
��32, 0�
x � � 32
�� 34��� 4
3 x� � �� 34��2�
�43x � 2
0 �43x � 2
−5 4
−3
3
y �43x � 2
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� 68.
Intercept:
The solution to is the same as the x-intercept of They are both x � �3.y � 10 � 2�x � 2�.
0 � 10 � 2�x � 2�
��3, 0�
x � �3
�2x � 6
0 � 6 � 2x
0 � 10 � 2x � 4
0 � 10 � 2�x � 2�
−12 6
−2
10
y � 10 � 2�x � 2�
69.
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� 70.
There is no x-intercept.
0 � �9611
0 � 6x �9611 � 6x
0 � 6x � 6�1611 � x�
−8 10
−10
2
y � 6x � 6�1611 � x�
Section 1.2 Linear Equations in One Variable 85
72.
The x-intercept is the y-intercept is �0, 16�.�163 , 0�;
163 � x
y � 16 �16 � �3x
y � 16 � 3�0� 0 � 16 � 3x
y � 16 � 3x y � 16 � 3x
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� 74.
The x-intercept is the y-intercept is �0, �1�.�1, 0�;
1 � x
y � �1 0 � �1 � x
y � 5 � �6 � 0� 0 � 5 � �6 � x�
y � 5 � �6 � x� y � 5 � �6 � x�
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 76.
The x-intercept is the y-intercept is �0, �125 �.�3, 0�;
y � �125 x � 3
�5y � 12 4x � 12
4�0� � 5y � 12 4x � 5�0� � 12
4x � 5y � 12 4x � 5y � 12
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
78.
The x-intercept is the y-intercept is �0, 52�.��158 , 0�;
x � �158
38
�8x3
�38
� ��5�
8x3
� �5
8x3
� 5 � 2�0� � 0
8x3
� 5 � 2y � 0
y �52
�2y � �5
8�0�
3� 5 � 2y � 0
8x3
� 5 � 2y � 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 � 5x
86 Chapter 1 Equations, Inequalities, and Mathematical Modeling
80.
The x-intercept is the y-intercept is �0, 1.13�.�1.36, 0�;
x � 1.36
x �3.42.5
2.5x � 3.4
3�0� � 2.5x � 3.4 � 0
3y � 2.5x � 3.4 � 0
y � 1.13
y �3.43
3y � 3.4
3y � 2.5�0� � 3.4 � 0
3y � 2.5x � 3.4 � 0
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 82.
1 � 4b
2 � a� x, a � �2
1 � 4b � x�2 � a�
1 � 4b � 2x � ax
4 � 2x � 4b � ax � 3
4 � 2�x � 2b� � ax � 3
83.
x �5
4 � a, a � �4
x�4 � a� � 5
4x � ax � 5
6x � ax � 2x � 5 84.
x �7
a � b, a � �b
x�a � b� � 7
ax � bx � 7
5 � ax � 12 � bx
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 86.
x �30b � 43a � 15
, a � �5
x�15 � 3a� � 30b � 4
�15x � 30b � 12 � 8 � 3ax
�5�3x � 6b� � 12 � 8 � 3ax
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 88.
x � �8a
, a � 0
ax � �8
45
x � ax �45
x � 2 � 10
45
x � ax � 2�25
x � 1� � 10
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 90.
1.326 x
20.9074 � 15.77x
2.763 � 9.45x � 23.1444 � 6.32x � 5
2.763 � 4.5�2.1x � 5.1432� � 6.32x � 5
Section 1.2 Linear Equations in One Variable 87
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
92. Multiply both sides by
�0.342 x
�38.1 � 111.3x
3x � 38.1 � 114.3x
3x � 6�6.350� � 18�6.350�x
6.350x. 3
6.350�
6
x� 18
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� 94.
x � 10 centimeters
200 � 20x
248 � 48 � 8x � 12x
248 � 2�24� � 2�4x� � 2�6x�
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
96. (a)
(b)
The earned income E is $6800.
E � 6800
12E � 3400
For S � 6600: 6600 � 10,000 �12E
S � 10,000 �12E
T � 10,000 �12E, 0 ≤ E ≤ 20,000
T � E � �10,000 �12E�
T � E � S (c)
Earned income: $7600
(d)
For
The subsidy is $7500.
� 10,000 � 2500 � 7500
S � 10,000 �12E � 10,000 �
12�5000�
5000 � E
2500 �12E
12,500 � 10,000 �12ET � 12,500:
T � 10,000 �12E
7600 � E
3800 �12E
13,800 � 10,000 �12E
88 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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:
98. (a) The number reached 33 million in 1995.
(b) Part (a) can be solved graphically by graphing and on the same axes and finding the intersection, Part (a) can be solved algebraically by solving for t.0.39t � 31.0 � 33t � 5 ⇒ 1995.
y � 33y � 0.39t � 31.0
99.
m � 23,437.5 miles
75000.32
� m
7500 � 0.32m
10,000 � 0.32m � 2500 100.
t � 28 hours
0.25t � 7
1 � �0.25t � 8
y � �0.25t � 8
101. False.
This is a quadratic equation. The equation cannot be written in the form ax � b � 0.
x�3 � x� � 10 ⇒ 3x � x2 � 10 102. False. If both sides of the equation are graphed, you cansee that they intersect, which means the equation has areal solution.
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
104. To transform an equation into an equivalent equation,you should first remove symbols of grouping, combinelike terms, and reduce fractions. Then, as needed, youmay add (or subtract) the same quantity to (from) bothsides of the equation, multiply (divide) both sides ofthe equation by the same nonzero quantity, or inter-change the two sides of the equation.
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
Section 1.2 Linear Equations in One Variable 89
106.
The solution to is in the interval
The solution is in the interval
The solution is in the interval 8.16 < x < 8.17.
8.1 < x < 8.2.
8 < x < 9.0.3�x � 1.5� � 2 � 0
0.3�x � 1.5� � 2 � 0
x 6 7 8 9 10
�0.65 �0.35 �0.05 0.25 0.550.3�x � 1.5� � 2
x 8.1 8.2 8.3 8.4
�0.02 0.01 0.04 0.070.3�x � 1.5� � 2
x 8.14 8.15 8.16 8.17 8.18
�0.008 �0.005 �0.002 0.001 0.0040.3�x � 1.5� � 2
107.x2 � 5x � 36
2x2 � 17x � 9�
�x � 9��x � 4��2x � 1��x � 9� �
x � 42x � 1
, x � �9 108.
cannot be simplified.
x2 � 49x3 � x2 � 3x � 21
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 � 5 110.
Intercepts: �0, �92�, ��9, 0�
x2
−6
4
4
−2
−8
2
−10
−2−4
y
y � �12x �
92
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� 112.
Intercepts: �5, 0�, �0, �5�
x
6
−1
1
−2
−1−2 1 2 3 4 5 6
2
3
4
5
y
y � �5 � x
Section 1.3 Modeling with Linear Equations
90 Chapter 1 Equations, Inequalities, and Mathematical Modeling
■ 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 2.
A number decreased by 10The difference of a number and 10
t � 10 3.
The ratio of a number and 5The quotient of a number and 5A number divided by 5
u
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
Section 1.3 Modeling with Linear Equations 91
4.
The product of and a number Two-thirds of a number
23
23
x 5.
The difference of a number and 4 is divided by 5. A number decreased by 4 is divided by 5.
y � 4
56.
The sum of a number and 10 isdivided by 7. A number increased by 10 is divid-ed by 7.
z � 107
7.
The product of and the sum of a number and 2Negative 3 is multiplied by a number increased by 2.
�3
�3�b � 2� 8.
The product of and thedifference of a number and 1 is divided by 8. Negative 5 is multiplied by anumber decreased by 1 and isdivided by 8.
�5
�5�x � 1�8
9.
The difference of a number and 5 is multiplied by 12 times thenumber.12 is multiplied by a number andthat product is multiplied by thenumber decreased by 5.
12x�x � 5�
10.
The product of the sum of a number and 4 and the difference of 3 and a number is divided by the product of 2 and a number.A number increased by 4 is multiplied by 3 decreased by a number and the product is divided by 2 multiplied by a number.
�q � 4��3 � q�2q
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,
��
12. Verbal Model: Product (first number) (second number)
Labels: Product first number second number
Expression: P � n�n � 1� � n2 � n
� n � 1� n,� P,
��
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,
�
14. Verbal Model: (Sum) (first even number) (second even number)
Labels: Sum first even number second even number
Expression: S � �2n�2 � �2n � 2�2 � 4n2 � 4n2 � 8n � 4 � 8n2 � 8n � 4
� 2n � 2� 2n,� S,
22 ��
15. Verbal Model: (Distance) (rate) (time)
Labels: Distance rate time
Expression: d � 50t
� t� 50 mph,� d,
�� 16. Verbal Model: (Distance) (rate) (time)
Labels: Distance km, rate time
Expression: t � 200�r200 � rt,
� t� r,� 200
��
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% �
92 Chapter 1 Equations, Inequalities, and Mathematical Modeling
18. Verbal Model: (Sale price) (list price) (discount)
Labels: Sale price list pricediscount
Expression: S � L � 0.2L � 0.8L
� 0.2L� L,� S,
�� 19. Verbal Model:
Labels: Perimeter widthlength (width)
Expression: P � 2x � 2�2x� � 6x
� 2x� 2� x,� P,
Perimeter � 2�width� � 2�length�
20. Verbal Model: (Area) (height)
Labels: Area base 20 inches, height inches
Expression: A �12�20�h � 10h
� h�� A,
�12(base)
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,
��
22. Verbal Model: (Revenue) (price)(number of units)
Labels: Revenue price 3.59, number of units
Expression: R � 3.59x
� x�� R,
�
23. Verbal Model: Thirty percent of the list price L
Expression: 0.30L
24. Verbal Model: 35% of q
Expression: 0.35q
25. Verbal Model: percent of 500 that is represented by thenumber N
Equation: , p is in decimal formN � p�500�
26. as a percent of
�S2 � S1�S1
�100�
S1�S2 � S1�
27.
A � 4x � 8x � 12x
Area � Area of top rectangle � Area of bottom rectangle
4
42x
x
x
8
28.
Area �12b�2
3b � 1� �13b2 �
12b
Area �12�base��height�
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,
��
Section 1.3 Modeling with Linear Equations 93
30. Verbal Model: (Sum) (first number) (second number) (third number)
Labels: Sum first number second number third number
Equation:
Answer: (second number), and (third number)n � 2 � 269n � 267, n � 1 � 268
267 � n
801 � 3n
804 � 3n � 3
804 � n � n � 1 � n � 2
� n � 2� n � 1,� n,� 804,
���
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,
��
32. Verbal Model: (Difference) (number) (one-fifth of number)
Labels: Difference number one-fifth of number
Equation:
Answer: The numbers are 95 and 15 � 95 � 19.
95 � n
76 �45n
76 � n �15n
�15n� n,� 76,
��
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��
34. Verbal Model: Difference (reciprocal of smaller number) (reciprocal of larger number) (reciprocal of smaller number)
Labels: Smaller number larger number difference
Equation: Multiply both sides by
Answer: The numbers are 3 and n � 1 � 4.
n � 3
n � 1 � 4n � 4 � 4n
n � 1 � 4�n � 1� � 4n
4n�n � 1�1
4n� 4n�n � 1�
1
n� 4n�n � 1�
1
n � 1
4n�n � 1�. 1
4n�
1
n�
1
n � 1
�1
4n� n � 1,� n,
�14
��
94 Chapter 1 Equations, Inequalities, and Mathematical Modeling
35.
� 13.5
� 0.30�45�
� �30%��45�
x � percent � number 36.
� 630
� �1.75��360�
� �175%��360�
x � percent � number 37.
p � 0.27 � 27%
4321600 � p
432 � p�1600�
432 � percent � 1600
38.
p � 1.35 � 135%
459
340� p
459 � p�340�
459 � percent � 340 39.
x � 2400
12
0.005� x
12 � 0.005x
12 �12
% � number 40.
number �70
40%�
70
0.40� 175
70 � 40% � 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
42. (a) Total income billion
Corporate taxes:
Income tax:
Social Security:
Other:
(b)
Interest on debt:
Health and human services:
Defense department:
Other:
(c) There is a deficit of billion.1853.2 � 2011.0 � �$157.8
173.52011
� 0.0863 � 8.6%
348.62011
� 0.1734 � 17.3%
1317.92011
� 0.65535 � 65.5%
1712011
� 0.085 � 8.5%
Defense Department17.3%
Interest on debt8.5%
Other 8.6%
Health and HumanServices 65.5%
Total expenses � 171.0 � 1317.9 � 348.6 � 173.5 � $2011 billion
146.11853.2
� 0.079 or 7.9%
700.81853.2
� 0.378 or 37.8%
858.31853.2
� 0.463 or 46.3%
Other7.9%
Corporation taxes8.0%
Social Security37.8%
Income tax46.3%
148.01853.2
� 0.080 or 8%
� 148.0 � 858.3 � 700.8 � 146.1 � $1853.2
Section 1.3 Modeling with Linear Equations 95
44. Verbal Model:
Labels:
Equation:
The percent increase in the price of a half-gallon of ice cream from 1990 to 2002 is 48%.
p �1.222.54
� 0.48
1.22 � 2.54p
3.76 � 2.54p � 2.54
1990 price � $2.54percent increase � p,2002 price � $3.76,
�2002 price� � �percent increase��1990 price� � 1990 price
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�
46. Verbal Model:
Labels:
Equation:
decrease
The percent decrease in the price of a personal computer from 1990 to 2002 is 18.6%.
p ��1951050
� �0.1857 � 18.6%
�195 � 1050p
855 � 1050p � 1050
1990 price � $1050percent increase � p,2002 price � $855,
�2002 price� � �percent increase��1990 price� � 1990 price
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,
��
96 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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
50. (a)
w
h
(b)
� 3.24w
� 2�0.62w� � 2w
P � 2h � 2w
h � 0.62w (c)
Width:
Height:
Dimensions: 0.38 m � 0.62 m
h � 0.62w � 0.38 meters
w � 0.62 meters
0.62 � w
2 � 3.24w
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
52. Verbal Model: (Average)
Labels: Average test #1 test #2 test #3 test #4
Equation:
You must score 187 out of 200 on the last test to get an A in the course.
187 � x
450 � 263 � x
450 � 87 � 92 � 84 � x
90 �87 � 92 � 84 � x
5
� x� 84,� 92,� 87,� 90,
��test #1� � �test #2� � �test #3� � �test #4�
5
53.
Total time �total distance
rate�
300 kilometers100 kilometers�hour
� 3 hours
Rate �distance
time�
50 kilometers12 hour
� 100 kilometers�hour
48. Verbal Model: (Total) (coworker’s paycheck)
Labels: Total $645, coworker’s paycheck salesperson’s paycheck
Equation:
Coworker’s paycheck is $348.65 and salesperson’s paycheck is 0.85x � $296.35.
348.65 � x
645 � 1.85x
645 � 0.85x � x
� x � 0.15x � 0.85x� x,�
� �salesperson’s paycheck) �
Section 1.3 Modeling with Linear Equations 97
54.
hour (or 20 minutes) must elapse before the two cars are 5 miles apart.13
t �13 hour
5 � 55t � 40t � 15t
5 � d2 � d1
(Distance between cars) � (second distance) � (first distance)
d2 � 55 mph � t
d1 � 40 mph � t
Distance � rate � time
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�
56. Verbal Model: (Distance traveled by first car) (distance traveled by second car)
Labels: Distance traveled by first car distance traveled by second car
Equation:
The second car catches up to the first car after 2.75 hours (or 2 hours and 45 minutes). The distance traveled is miles. Thus, the second car catches up before the first car arrives at the game.45�2.75� � 123.75
t � 2.75
�10t � �27.5
45t � 55t � 27.5
45t � 55�t �12�
� 55�t �12�� 45t,
�
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
98 Chapter 1 Equations, Inequalities, and Mathematical Modeling
58. Verbal Model: (Distance) (rate)(time time
Labels: Distance miles, rate
,
Equation:
The average speed for the round trip was approximately46.3 miles per hour.
46.3 � r
400 � r�1600
440�
2200
440 � �3800
440r
400 � r�200
55�
200
40 �
time2 �distance
rate2
�200
40 hours
time1 �distance
rate1
�200
55 hours
� r,� 2 � 200 � 400
2)1 �� 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
rate
60. Verbal Model: (Distance) (rate)(time)
Labels: Distance meters
Rate meters per second
Time
Equation:
Light from the sun travels to the earth in 500 seconds orapproximately 8.33 minutes.
500 � t
1.5 � 1011 � �3.0 � 108�t
� t
� 3.0 � 108
� 1.5 � 1011
�61. 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
62. Verbal Model:
Labels: Height of tree height of tree’s shadow meters,height of lamppost meters, height of lamppost’s shadow meter
Equation:
The tree is meters tall.2113
h �8�2�0.75
� 2113
h
8�
2
0.75
� 0.75� 2� 8� h,
�height of tree��height of tree's shadow�
��height of lamppost�
�height of lamppost's shadow�
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�
Section 1.3 Modeling with Linear Equations 99
64. (a)
h
30 ft5 ft
6 ft
(b) Verbal Model:
Labels: Height of pole height of pole’s shadow feet,height of person feet, height of person’s shadow feet
Equation:
The pole is 42 feet tall.
h �6
5� 35 � 42
h
35�
6
5
� 5� 6� 30 � 5 � 35� h,
�height of pole��height of pole’s shadow�
��height of person�
�height of person’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�
66. Verbal Model: (interest from 5%) (total interest)
Labels: Amount invested at amount invested at 5%
interest from interest from total annual interest $580
Equation:
The smallest amount that can be invested at 5% is 12,000 � 4000 � $8000.
x � 4000
�0.005x � �20
0.045x � 600 � 0.05x � 580
0.045x � 0.05�12,000 � x� � 580
�5% � �12,000 � x��0.05�,412% � x�0.045�,
� 12,000 � x,412% � x,
��Interest from 412%� �
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� � �
68. Verbal Model: (Profit from dogwood trees) (profit from red maple trees) (total profit)
Labels: Inventory of dogwood trees inventory of red maple trees
profit from dogwood trees profit from red maple trees
total profit
Equation:
The amount invested in dogwood trees was $7500 and the amount invested in red maple trees was 20,000 � 7500 � $12,500.
x � 7500
0.08x � 600
0.25x � 3400 � 0.17x � 4000
0.25x � 0.17�20,000 � x� � 4000
� 0.20�20,000� � 4000
� 0.17�20,000 � x�,� 0.25x,
� 20,000 � x,� x,
��
100 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 ��
��
70. (a) Verbal Model:
Labels: Amount in Solution 1 amount in Solution 2amount in final solution
Equation:
Use 25 gallons of Solution 1 and gallons of Solution 2.
(b) Verbal Model:
Labels: Amount in Solution 1 amount in Solution 2amount in final solution
Equation:
Use 4 liters of Solution 1 and liter of Solution 2.
(c) Verbal Model:
Labels: Amount in Solution 1 amount in Solution 2amount in final solution
Equation:
Use 5 quarts of Solution 1 and quarts of Solution 2.
(d) Verbal Model:
Labels: Amount in Solution 1 amount in Solution 2amount in final solution
Equation:
Use 18.75 gallons of Solution 1 and gallons of Solution 2.25 � 18.75 � 6.25
x � 18.75
�0.20x � �3.75
0.70x � 22.5 � 0.90x � 18.75
0.70x � 0.90�25 � x� � 0.75�25�
� 0.75�25�� 0.90�25 � x�,� 0.70x,
�Amount inSolution 1� � �amount in
Solution 2� � � amount infinal solution�
10 � 5 � 5
x � 5
�0.30x � �1.5
0.15x � 4.5 � 0.45x � 3
0.15x � 0.45�10 � x� � 0.30�10�
� 0.30�10�� 0.45�10 � x�,� 0.15x,
�Amount inSolution 1� � �amount in
Solution 2� � � amount infinal solution�
5 � 4 � 1
x � 4
�0.25x � �1
0.25x � 2.5 � 0.50x � 1.5
0.25x � 0.50�5 � x� � 1.5
� 0.30�5�� 0.50�5 � x�,� 0.25x,
�Amount inSolution 1� � �amount in
Solution 2� � � amount infinal solution�
100 � 25 � 75
x � 25
�0.20x � �5
0.10x � 30 � 0.30x � 25
0.10x � 0.30�100 � x� � 0.25�100�
� 0.25�100�� 0.30�100 � x�,� 0.10x,
�Amount inSolution 1� � �amount in
Solution 2� � � amount infinal solution�
Section 1.3 Modeling with Linear Equations 101
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�
72. Verbal Model:
Labels: Amount of gasoline in Solution 1 amount of gasoline in Solution amount of gasoline in final solution
Equation:
Multiply both sides by 33(41).
Approximately 0.48 gallon of gasoline must be added.
x � 0.48
33x � 16
2624 � 1353x � 2640 � 1320x
6433 � x �
8041 �
4041x
6433 � x �
4041�2 � x�
�4041�2 � x�
2 � x,�3233�2�,
�Amount of gasolinein Solution 1 � � �amount of gasoline
in Solution 2 � � �amount of gasolinein final solution �
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 ��
��
74. Verbal Model: (Fixed costs) (variable cost per unit)(number of units) (total costs)
Labels: Fixed costs variable costs total costs
Equation:
The company can produce no more than 8823 units with $85,000 budgeted monthly for costs.
x � 8823.53
8.50x � 75,000
10,000 � 8.50x � 85,000
� $85,000� 8.50x,� $10,000,
��
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�
102 Chapter 1 Equations, Inequalities, and Mathematical Modeling
77.
2Ab
� h
2A � bh
A �12
bh
78.
b �2Ah
� a
2Ah
� a � b
2A � �a � b�h
A �12
�a � b�h 79.
S
1 � R� C
S � C�1 � R�
S � C � RC 80.
A � P
Pt� r
A � P � Prt
A � P � Prt
81.
3V
4�a2 � b
�3�4�V
�a2 � b
34
V � �a2b
V �43
�a2b 82.
3V � �h3
3�h2� r
3V � �h3 � 3�rh2
3V � 3�rh2 � �h3
3V � �h2�3r � h�
3V �1
3�h2�3r � h� 83.
2�h � v0t�
t2� a
2�h � v0t� � at2
h � v0t �1
2at2
h � v0t �1
2at2
84.
�R1 is the reciprocal of 1
R1
.�
R1 ��n � 1� f R2
R2 � �n � 1� f
R2 � �n � 1� f
�n � 1� f R2
�1
R1
1
�n � 1� f�
1
R2
�1
R1
1
�n � 1� f�
1
R1
�1
R2
1
f� �n � 1�� 1
R1
�1
R2�
76. Verbal Model: (Volume) (length)(width)(height)
Labels:
length 12 feet, width 3 feet,height
Equation:
The depth of the water is 0.26 foot.
0.26 � h
9.3576 � 36h
9.3576 � �12��3�h
� h��
� 9.3576 cubic feet, Volume � 70 gallons � 0.13368 cubic feet
�
86.
2S � n�n � 1�d
2n� a
2S � n�n � 1�d
n� 2a
2Sn
� �n � 1�d � 2a
2Sn
� 2a � �n � 1�d
S �n2
�2a � �n � 1�d�85.
CC2
C2 � C� C1
C2 � C
CC2�
1C1
1C
�1
C2�
1C1
1C
�1
C1�
1C2
C �1
1C1
�1
C2
Section 1.3 Modeling with Linear Equations 103
91.
r � 34.47�
� 1.12 inches
17.88
4�� r3
17.88 � 4�r3
5.96 �43
�r3
V �43
�r3 92.
h �V
�r2 �603.2� �2�2 � 48 feet
V � �r2h
93.
When
59�64.4 � 32� � 18�C.
F � 64.4�,
C �59�F � 32� 94.
When 59�39.1 � 32� � 3.9�C.
F � 39.1�,
C �59�F � 32� 95.
When
95�50� � 32 � 122�F.
C � 50�,
F �95C � 32
96.
When
95
��30� � 32 � �22�F.
C � �30�,
F �95
C � 32 97. False, it should be written as
z3 � 8z2 � 9
.
98. False
� 860.290 in.3
�43
� �5.9�3
Sphere: V �43
�r3
� 857.375 in.3 � 9.53
Cube: V � s3
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
100. Answers will vary. Sample answer: x � 7 � 4
101. �5x4��25x2��1 �5x4
25x2�
x2
5, x 0 102. 150s2t3 � 25 � 6s2t2t � 5st6t
90.
feet from the person x � 32
3
750x � 2750
200x � 2750 � 550x
200x � 550�5 � x�
L � 5 feet
W2 � 550 pounds
W1 � 200 pounds
W1x � W2�L � x�
89.
x � 6 feet from 50-pound child.
125x � 750
50x � 750 � 75x
50x � 75�10 � x�
W1x � W2�L � x�88.
r �S � a
S � L
r�S � L� � S � a
Sr � rL � S � a
Sr � S � rL � a
S�r � 1� � rL � a
S �rL � a
r � 187.
L � a � d
d� n
L � a � d � nd
L � a � nd � d
L � a � �n � 1�d
104 Chapter 1 Equations, Inequalities, and Mathematical Modeling
103. 3
x � 5�
2
5 � x�
3
x � 5�
��1��2�x � 5
�3 � 2
x � 5�
1
x � 5
104.
��2x2 � 30x � 45x�x � 3��x � 3�
�5x2 � 45 � 3x2 � 10x2 � 30x
x�x � 3��x � 3�
�5�x2 � 9� � 3x2 � 10x2 � 30x
x�x � 3��x � 3�
5
x�
3x
x2 � 9�
10
x � 3�
5�x � 3��x � 3� � 3x2 � 10x�x � 3�x�x � 3��x � 3�
105.10
7�3�
107�3
��3�3
�10�3
21106. �
4 3�363�216
�4 3�36
6�
2 3�62
3 4
3�6�
43�6
�3�363�36
107.
� �11 � �6
� ���6 � �11 �
�5�6 � 5�11
�5
5
�6 � �11�
5
�6 � �11��6 � �11
�6 � �11108.
�42�10 � 14
89
�14�3�10 � 1�
89
�14�3�10 � 1�
9�10� � 1
�14�3�10 � 1��3�10�2 � �1�2
14
3�10 � 1�
143�10 � 1
�3�10 � 1
3�10 � 1
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
Section 1.4 Quadratic Equations and Applications 105
1.
General form: 2x2 � 8x � 3 � 0
2x2 � 3 � 8x 2.
General form: x2 � 16x � 0
x2 � 16x 3.
General form: x2 � 6x � 6 � 0
x2 � 6x � 9 � 3
�x � 3�2 � 3
4.
General form:
�3x2 � 42x � 134 � 0
13 � 3x2 � 42x � 147 � 0
13 � 3�x2 � 14x � 49� � 0
13 � 3�x � 7�2 � 0 5.
General form: 3x2 � 90x � 10 � 0
3x2 � 10 � 90x
15�3x2 � 10� � 18x 6.
General form: 4x2 � 2x � 1 � 0
��1���4x2 � 2x � 1� � �1�0�
�4x2 � 2x � 1 � 0
x2 � 2x � 5x2 � 1
x�x � 2� � 5x2 � 1
7.
x � 0 or x � �12
3x � 0 or 2x � 1 � 0
3x�2x � 1� � 0
6x2 � 3x � 0 8.
3x � 1 � 0 ⇒ x �13
3x � 1 � 0 ⇒ x � �13
�3x � 1��3x � 1� � 0
9x2 � 1 � 0 9.
x � 4 or x � �2
x � 4 � 0 or x � 2 � 0
�x � 4��x � 2� � 0
x2 � 2x � 8 � 0
10.
x � 1 � 0 ⇒ x � 1
x � 9 � 0 ⇒ x � 9
�x � 9��x � 1� � 0
x2 � 10x � 9 � 0 11.
x � �5
x � 5 � 0
�x � 5�2 � 0
x2 � 10x � 25 � 0 12.
2x � 3 � 0 ⇒ x � �32
�2x � 3��2x � 3� � 0
4x2 � 12x � 9 � 0
13.
x � 3 or x � �12
3 � x � 0 or 1 � 2x � 0
�3 � x��1 � 2x� � 0
3 � 5x � 2x2 � 0 14.
x � 11 � 0 ⇒ x � 11
2x � 3 � 0 ⇒ x � �32
�2x � 3��x � 11� � 0
2x2 � 19x � 33 � 0
2x2 � 19x � 33 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
16.
x � 2 � 0 ⇒ x � 2
x � 6 � 0 ⇒ x � 6
�x � 6��x � 2� � 0
x2 � 8x � 12 � 0
��1���x2 � 8x � 12� � ��1��0�
�x2 � 8x � 12 � 0
�x2 � 8x � 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 18.
x � 8 � 0 ⇒ x � �8
x � 16 � 0 ⇒ x � 16
�x � 16��x � 8� � 0
x2 � 8x � 128 � 0
18x2 � x � 16 � 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;
106 Chapter 1 Equations, Inequalities, and Mathematical Modeling
19.
x � �a
x � a � 0
�x � a�2 � 0
x2 � 2ax � a2 � 0 20.
x � �a � b� � 0 ⇒ x � �a � b
x � �a � b� � 0 ⇒ x � �a � b
�x � �a � b���x � �a � b�� � 0
��x � a� � b���x � a� � b� � 0
�x � a�2 � b2 � 0 21.
x � ±7
x2 � 49
22.
x � ±�169 � ±13
x2 � 169 23.
x � ±�11
x2 � 11 24.
x � ±�32 � ±4�2
x2 � 32
25.
x � ±3�3
x2 � 27
3x2 � 81 26.
x � ±�4 � ±2
x2 � 4
9x2 � 36 27.
x � 16 or x � 8
x � 12 ± 4
x � 12 � ±4
�x � 12�2 � 16
28.
x � �13 ± 5 � �8, �18
x � 13 � ±5
x � 13 � ±�25
�x � 13�2 � 25 29.
x � �2 ± �14
x � 2 � ±�14
�x � 2�2 � 14 30.
x � 5 ± �30
x � 5 � ±�30
�x � 5�2 � 30
31.
x �1 ± 3�2
2
2x � 1 ± 3�2
2x � 1 � ±�18
�2x � 1�2 � 18 32.
x ��7 ± 2�11
4� �
74
±�11
2
4x � �7 ± 2�11
4x � 7 � ±�44
�4x � 7�2 � 44
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 34.
or
or
The only solution to the equation is x � �92.
x � �92
2x � �9
x � 5 � �x � 4 5 � 4
x � 5 � ��x � 4�x � 5 � ��x � 4�
x � 5 � ± �x � 4�
�x � 5�2 � �x � 4�2
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 36.
x � 3 or x � �1
x � 1 ± 2
x � 1 � ±�4
�x � 1�2 � 4
x2 � 2x � ��1�2 � 3 � ��1�2
x2 � 2x � 3
x2 � 2x � 3 � 0 37.
x � �6 ± �11
x � 6 � ±�11
�x � 6�2 � 11
x2 � 12x � 62 � �25 � 62
x2 � 12x � �25
x2 � 12x � 25 � 0
Section 1.4 Quadratic Equations and Applications 107
38.
x � �4 ± �2
x � 4 � ±�2
�x � 4�2 � 2
x2 � 8x � 42 � �14 � 16
x2 � 8x � �14
x2 � 8x � 14 � 0 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 40.
x �2
3± �2
x �2
3� ±�2
�x �2
3�2
� 2
�x �2
3�2
�18
9
x2 �4
3x � ��
2
3�2
�14
9�
4
9
x2 �4
3x �
14
9
9x2 � 12x � 14
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 42.
No real solution
�x �12�2
� �34
x2 � x �14 � �1 �
14
x2 � x � 1 � 0
�x2 � x � 1 � 0
45.
�1
�x � 1�2 � 4
1
x2 � 2x � 5�
1x2 � 2x � 12 � 12 � 5
46.
�1
�x � 6�2 � 17
1
x2 � 12x � 19�
1�x2 � 12x � ��6�2� � 19 � 36
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 44.
x �1
2± 5 �
11
2, �
9
2
x �1
2� ±�25
�x �1
2�2
� 25
�x �1
2�2
�100
4
x2 � x � ��1
2�2
�99
4�
1
4
x2 � x �99
4
4x2 � 4x � 99 � 0
108 Chapter 1 Equations, Inequalities, and Mathematical Modeling
47.
�4
�x � 2�2 � 7
4
x2 � 4x � 3�
4x2 � 4x � 4 � 4 � 3
48.
�5
�x �252 �2
� 5814
5
x2 � 25x � 11�
5
x2 � 25x � �252 �2
� 11 �6254
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�
50.
�1
�25 � �x � 3�2
�1
�16 � �x2 � 6x � 32� � 9
1
�16 � 6x � x2�
1
�16 � 1�x2 � 6x�
51. (a)
(b) The x-intercepts are ��1, 0� and ��5, 0�.
−10 5
−6
4
y � �x � 3�2 � 4 (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
52. (a)
(b) The x-intercepts are �5, 0� and �3, 0�.
−1 8
2
4
y � �x � 4�2 � 1 (c)
(d) The x-intercepts of the graph are solutions to the equation0 � �x � 4�2 � 1.
x � 4 ± 1 � 5, 3
x � 4 � ±�1
�x � 4�2 � 1
0 � �x � 4�2 � 1
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
54. (a)
(b) The x-intercepts are �5, 0� and �11, 0�.
0 18
−2
10
y � 9 � �x � 8�2 (c)
(d) The x-intercepts of the graph are solutions to the equation0 � 9 � �x � 8�2.
x � 8 ± 3 � 11, 5
x � 8 � ±�9
�x � 8�2 � 9
0 � 9 � �x � 8�2
Section 1.4 Quadratic Equations and Applications 109
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
56. (a)
(b) The x-intercepts are and .�12, 0���1
2, 0�
−3 3
−2
2
y � 4x2 � 1 (c)
(d) The x-intercepts of the graph are solutions to the equation0 � 4x2 � 1.
x � ±�14 � ±1
2
x2 �14
4x2 � 1
0 � 4x2 � 1
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
58. (a)
(b) The x-intercepts are and ��3, 0�.�8, 0�
−5 10
−35
5
y � x2 � 5x � 24 (c)
(d) The x-intercepts of the graph are solutions to the equation0 � x2 � 5x � 24.
x � 3 � 0 ⇒ x � �3
x � 8 � 0 ⇒ x � 8
�x � 8��x � 3� � 0
0 � x2 � 5x � 24
59.
No real solution
b2 � 4ac � ��5�2 � 4�2��5� � �15 < 0
2x2 � 5x � 5 � 0 60.
Two real solutions
� 16 � 20 � 36 > 0
b2 � 4ac � ��4�2 � 4��5��1�
�5x2 � 4x � 1 � 0
61.
Two real solutions
b2 � 4ac � ��1�2 � 4�2���1� � 9 > 0
2x2 � x � 1 � 0 62.
One repeated solution
� 16 � 16 � 0
b2 � 4ac � ��4�2 � 4�1��4�
x2 � 4x � 4 � 0
110 Chapter 1 Equations, Inequalities, and Mathematical Modeling
63.
No real solution
b2 � 4ac � ��5�2 � 4�13��25� � �
253 < 0
13x2 � 5x � 25 � 0 64.
One repeated solution
� 64 � 64 � 0
b2 � 4ac � ��8�2 � 4�47 ��28�
47x2 � 8x � 28 � 0
65.
Two real solutions
b2 � 4ac � �1.2�2 � 4�0.2���8� � 7.84 > 0
0.2x2 � 1.2x � 8 � 0 66.
Two real solutions
� 5.76 � 298.8 � 304.56 > 0
b2 � 4ac � �2.4�2 � 4��8.3��9�
9 � 2.4x � 8.3x2 � 0
67.
��1 ± 3
4�
12
, �1
��1 ± �12 � 4�2���1�
2�2�
x ��b ± �b2 � 4ac
2a
2x2 � x � 1 � 0 68.
�1 ± 3
4� 1, �
1
2
�1 ± �1 � 8
4
����1� ± ���1�2 � 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
70.
�20 ± 10
50�
3
5,
1
5
�20 ± �400 � 300
50
����20� ± ���20�2 � 4�25��3�
2�25�
x ��b ± �b2 � 4ac
2a
25x2 � 20x � 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
73.
��14 ± 2�5
2� �7 ± �5
��14 ± �142 � 4�1��44)
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 14x � 44 � 0 74.
� �3 ± �13
��6 ± 2�13
2
��6 ± �36 � 16
2
��6 ± �62 � 4�1���4�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 6x � 4 � 0
6x � 4 � x2 75.
��8 ± 4�5
2� �4 ± 2�5
��8 ± �82 � 4�1���4�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 8x � 4 � 0
72.
�10 ± 2�3
2� 5 ± �3
�10 ± �100 � 88
2
����10� ± ���10�2 � 4�1��22�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 10x � 22 � 0
Section 1.4 Quadratic Equations and Applications 111
76.
�1
2±�5
2
�1 ± �1 � 4
2
����1� ± ���1�2 � 4�1���1�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � x � 1 � 0
4x2 � 4x � 4 � 0 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 78.
�5
4±�3
4
�20 ± �400 � 352
16
����20� ± ���20�2 � 4�8��11�
2�8�
x ��b ± �b2 � 4ac
2a
8x2 � 20x � 11 � 0
16x2 � 22 � 40x
79.
� �43
��24 ± 0
18
��24 ± �242 � 4�9��16�
2�9�
x ��b ± �b2 � 4ac
2a
9x2 � 24x � 16 � 0 80.
� �1
3±�11
6
��24 ± ��144��11�
72
��24 ± �576 � 1008
72
��24 ± �242 � 4�36���7�
2�36�
x ��b ± �b2 � 4ac
2a
36x2 � 24x � 7 � 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
82.
�5
4±�5
2
�40 ± 16�5
32
�40 ± �1600 � 320
32
����40� ± ���40�2 � 4�16��5�
2�16�
x ��b ± �b2 � 4ac
2a
16x2 � 40x � 5 � 0 83.
��28 ± 0
�98�
27
��28 ± �282 � 4��49���4�
2��49�
x ��b ± �b2 � 4ac
2a
�49x2 � 28x � 4 � 0
28x � 49x2 � 4 84.
��3 ± �13
2� �
3
2±�13
2
��3 ± �32 � 4�1���1�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 3x � 1 � 0
3x � x2 � 1 � 0
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 86.
� �8
5±�3
5
� �8
5±
10�3
50
��80 ± �6400 � 6100
50
��80 ± �802 � 4�25��61�
2�25�
h ��b ± �b2 � 4ac
2a
25h2 � 80h � 61 � 0 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
112 Chapter 1 Equations, Inequalities, and Mathematical Modeling
88.
� �7 ± �13
��14 ± �52
2
��14 ± �142 � 4�1��36�
2�1�
z ��b ± �b2 � 4ac
2a
z2 � 14z � 36 � 0
z2 � 12z � 36 � �2z
�z � 6�2 � �2z 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
90.
�686 ± 196�6
25
�1372 ± �921,984
50
����1372� ± ���1372�2 � 4�25��9604�
2�25�
x ��b ± �b2 � 4ac
2a
25x2 � 1372x � 9604 � 0
2549
x2 � 28x � 196 � 0
2549
x2 � 20x � 196 � 8x
�57
x � 14�2
� 8x 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
92.
� 1.400, �0.150
�2.50 ± �9.61
4
����2.50� ± ���2.50�2 � 4�2���0.42�
2�2�
x ��b ± �b2 � 4ac
2a
2x2 � 2.50x � 0.42 � 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
94.
� 2.137, 18.063
��0.101 ± �0.006341
�0.01
��0.101 ± ��0.101�2 � 4��0.005���0.193�
2��0.005�
x ��b ± �b2 � 4ac
2a
�0.005x2 � 0.101x � 0.193 � 0 95.
x � 1.687, �0.488
x �506 ± ���506�2 � 4�422���347�
2�422�
422x2 � 506x � 347 � 0
Section 1.4 Quadratic Equations and Applications 113
96.
� 0.672, �0.968 ��326 ± �3,252,276
2200
��326 ± ��326�2 � 4�1100���715�
2�1100�
x ��b ± �b2 � 4ac
2a
1100x2 � 326x � 715 � 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
98.
� �2.995, 2.971 �0.08 ± �369.031
�6.44
����0.08� ± ���0.08�2 � 4��3.22��28.651�
2��3.22�
x ��b ± �b2 � 4ac
2a
�3.22x2 � 0.08x � 28.651 � 0 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
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
104. Complete the square.
x � �32 ± �3
x �32 � ±�3
�x �32�2
� 3
x2 � 3x � �32�2
�34 �
94
x2 � 3x �34 � 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
106. Factor.
ax � b � 0 ⇒ x �b
a
ax � b � 0 ⇒ x � �b
a
�ax � b��ax � b� � 0
a2x2 � b2 � 0
100. Factor.
x � �3
x � 0 or x � 3 � 0
x�x � 3� � 0
11�x2 � 3x� � 0
11x2 � 33x � 0 101.
x � 6 or x � �12
x � 3 � 9 or x � 3 � �9
x � 3 � ±9
�x � 3�2 � 81 Extract square roots.
102. Extract square roots.
x � 7
x � 7 � 0
�x � 7�2 � 0
x2 � 14x � 49 � 0
114 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 � 7 108. Factor.
x � 1 x � �1
x � 1 � 0 or x � 1 � 0
�x � 1��x � 1� � 0
4�x2 � 1� � 0
4x2 � 4 � 0
4x2 � 2x � 4 � 2x � 8
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
110. Total fencing:
Total area:
For
There are two solutions: meters and meters or meters and meters.y � 10x � 17.5y �
703
x � 7.5
x � 7.5
2x�703 � � 350
y �703
:
�y � 10 � 0 ⇒ y � 10
3y � 70 � 0 ⇒ y �703
�3y � 70���y � 10� � 0
�3y2 � 100y � 700 � 0
100y � 3y2 � 700 � 0
12
�100y � 3y2� � 350 � 0
2�100 � 3y4 �y � 350
x �100 � 3y
4
2xy � 350
4x � 3y � 100
For
x � 17.5
2x�10� � 350
y � 10:
111.
Since x must be positive, we have inches. Thedimensions 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 112. Volume:
Original size of material:
Side: centimeters
Dimensions: 14 centimeters 14 centimeters�
x � 4 � 10 � 4 � 14
x � ±10
x2 � 100
2x2 � 200
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�
Section 1.4 Quadratic Equations and Applications 115
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
114. Original arrangement: x rows, y seats per row,
New arrangement: rows, seats per row
Originally, there were 8 rows of seats with seats per row.72
8 � 9
x � 6 � 0 ⇒ x � �6
x � 8 � 0 ⇒ x � 8
�x � 8��x � 6� � 0
x2 � 2x � 48 � 0
3x2 � 6x � 144 � 0
72x � 3x2 � 144 � 6x � 72x
�x � 2��72 � 3x� � 72x
x�x � 2��72x
� 3� � 72x
�x � 2��72x
� 3� � 72
�x � 2��y � 3� � 72
�y � 3��x � 2�
y �72x
xy � 72, 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
116. (a)
Since the object was dropped, and the initialheight is Thus,
(b) feet
(c)
It will take the coin about 7.84 seconds to strike the ground.
t ��98416
��246
2� 7.84
t2 �98416
16t2 � 984
0 � �16t2 � 984
s � �16�4�2 � 984 � 728
s � �16t2 � 984.s0 � 984.v0 � 0,
s � �16t2 � v0t � s0 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
116 Chapter 1 Equations, Inequalities, and Mathematical Modeling
118. (a)
Since the object was dropped, and the initial height is Thus,
(b)
(c) The object reaches the ground between seconds and seconds. Numerical approximations will vary, though 10.7 seconds is a reasonable estimate.
(d)
It will take the object about 10.65 seconds to reach the ground.
(e)
The zero of the graph is at seconds.t � 10.65
00 12
2000
t ��181516
�11�15
4� 10.65
t2 �181516
16t2 � 1815
0 � �16t2 � 1815
t � 12t � 10
s � �16t2 � 1815.s0 � 1815.v0 � 0,
s � �16t2 � v0t � s0
Time, t 0 2 4 6 8 10 12
Height, s 1815 1751 1559 1239 791 215 �489
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
120. (a)
The median income reached and surpassed $40,000in 1998, x � 8.65.
5 1330,000
45,000 (b) Substituting into the model gives $40,003.46.Substituting into the model gives $39,940.31.
(c) For 2008, and For 2013, and No. The model shows the median income decreasingafter 2002.
I � $30,609.28.t � 23I � $40,108.68.t � 18
x � 8.6x � 8.65
Section 1.4 Quadratic Equations and Applications 117
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
122. Model:
Labels:
Equation:
Each side of the equilateral triangle is approximately inches long.11.55
s ��4003
�20�3
3� 11.55 inches
s2 �4003
34
s2 � 100
100 �s2
4� s2
102 � �s2�
2� s2
half of side �s2
side � s,height � 10 inches, s
s
10
12
�height�2 � �half of side�2 � �side�2
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�
124. (a) Model:
Labels:
Equation:
(b) When
The boat is approximately feet from the dock when there is 75 feet of rope out.73.5
x � �5400 � 30�6 � 73.5 feet
x2 � 5625 � 225 � 5400
152 � x2 � 752l � 75:
152 � x2 � l2
length of rope � ldistance to dock � x,winch � 15,
�winch�2 � �distance to dock�2 � �length of rope�2
118 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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
126.
x �150,000 ± �150,0002 � 4�550,000,000�
2� 3761 units or 146,239 units
x2 � 150,000x � 550,000,000 � 0
�4x2 � 600,000x � 2,200,000,000 � 0
60x � 0.0004x2 � 220,000 � 0
x�60 � 0.0004x� � 220,000
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 � 14,000 128.
Choosing the positive value for x, 100 units can beproduced for a cost of $11,500.
x � 100, �130
x ��15 ± �152 � 4(0.5���6500�
2�0.5�
0 � 0.5x2 � 15x � 6500
11,500 � 0.5x2 � 15x � 5000
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 130.
Choosing the positive value for x, 48 units canbe produced for a cost of $896.
x � 48, �8
x ����10� ± ���10�2 � 4�1
4���96�2�1
4�
0 � �96 � 10x �x2
4
896 � 800 � 10x �x2
4
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
Section 1.4 Quadratic Equations and Applications 119
132. (a)
Because choose 16.8�C.10 ≤ x ≤ 25,
� �13.1, 16.8
x �1.65 ± ���1.65�2 � 4�0.45���99.25�
2�0.45�
0 � 0.45x2 � 1.65x � 99.25
150 � 0.45x2 � 1.65x � 50.75 (b)
Oxygen consumption is increased by a factor of approximately 2.5.
197.75 � 79.25 � 2.5
x � 20: 0.45�20�2 � 1.65�20� � 50.75 � 197.75
x � 10: 0.45�10�2 � 1.65�10� � 50.75 � 79.25
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
134. (a)
so:
x � 1 � 16.5078 ft
x � 15.5078 ft
4x2 � 4x � 1024 � 0
4x�x � 1� � 1024
V � 1024 ft3,
4
x + 1
x
(b) weighs approximately 62.4 lbs., so
(c)
At 5 gallons per minute:
7660.08 gal5 galsmin
� 1532.02 min � 25.5 hours
1024 ft3
0.13368 ft3�gal� 7660.08 gal
1024 ft3 � 62.4 lbs�ft3 � 63,897.6 lbs.
1 ft3
135. False
so, the quadratic equation has no real solutions.
b2 � 4ac � ��1�2 � 4��3���10� < 0,
136. False. The product must equal zero for the Zero-Factorproperty to be used.
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
138. (a) Neither. A linear equation has at most one real solution, and a quadratic equation has at most two real solutions.
(b) Both
(c) Quadratic
(d) Neither
120 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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
140. (a)
ax � b � 0 ⇒ x � �b
a
x � 0
x�ax � b� � 0
ax2 � bx � 0 (b)
x � 1 � 0 ⇒ x � 1
ax � 0 ⇒ x � 0
ax�x � 1� � 0
ax2 � ax � 0
144. is a factor.
is a factor.
30x2 � 7x � 2 � 0
�6x � 1��5x � 2� � 0
x � �25 ⇒ 5x � �2 ⇒ 5x � 2
x �16 ⇒ 6x � 1 ⇒ 6x � 1
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
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 6 142.
x2 � 15x � 44 � 0
�x � 4��x � 11� � 0
�x � ��4���x � ��11�� � 0
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
146. so:
x2 � 6x � 4 � 0
�x � 3 � �5 ��x � 3 � �5 � � 0
�x � ��3 � �5 ���x � ��3 � �5 �� � 0
x � �3 � �5, x � �3 � �5,
147. by the Associative Property ofMultiplication.�10x�y � 10�xy� 148. by the Distributive Property.�4�x � 3� � �4x � 12
Section 1.5 Complex Numbers 121
149. by the Additive Inverse Property.7x 4 � ��7x 4� � 0
159. Answers will vary.
150. by the AssociativeProperty of Addition.�x � 4� � x3 � x � �4 � x3�
151.
� x2 � 3x � 18
�x � 3��x � 6� � x2 � 6x � 3x � 18 152.
� x2 � 9x � 8
�x � 8��x � 1� � x2 � x � 8x � ��8���1�
153.
� x3 � 3x2 � 2x � 8
� x3 � x2 � 2x � 4x2 � 4x � 8
�x � 4��x2 � x � 2� � x�x2 � x � 2� � 4�x2 � x � 2� 154.
� x3 � 3x2 � 50x � 36
�x � 9��x2 � 6x � 4� � x3 � 6x2 � 4x � 9x2 � 54x � 36
155.
� x2�x � 3��x2 � 3x � 9�
� x2�x3 � 33�
x5 � 27x2 � x2�x3 � 27� 156.
� x�x � 7��x � 2�
x3 � 5x2 � 14x � x�x2 � 5x � 14�
157.
� �x � 5��x2 � 2�
x3 � 5x2 � 2x � 10 � x2�x � 5� � 2�x � 5� 158.
� �x � 2��x � 2��2x � 1�
� �x2 � 4��2x � 1�
� x2�2x � 1� � 4�2x � 1�
2x3 � x2 � 8x � 4 � �2x3 � x2� � �8x � 4�
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
Vocabulary Check
1. (a) iii (b) i (c) ii 2.
3. principal square 4. complex conjugates
��1; �1
122 Chapter 1 Equations, Inequalities, and Mathematical Modeling
1.
b � 6
a � �10
a � bi � �10 � 6i 2.
b � 4
a � 13
a � bi � 13 � 4i 3.
b � 3 � 8 ⇒ b � 5
a � 1 � 5 ⇒ a � 6
�a � 1� � �b � 3�i � 5 � 8i
4.
a � 0
a � 6 � 6
b � �52
2b � �5
�a � 6� � 2bi � 6 � 5i 5. 4 � ��9 � 4 � 3i 6. 3 � ��16 � 3 � 4i
7.
� 2 � 3�3 i
2 � ��27 � 2 � �27 i 8. 1 � ��8 � 1 � 2�2i 9. ��75 � �75 i � 5�3 i
10. ��4 � 2i 11. 8 � 8 � 0i � 8 12. 45
13.
� �1 � 6i
�6i � i2 � �6i � 1 14.
� 4 � 2i
�4i2 � 2i � �4��1� � 2i 15.
� 0.3i
��0.09 � �0.09 i
16. ��0.0004 � 0.02i 17. �5 � i� � �6 � 2i� � 11� i 18. �13 � 2i�� ��5 � 6i� � 8 �4i
19.
� 4
�8 � i� � �4 � i� � 8 � i � 4 � i 20.
� �3 � 11i
�3 � 2i� � �6 � 13i� � 3 � 2i � 6 � 13i
21.
� 3 � 3�2 i
��2 � ��8 � � �5 � ��50 � � �2 � 2�2 i � 5 � 5�2 i
22.
� 4
�8 ���18 � � �4 � 3�2 i� � 8 � 3�2 i � 4 � 3�2 i 23.
� �14 � 20i
13i � �14 � 7i� � 13i � 14 � 7i
24. 22 � ��5 � 8i� � 10i � 17 � 18i 25.
� 16 �
76i
� �96 �
156 i �
106 �
226 i
��32 �
52i� � �5
3 �113 i� � �
32 �
52i �
53 �
113 i
26. �1.6 � 3.2i� � ��5.8 � 4.3i� � �4.2 � 7.5i 27.
� 3 � i � 2 � 5 � i
�1 � i��3 � 2i� � 3 � 2i � 3i � 2i2
28.
� 12 � 22i � 6 � 6 � 22i
�6 � 2i��2 � 3i� � 12 � 18i � 4i � 6i2 29.
� 12 � 30i
6i�5 � 2i� � 30i � 12i2 � 30i � 12
30.
� 32 � 72i
�8i�9 � 4i� � �72i � 32i2 31.
� 14 � 10 � 24
��14 � �10 i���14 � �10 i� � 14 � 10i2
Section 1.5 Complex Numbers 123
32.
� 3 � 15 � 18
� 3 � 15��1�
��3 � �15 i���3 � �15 i� � 3 � 15i2 33.
� �9 � 40i
� 16 � 40i � 25
�4 � 5i�2 � 16 � 40i � 25i 2
34.
� �5 � 12i
� 4 � 9 � 12i
�2 � 3i�2 � 4 � 12i � 9i2 35.
� �10
� 4 � 12i � 9 � 4 � 12i � 9
�2 � 3i�2 � �2 � 3i�2 � 4 � 12i � 9i2 � 4 � 12i � 9i2
36.
� �8i
� 1 � 4i � 4i2 � 1 � 4i � 4i2
�1 � 2i�2 � �1 � 2i�2 � 1 � 4i � 4i2 � �1 � 4i � 4i2� 37. The complex conjugate of is
�6 � 3i��6 � 3i� � 36 � �3i�2 � 36 � 9 � 45
6 � 3i.6 � 3i
38. The complex conjugate of is
� 193
� 49 � ��144�
�7 � 12i��7 � 12i� � 49 � 144i2
7 � 12i.7 � 12i 39. The complex conjugate of is
� 1 � 5 � 6
��1 � �5i���1 � �5i� � ��1�2 � ��5i�2
�1 � �5i.�1 � �5i
40. The complex conjugate of is
� 11
� 9 � ��2�
��3 � �2 i���3 � �2 i� � 9 � 2i2
�3 � �2 i.�3 � �2 i 41. The complex conjugate of is
�2�5i���2�5i� � �20i2 � 20
�2�5i.��20 � 2�5i
42. The complex conjugate of is
��15 i����15 i� � �15i2 � ���15� � 15
��15 i.��15 � �15 i 43. The complex conjugate of is
��8���8� � 8
�8.�8
44. The complex conjugate of is
� 9 � 4�2
�1 � �8��1 � �8� � 1 � 2�8 � 8
1 � �8.1 � �8 45.5i
�5i
��i�i
��5i
1� �5i
46. �14
2i�
�2i
�2i�
28i
�4i 2�
28i
4� 7i 47.
�2�4 � 5i�16 � 25
�8 � 10i
41�
841
�1041
i
2
4 � 5i�
24 � 5i
�4 � 5i4 � 5i
48.5
1 � i�
1 � i
1 � i�
5 � 5i
1 � i2�
5 � 5i
2�
5
2�
5
2i 49.
�9 � 6i � i 2
9 � 1�
8 � 6i10
�45
�35
i
3 � i3 � i
�3 � i3 � i
�3 � i3 � i
50.
�20 � 5i
5�
20
5�
5
5i � 4 � i
6 � 7i
1 � 2i�
1 � 2i
1 � 2i�
6 � 12i � 7i � 14i2
1 � 4i251.
��6i � 5i 2
1� �5 � 6i
6 � 5i
i�
6 � 5ii
��i�i
124 Chapter 1 Equations, Inequalities, and Mathematical Modeling
52.8 � 16i
2i�
�2i�2i
��16i � 32i2
�4i2� 8 � 4i
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
54.
�60 � 25i
169�
60169
�25
169i
��25i � 60i2
25 � 144i2
�5i
�5 � 12i�
�5 � 12i�5 � 12i
5i
�2 � 3i�2 �5i
4 � 12i � 9i2
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�
56.
�12
5�
9
5i
�12 � 9i
5
�4i � 2i2 � 10 � 5i
4 � i2
2i
2 � i�
5
2 � i�
2i�2 � i��2 � i��2 � i�
�5�2 � i�
�2 � i��2 � 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� 58.
�5
17�
20
17i
�5 � 20i
1 � 16i2
�5
1 � 4i�
1 � 4i
1 � 4i
�4 � i � 4i � i2 � 3i
4i � i2
1 � i
i�
3
4 � i�
�1 � i��4 � i� � 3i
i�4 � i�
59.
� �2�3
��6 � ��2 � ��6i���2i� � �12i 2 � �2�3 ���1� 60.
� �50 i2 � 5�2��1� � �5�2
��5 � ��10 � ��5 i���10 i�
61. ���10 �2� ��10 i �2
� 10i2 � �10 62. ���75 �2� ��75 i�2
� 75i2 � �75
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�
Section 1.5 Complex Numbers 125
64.
� �2 � 4�6i
� 4 � 6 � 4�6i
� 4 � 2�6i � 2�6i � 6��1�
� 4 � 2�6i � 2�6i � 6i2
�2 � ��6 �2� �2 � �6i��2 � �6i� 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
66.
� �3 ± i
��6 ± ��4
2
x ��6 ± �62 � 4�1��10�
2�1�
a � 1, b � 6, c � 10x2 � 6x � 10 � 0; 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
68.
�1
3±
36i
18�
1
3± 2i
�6 ± ��1296
18
x ����6� ± ���6�2 � 4�9��37�
2�9�
9x2 � 6x � 37 � 0; a � 9, b � �6, c � 37 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
70.
�1
8±�11
8i
�4 ± 4�11 i
32
�4 ± ��176
32
t ����4� ± ���4�2 � 4�16��3�
2�16�
16t2 � 4t � 3 � 0; a � 16, b � �4, c � 3 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
72.
�37
±�3414
i
�12 ± 2i�34
28
�12 ± ��136
28
x ����12� ± ���12�2 � 4�14��5�
2�14�
14x2 � 12x � 5 � 0; a � 14, b � �12, c � 5
78
x2 �34
x �5
16� 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
126 Chapter 1 Equations, Inequalities, and Mathematical Modeling
74.
�3 ± ��207
9�
3 ± 3i�239
�13
±�23
3i
x ����3� ± ���3�2 � 4�4.5��12�
2�4.5�
4.5x2 � 3x � 12 � 0; a � 4.5, b � �3, c � 12 75.
� �1 � 6i
� 6i � 1
� �6��1�i � ��1�
�6i 3 � i 2 � �6i2i � i2
76. 4i2 � 2i3 � �4 � 2i 77.
� �5i
� �5��1���1�i
�5i5 � �5i2i2i 78. ��i�3 � ��1��i3� � ��1���i� � i
79.
� �375�3 i
� 125�3�3 � ��1�i
� 53��3 �3i 3
���75 �3 � �5�3 i�3 80. ���2 �6� ��2 i�6
� 8i 6 � 8i4i2 � �8
81.1
i3�
1
�i�
1
�i�
i
i�
i
�i2�
i
1� i 82.
1
�2i�3�
1
8i3�
1
�8i�
8i
8i�
8i
�64i2�
1
8i
84. (a)
(b)
(c)
� 8
� �1 � 3�3 i � 9 � 3�3 i
� �1 � 3�3 i � 9i2 � 3�3 i3
��1 � �3 i�3 � ��1�3 � 3��1�2 ���3 i� � 3��1����3 i� 2 � � � �3 i�3
� 8
� �1 � 3�3 i � 9 � 3�3i
� �1 � 3�3 i � 9i2 � 3�3 i3
��1 � �3i�3 � ��1�3 � 3��1�2 ��3i� � 3��1���3i�2 � ��3i� 3
�2�3 � 8
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 86. (a)
(b)
(c)
(d) i67 � �i 4�16�i 3� � �1���i� � �i
i50 � �i 4�12�i 2� � �1���1� � �1
� i
� �1�6i
i 25 � �i 4�6 � i
� 1
� �1�10
i 40 � �i 4�10
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
Section 1.5 Complex Numbers 127
87. False, if then
That is, if the complex number is real, the number equalsits conjugate.
a � bi � a � bi � a.b � 0 88. True
56 � 56
36 � 6 � 14 �?
56
��i�6 �4� ��i�6 �2
� 14 �?
56
x 4 � x2 � 14 � 56
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�
90. ��6��6 � �6i�6i � 6i2 � �6
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
92.
The complex conjugate of this sum is
The sum of the complex conjugates is
Thus, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates.
�a1 � b1i� � �a2 � b2i� � �a1 � a2� � �b1 � b2�i.
�a1 � a2� � �b1 � b2�i.
�a1 � b1i� � �a2 � b2i� � �a1 � a2� � �b1 � b2�i
93. �4 � 3x� � �8 � 6x � x2� � �x2 � 3x � 12
94.
� x3 � x2 � 2x � 6
�x3 � 3x2� � �6 � 2x � 4x2� � x3 � 3x2 � 6 � 2x � 4x2 95.
� 3x2 �232 x � 2
�3x �12��x � 4� � 3x2 � 12x �
12x � 2
96.
� 4x2 � 20x � 25
�2x � 5�2 � �2x�2 � 2�2x��5� � �5�2 97.
x � �31
�x � 31
�x � 12 � 19 98.
x � 14
�3x � �42
8 � 3x � �34
99.
x �272
2x � 27
2x � 27 � 0
20x � 24 � 18x � 3 � 0
4�5x � 6� � 3�6x � 1� � 0 100.
x �40
�30� �
43
�30x � 40
5x � 15x � 55 � 20x � 15
5�x � �3x � 11�� � 20x � 15 101.
a �1
2�3V
�b�
�3V�b
2�b
� 3V
4�b� a
3V
4�b� a2
3V � 4�a2b
V �4
3�a2b
128 Chapter 1 Equations, Inequalities, and Mathematical Modeling
102.
r ���m1m2
F�
��m1m2
�F��F�F
���m1m2F
F
r2 � �m1m2
F
F � �m1m2
r2103. 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 � #
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,
1.
2x2 � 9 � 0 ⇒ x � ±3�2
2
2x2 � 0 ⇒ x � 0
2x2�2x2 � 9� � 0
4x4 � 18x2 � 0
2.
2x � 5 � 0 ⇒ x �52
2x � 5 � 0 ⇒ x � �52
5x � 0 ⇒ x � 0
5x�2x � 5��2x � 5� � 0
5x�4x2 � 25� � 0
20x3 � 125x � 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
4.
x2 � 2x � 4 � 0 ⇒ x � 1 ± �3 i
x � 2 � 0 ⇒ x � �2
x2 � 2x � 4 � 0 ⇒ x � �1 ± �3 i
x � 2 � 0 ⇒ x � 2
�x � 2��x2 � 2x � 4��x � 2��x2 � 2x � 4� � 0
�x3 � 8��x3 � 8� � 0
x6 � 64 � 0
Vocabulary Check
1. polynomial 2. extraneous 3. quadratic type
Section 1.6 Other Types of Equations 129
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
6.
� �43
±4�3 i
3
� �43
±24�3 i
18
��2418
±��1728
18
��24 ± ��1728
18
9x2 � 24x � 64 � 0 ⇒ x ��24 ± ��24�2 � 4�9��64�
2�9�
3x � 8 � 0 ⇒ x �8
3
�3x � 8��9x2 � 24x � 64� � 0
27x3 � 512 � 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 8.
3x � 4 � 0 ⇒ x �43
x2 � 0 ⇒ x � 0
x2�3x � 4�2 � 0
x2�9x2 � 24x � 16� � 0
9x4 � 24x3 � 16x2 � 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 10.
x2 � 3 � 0 ⇒ x � ±�3 i
x � 2 � 0 ⇒ x � �2
�x � 2��x2 � 3� � 0
x2�x � 2� � 3�x � 2� � 0
x3 � 2x2 � 3x � 6 � 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
130 Chapter 1 Equations, Inequalities, and Mathematical Modeling
12.
x � 2 � 0 ⇒ x � �2
x2 � 2x � 4 � 0 ⇒ x � �1 ± �3 i
x � 2 � 0 ⇒ x � 2
�x � 2��x2 � 2x � 4��x � 2� � 0
�x3 � 8��x � 2� � 0
x3�x � 2� � 8�x � 2� � 0
x4 � 2x3 � 8x � 16 � 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
14.
x � 2 � 0 ⇒ x � 2
x � 2 � 0 ⇒ x � �2
x2 � 9 � 0 ⇒ x � ±3i
�x2 � 9��x � 2��x � 2� � 0
�x2 � 9��x2 � 4� � 0
x4 � 5x2 � 36 � 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
16.
t2 � 1 � 0 ⇒ t � ± i
6t � �7 � 0 ⇒ t ��7
6
6t � �7 � 0 ⇒ t � ��7
6
�6t � �7 ��6t � �7 ��t2 � 1� � 0
�36t2 � 7��t2 � 1� � 0
36t4 � 29t2 � 7 � 0
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
18.
x2 � x � 1 � 0 ⇒ x �1 ± �3 i
2
x � 1 � 0 ⇒ x � �1
x2 � 3�2x � � 3�2�2� 0 ⇒ x �
1 ± �3 i
22�3
x � 3�2 � 0 ⇒ x � � 3�2
�x � 3�2 ��x2 � 3�2x � �3�2 �2��x � 1��x2 � x � 1� � 0
�x3 � 2��x3 � 1� � 0
x6 � 3x3 � 2 � 0
Section 1.6 Other Types of Equations 131
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 20.
Let
x
x � 1� �
3
2 ⇒ x � �
3
5
x
x � 1�
2
3 ⇒ x � 2
2u � 3 � 0 ⇒ u � �3
2
3u � 2 � 0 ⇒ u �2
3
�3u � 2��2u � 3� � 0
6u2 � 5u � 6 � 0
u � x��x � 1�.
6� x
x � 1�2
� 5� x
x � 1� � 6 � 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 22.
Let
is not a solution.
�x �32 ⇒ x �
94
�x � �13
2u � 3 � 0 ⇒ u �32
3u � 1 � 0 ⇒ u � �13
�3u � 1��2u � 3� � 0
6u2 � 7u � 3 � 0
u � �x.
6x � 7�x � 3 � 0
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
24.
t � �6427
3t1�3 � 4 � 0 ⇒ t1�3 � �43
�3t1�3 � 4�2 � 0
9t2�3 � 24t1�3 � 16 � 0
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 solutionsto 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
132 Chapter 1 Equations, Inequalities, and Mathematical Modeling
26. (a)
(b) x-intercepts:
(c)
(d) The x-intercepts and the solutions are the same.
x � 0, 32, 6
0 � x � 6 ⇒ x � 6
0 � 2x � 3 ⇒ x �32
0 � x2 ⇒ x � 0
� x2(2x � 3)(x � 6)
� x2(2x2 � 15x � 18)
0 � 2x4 � 15x3 � 18x2
y � 2x4 � 15x3 � 18x2
�0, 0�, �32, 0�, �6, 0�
−3 9
−200
40 27.
(a)
(b) x-intercepts:
(c)
(d) The x-intercepts of the graph are the same as thesolutions 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
�±1, 0�, �±3, 0�
−5 5
20
−20
y � x4 � 10x2 � 9
28. (a)
(b) x-intercepts:
(d) The x-intercepts and the solutions are the same.
��5, 0�, �5, 0���2, 0�, �2, 0�,
−10 10
150
−150
(c)
x � ±5, ±2
0 � x � 5 ⇒ x � 5
0 � x � 5 ⇒ x � �5
0 � x � 2 ⇒ x � 2
0 � x � 2 ⇒ x � �2
� �x � 2��x � 2��x � 5��x � 5�
� �x2 � 4��x2 � 25�
0 � x4 � 29x2 � 100
y � x4 � 29x2 � 100
29.
x � 50
2x � 100
�2x � 10
�2x � 10 � 0 30.
x �916
16x � 9
4�x � 3
4�x � 3 � 0 31.
x � 26
x � 10 � 16
�x � 10 � 4
�x � 10 � 4 � 0
32.
x � �4
5 � x � 9
�5 � x � 3
�5 � x � 3 � 0 33.
x � �16
2x � �32
2x � 5 � �27
33�2x � 5 � �3
3�2x � 5 � 3 � 0 34.
x �1243
3x � 124
3x � 1 � 125
3�3x � 1 � 5
3�3x � 1 � 5 � 0
Section 1.6 Other Types of Equations 133
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 36.
0 � x � 2 ⇒ x � �2
0 � x � 3 ⇒ x � 3
0 � �x � 3��x � 2�
0 � x2 � x � 6
31 � 9x � 25 � 10x � x2
�31 � 9x � 5 � x
x � �31 � 9x � 5
37.
x � 0
�2x � 0
x � 1 � 3x � 1
�x � 1 � �3x � 1 38.
No solution
5 � �5
x � 5 � x � 5
�x � 5 � �x � 5
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 40.
36 � x
16 � x � 20
4 � �x � 20
�80 � �20�x � 20
x � 100 � 20�x � 20 � x � 20
��x �2� �10 � �x � 20�2
�x � 10 � �x � 20
�x � �x � 20 � 10
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 42.
x � 1 � 0 ⇒ x � �1, extraneous
x � 3 � 0 ⇒ x � 3
�x � 3��x � 1� � 0
x2 � 2x � 3 � 0
x2 � 2x � 3
x � �2x � 3
2x � 2�2x � 3
4�x � 1� � 1 � 2�2x � 3 � 2x � 3
�2�x � 1 �2� �1 � �2x � 3 �2
2�x � 1 � 1 � �2x � 3
2�x � 1 � �2x � 3 � 1
134 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 44.
x � 7 � 0 ⇒ x � 7
25x � 79 � 0 ⇒ x �7925, extraneous
�25x � 79��x � 7� � 0
25x2 � 254x � 553 � 0
100x2 � 1016x � 2212 � 0
100x2 � 800x � 1600 � 36�6x � 17�
�10x � 40�2 � �6�6x � 17 �2
10x � 40 � 6�6x � 17
16�x � 3� � 9 � 6�6x � 17 � 6x � 17
�4�x � 3 �2� �3 � �6x � 17 �2
4�x � 3 � 3 � �6x � 17
4�x � 3 � �6x � 17 � 3
45.
x � 5 � 4 � 9
x � 5 � 3�64
�x � 5�3 � 82
�x � 5�3�2 � 8 46.
x � �3 � 4 � 1
x � 3 � 3�64
�x � 3�3 � 64
�x � 3�3 � 82
�x � 3�3�2 � 8 47.
x � �3 ± 16�2
x � 3 � ±�512
x � 3 � ±�83
�x � 3�2 � 83
�x � 3�2�3 � 8
48.
x � �2 ± 27 � �29, 25
x � 2 � ±�729
�x � 2�2 � 93
�x � 2�2�3 � 9 49.
x � ±�14
x2 � 14
x2 � 5 � 9
x2 � 5 � 3�272
�x2 � 5�3 � 272
�x2 � 5�3�2 � 27
50.
x ����1� ± ���1�2 � 4�1���31�
2�1� �1 ± �125
2�
1 ± 5�52
x2 � x � 31 � 0
x2 � x � 22 � 9
x2 � x � 22 � 272�3
�x2 � x � 22�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 52.
5x � 3 � 0 ⇒ x �35
x � 1 � 0 ⇒ x � 1
2x � 0 ⇒ x � 0
2x�x � 1�1�3�5x � 3� � 0
2x�x � 1�1�3�2x � 3�x � 1�� � 0
2x�2x�x � 1�1�3 � 3�x � 1�4�3� � 0
4x2�x � 1�1�3 � 6x�x � 1�4�3 � 0
Section 1.6 Other Types of Equations 135
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 54. (a)
(b) x-intercept:
(c)
(d) The x-intercept and the solution are the same.
x �32
0 � 2x � 3 ⇒ x � 32
0 � 2x � 5 ⇒ x � �52, extraneous
0 � �2x � 5��2x � 3�
0 � 4x2 � 4x � 15
15 � 4x � 4x2
�15 � 4x � 2x
0 � 2x � �15 � 4x
y � 2x � �15 � 4x
�32, 0�
−4 8
−5
3
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 56. (a)
(b) x-intercept:
(c)
(d) The x-intercept and the solution are the same.
x � 4
0 � �x � 2 ⇒ x � 4
0 � 3�x � 2 ⇒ x �4
9, extraneous
0 � �3�x � 2���x � 2�0 � 3x � 4 � 4�x
0 � �x�3�x �4
�x� 4�
0 � 3�x �4
�x� 4
y � 3�x �4
�x� 4
(4, 0)
−2 10
−4
4
136 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
258.
x � 2 � 0 ⇒ x � 2
x � 12 � 0 ⇒ x � �12
�x � 12��x � 2� � 0
x2 � 10x � 24 � 0
24 � 10x � x2
�6x�4
x� �6x�
5
3� �6x�
x
6
4
x�
5
3�
x
6
59.
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 60.
x � 3 � 0 ⇒ x � �3
x � 1 � 0 ⇒ x � 1
�x � 1��x � 3� � 0
x2 � 2x � 3 � 0
4x � 8 � 3x � 3 � x2 � 3x � 2
4�x � 2� � 3�x � 1� � �x � 1��x � 2�
4
x � 1�
3
x � 2� 1
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 62.
x � 1 � 0 ⇒ x � �1
4x � 3 � 0 ⇒ x �3
4
�4x � 3��x � 1� � 0
4x2 � x � 3 � 0
4x2 � x � 3
�x�4x � �x�1 � �x�3
x
4x � 1 �3
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 64.
x � 1 � 0 ⇒ x � 1
x � 1 � 0 ⇒ x � �1
�x � 1��x � 1� � 0
x2 � 1 � 0
x2 � 3x � 2 � 3x � 3 � 0
�x � 2��x � 1) � 3�x � 1� � 0
3�x � 2�x � 1
3� 3�x � 2�
x � 1
x � 2� 0
x � 1
3�
x � 1
x � 2� 0
Section 1.6 Other Types of Equations 137
65.
��2x � 1� � 5 ⇒ x � �2
2x � 1 � 5 ⇒ x � 3
2x � 1 � 5 66.
�3x � 2 � 7 ⇒ x � �3
��3x � 2� � 7
3x � 2 � 7 ⇒ x �53
3x � 2 � 7
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
68.
First equation:
The solutions to the original equation are x � ±3 and x � �6.
x � 6 � 0 ⇒ x � �6
x � 3 � 0 ⇒ x � 3
�x � 3��x � 6� � 0
x2 � 3x � 18 � 0
x2 � 6x � 3x � 18
x2 � 6x � 3x � 18
Second equation:
0 � x � 6 ⇒ x � �6
0 � x � 3 ⇒ x � �3
0 � �x � 3��x � 6�
0 � x2 � 9x � 18
��x2 � 6x� � 3x � 18
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
70.
First equation:
The solutions to the original equation are is extraneous.x � 10 and x � �1. x � 1
0 � x � 10 ⇒ x � 10
0 � x � 1 ⇒ x � 1
0 � �x � 1��x � 10�
0 � x2 � 11x � 10
x � 10 � x2 � 10x
x � 10 � x2 � 10x
Second equation:
0 � x � 1 ⇒ x � �1
0 � x � 10 ⇒ x � 10
0 � �x � 10��x � 1�
0 � x2 � 9x � 10
��x � 10� � x2 � 10x
138 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 72.
(a)
(b) x-intercept:
(c)
(d) The x-intercept and the solution are the same.
x � 2
0 � x � 2 ⇒ x � 2
0 � �x � 2��x � 2�
0 � x2 � 4x � 4
0 � x2 � x � 9 � 5x � 5
0 � x�x � 1� � �x � 1�9
x � 1� 5�x � 1�
0 � x �9
x � 1� 5
0 � x �9
x � 1� 5
�2, 0�
−15 15
−30
20
y � x �9
x � 1� 5
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 74.
(a)
(b) x-intercepts:
(c)
OR
(d) The x-intercepts and the solutions are the same.
x � 2 � 3 ⇒ x � 5
3 � x � 20 � x � 2 � 3
�5, 0�, ��1, 0�
−7 11
−5
7
y � x � 2 � 3
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
x � 5, �1
⇒ x � �1 �x � 2 � 3
��x � 2� � 3
Section 1.6 Other Types of Equations 139
76.
x � 3��4.15 � �289.0945
14.16 �1.143
x � 3��4.15 � �289.0945
14.16 0.968
��4.15 ± �289.0945
14.16
x3 ��4.15 ± ��4.15�2 � 4�7.08���9.6�
2�7.08�
a � 7.08, b � 4.15, c � �9.6
7.08x6 � 4.15x3 � 9.6 � 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
78.
x � ��8 � �6.4
8 �3
�2.280
x � ��8 � �6.4
8 �3
�0.320
x1�3 ��8 ± �82 � 4�4��3.6�
2�4�
a � 4, b � 8, c � 3.6
4x2�3 � 8x1�3 � 3.6 � 0 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
80.
Any non-zero multiple of this equation would also havethese solutions.
x3 � 8x2 � 15x � 0
x�x2 � 8x � 15� � 0
x�x � 3��x � 5� � 0
�x � 0��x � 3��x � 5� � 0
0, 3, 5 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
82.
Any non-zero multiple of this equation would also havethese solutions.
40x2 � 37x � 4 � 0
40x2 � 32x � 5x � 4 � 0
x2 �45x �
18x �
440 � 0
�x �18��x �
45� � 0
�x � ��18���x � ��4
5�� � 0
�18, �4
5 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,
84.
Any non-zero multiple of this equation would also havethese solutions.
x2 � x�7 � 14 � 0
x2 � x�7 � 2x�7 � 2�7� � 0
�x � 2�7 ��x � �7 � � 0
2�7, ��7 85. and
One possible equation is:
Any non-zero multiple of this equation would also havethese 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,
140 Chapter 1 Equations, Inequalities, and Mathematical Modeling
86.
x 4 � 20x2 � 576 � 0
�x2 � 16��x2 � 36� � 0
�x � 4i��x � 4i��x � 6��x � 6� � 0
4i, �4i, 6, �6
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 �
88. Model:
Labels: Monthly rent
Number of students
Original cost per student
Cost per student
Equation:
The monthly rent is $900.
x � 900
x3
� 300
4x3
� x � 300
4x3
� 300 � x
�x3
� 75��4� � x
�x3
� 75
�x3
� 4
� x
�Cost perstudent � � �Number of
students � � �Monthlyrent � 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.
Equation:
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
t �145
x � 40�
1
5.
t �12
60�
145
x � 40
t � 145�x.x �
Time �Distance
Rate
Section 1.6 Other Types of Equations 141
90. Model:
Labels: Distance 1080
Original time
Original rate
Return time
Return rate
Equation:
The original rate was miles per hour.108020 � 54
t � 20 � 0 ⇒ t � 20
2t � 45 � 0 ⇒ t � �22.5
�2t � 45��t � 20� � 0
2t2 � 5t � 900 � 0
2700 � 6t2 � 15t � 0
2700
t� 6t � 15 � 0
1080 �2700
t� 6t � 15 � 1080
�1080
t� 6��t � 2.5� � 1080
�1080
t� 6
� t � 2.5
�1080
t
� t
�
�Rate� � �time� � �distance� 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
92. (a) Use
(b)
4��16
5 �1�80
� 1� � r 0.059 or 5.9%
�16
5 �1�80
� 1 �r
4
16
5� �1 �
r
4�80
32,000 � 10,000�1 �r
4�80
4��5
2�1�80
� 1� � r 0.046 or 4.6%
�5
2�1�80
� 1 �r
4
�5
2�1�80
� 1 �r
4
5
2� �1 �
r
4�80
25,000 � 10,000�1 �r
4�80
25,000 � 10,000�1 �r
4�4�20
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
142 Chapter 1 Equations, Inequalities, and Mathematical Modeling
94. (a) million when:
So the total voting-age population reached 200 million during 1998.
t � 8.7
2.011t � 17.55
182.45 � 3.189t � 200 � 5.2t
182.45 � 3.189t � 200�1.00 � 0.026t�
182.45 � 3.189t1.00 � 0.026t
� 200
P � 200 (b) For
The model predicts that the total voting-age populationwill reach 230 million during 2007. This value is reason-able but the model is reaching its limit since it soonbegins to rise very fast due to its asymptotic behavior.
t � 17
so 2.791t � 47.55
182.45 � 3.189t � 230 � 5.98
182.45 � 3.189t � 230�1.00 � 0.026t�
182.45 � 3.189t1.00 � 0.026t
� 230
P � 230:
95.
(a)
(b) when pounds per square inch.x 15T � 212�
T � 75.82 � 2.11x � 43.51�x, 5 ≤ x ≤ 40
96. When we have:
x � 26.25 � 26,250 passengers
5.25 � 0.2x
6.25 � 0.2x � 1
2.5 � �0.2x � 1
C � 2.5 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
98. When we have:
So the demand is 249,900 units when the price is $750.
x � 249,900
0.01x � 2499
2500 � 0.01x � 1
50 � �0.01x � 1
750 � 800 � �0.01x � 1
p � $750,
(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
Section 1.6 Other Types of Equations 143
100.
(a)
(b)
when is between 170 and 175 feet.hd � 200
d � 200 when h 173 feet.
02400
300
d � �1002 � h2
(c)
(d) Solving graphically or numerically yields an approximation. An exact solution is obtained algebraically.
h 173.2 feet
h � �30,000
30,000 � h2
40,000 � 10,000 � h2
200 � �1002 � h2
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 � 642�64 � h2�
�350�2 � �8�64 � h2 �2
350 � 8�64 � h2
102. Let the number of hours required for the faster person to complete the task alone. Then the number of hours needed for the slower person to complete the same task alone. In one hour, the faster person completes of the task, the slower person of the task, and together they complete of the task.
By completing the square, we have Choosing the positive value for x, we have the following times for each person:
Faster person: slower person: 9 � �65 17 hours 7 � �65 15 hours;
x � 7 ± �65.
0 � x2 � 14x � 16
8x � 16 � 8x � x2 � 2x
8�x � 2� � 8x � x�x � 2�
1x
�1
x � 2�
18
1�81��x � 2�1�x
x � 2 �x �
160 165 170 175 180 185
188.7 192.9 197.2 201.6 205.9 210.3d
h
8 9 10 11 12 13
284.3 302.6 321.9 341.8 362.5 383.6S
h
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
144 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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�
104. (a) Verbal Model: Total cost Cost underwater Distance underwater Cost overland Distance overland
Labels: Total cost:
Cost overland: $24 per foot
Distance overland in feet:
Cost underwater: $30 per foot
Distance underwater in feet:
Equation:
(b)
So, the total cost when is $1,123,424.95.
(c)
By the Quadratic Formula, we have miles or mile.
—CONTINUED—
x 0.382x 2
0 � 144x2 � 343.04x � 110.0816
256x2 � 343.04x � 114.9184 � 400x2 � 225
256x2 � 343.04x � 114.9184 � 25�16x2 � 9�
�16x � 10.72�2 � �5�16x2 � 9�2
16x � 10.72 � 5�16x2 � 9
138.72 � 128 � 16x � 5�16x2 � 9
138.72 � 16�8 � x� � 5�16x2 � 9
1,098,662.40 � 7920�16�8 � x� � 5�16x2 � 9� 1,098,662.40 � 126,720�8 � x� � 39,600�16x2 � 9
x � 3
1,123,424.95
633,600 � 489,824.95
� 126,720�5� � 39,600�153
C � 126,720�8 � 3� � 39,600�16�3�2 � 9
C � 126,720�8 � x� � 39,600�16x2 � 9
C � 24�5280�8 � x�] � 30�1320�16x2 � 9�� 1320�16x2 � 95280�x2 � �3�4�2
5280�8 � x�
C
����
Section 1.6 Other Types of Equations 145
104. —CONTINUED—
(d)
00 10
1,500,000 (e) The cost is a minimum when (The minimum costis $1,085,040.00.)
x � 1.
105.
v2s
R� g
v2s � gR
v2 �gRs
v ��gRs
106.
±�LCi2 � q � Q
LCi2 � q � Q2
LCi2 � Q2 � q
i2 �1
LC�Q2 � q�
i2 � �±� 1
LC�Q2 � q�2
i � ±� 1
LC�Q2 � q
107. False—See Example 7 on page 137. 108. False. has only one solution to check, 0.x � 0
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� 110. The distance between is 13.
Both and are a distance of 13from ��8, 0�.
�4, 5���20, 5�
x � 4 � 0 ⇒ x � 4
x � 20 � 0 ⇒ x � �20
�x � 20��x � 4� � 0
x2 � 16x � 80 � 0
x2 � 16x � 64 � 25 � 169
�x � 8�2 � 52 � 132
��x � 8�2 � �5 � 0�2 � 13
��8, 0� and �x, 5�
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� 112. The distance between is 17.
Both and are a distance of 17from ��8, 4�.
�7, 12��7, �4�
y � 4 ± 8 � �4, 12
y � 4 � ±8
�y � 4�2 � 64
225 � �y � 4�2 � 289
�15�2 � �y � 4�2 � 172
��7 � ��8��2 � �y � 4�2 � 17
��8, 4� and �7, y�
146 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 114. Isolate the absolute value by subtracting x from bothsides of the equation. The expression inside the absolutevalue signs can be positive or negative, so two separateequations must be solved. Each solution must bechecked since extraneous solutions may be included.
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 116. First isolate the radical by subtracting x from both sidesof the equation. Then, square both sides and solve theresulting equation using the Quadratic Formula. Eachsolution must be checked since extraneous solutions maybe included.
117.83x
�32x
�166x
�96x
�256x
118.
�x � 4
�x � 1�(x2 � 4�
�2x � 2 � x � 2
�x � 1��x2 � 4�
�2�x � 1� � �x � 2�
�x � 1��x � 2��x � 2�
2
x2 � 4�
1
x2 � 3x � 2�
2
�x � 2��x � 2��
1
�x � 1��x � 2�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
Section 1.7 Linear Inequalities in One Variable 147
120.
�125y
x, y � 0
25y2 �xy
5�
25y2
1�
5
xy121.
x � 11
x � 11 � 0
�x � 11�2 � 0
x2 � 22x � 121 � 0
122.
x � 20 � 0 ⇒ x � 20
x � 3 � 0 ⇒ x � �3
�x � 3��x � 20� � 0
x�x � 20� � 3�x � 20� � 0
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� 2. Interval:
(a) Inequality:
(b) The interval is bounded.
2 < x ≤ 10
�2, 10� 3. Interval:
(a) Inequality:
(b) The interval is unbounded.
x > 11
�11, ��
4. Interval:
(a) Inequality:
(b) The interval is unbounded.
�5 ≤ x < � or x ≥ �5
��5, �� 5. Interval:
(a) Inequality:
(b) The interval is unbounded.
x < �2
���, �2� 6. Interval:
(a) Inequality:or
(b) The interval is unbounded.
x ≤ �7�� < x ≤ 7
���, 7�
7.
Matches (b).
x < 3 8.
Matches (f).
x ≥ 5 9.
Matches (d).
�3 < x ≤ 4 10.
Matches (c).
0 ≤ x ≤ 92
Vocabulary Check
1. solution set 2. graph 3. negative
4. equivalent 5. double 6. union
148 Chapter 1 Equations, Inequalities, and Mathematical Modeling
11.
Matches (e).
�x� < 3 ⇒ �3 < x < 3 12.
Matches (a).
�x� > 4 ⇒ x > 4 or x < �4
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
14.
(a)
No, is nota solution.
x � 0
1 <� �3
2�0� � 1 <?
�3
x � 0
2x � 1 < �3
(b)
No, is nota solution.
x � �14
12 <� �3
2��14� � 1 <
?�3
x � �14 (c)
Yes, is asolution.
x � �4
�7 < �3
2��4� � 1 <?
�3
x � �4 (d)
No, is nota solution.
x � �32
�2 <� �3
2��32� � 1 <
?�3
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
16.
(a)
No, is not asolution.
x � 0
�1 <32
≤� 1
�1 <? 3 � 0
2≤? 1
x � 0
�1 <3 � x
2≤ 1
(b)
No, is not asolution.
x � �5
�1 < 4 ≤ 1
�1 <? 3 � 5
2≤? 1
x � �5 (c)
Yes, is a solution.x � 1
�1 < 1 ≤ 1
�1 <? 3 � 1
2≤? 1
x � 1 (d)
No, is not a solution.
x � 5
�1 <?
�1 ≤ 1
�1 <? 3 � 5
2≤? 1
x � 5
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
Section 1.7 Linear Inequalities in One Variable 149
18.
(a)
No, is nota solution.
x � �6
15 <� 15
�2��6� � 3� <?
15
x � �6
�2x � 3� < 15
(b)
Yes, is asolution.
x � 0
3 < 15
�2�0� � 3� <?
15
x � 0 (c)
No, is nota solution.
x � 12
21 <� 15
�2�12� � 3� <?
15
x � 12 (d)
Yes, is asolution.
x � 7
11 < 15
�2�7� � 3� <?
15
x � 7
19.
1 2 3 4 5
x
x < 3
14�4x� < 14�12�
4x < 12 20.
−2−6 −5 −4 −3
x
x < �4
10x < �40 21.
2 310−1
x
−2
32
x < 32
�12��2x� < ��1
2���3�
�2x > �3
22.
2− 5
−1−2−3−4
x
x < �156 or x < �
52
�6x > 15 23.
10 11 12 13 14
x
x ≥ 12
x � 5 ≥ 7 24.
3 4 5 6 7
x
x ≤ 5
x � 7 ≤ 12
25.
0 1 2 3 4
x
x > 2
�2x < �4
2x � 7 < 3 � 4x 26.
−1 0 1
1
2
2x
x ≥ 12
2x ≥ 1
3x � 1 ≥ 2 � x 27.
−2 −1 10 2
x
27
x ≥ 27
7x ≥ 2
2x � 1 ≥ 1 � 5x
28.
x
−1−2−3−4−5
x ≥ �3
�2x ≤ 6
6x � 4 ≤ 2 � 8x 29.
43 5 6 7
x
x < 5
4 � 2x < 9 � 3x
4 � 2x < 3�3 � x� 30.
−2
2
−1
−1
0 1x
x < �12
2x < �1
4x � 4 < 2x � 3
4�x � 1� < 2x � 3
31.
432 5 6
x
x ≥ 4
�14x ≤ �1
34x � 6 ≤ x � 7 32.
5 6 7 8 9
x
x < 7
�5x > �35
21 � 2x > 7x � 14
3 �27x > x � 2 33.
210 3 4
x
x ≥ 2
4x �12 ≥ 3x �
52
12�8x � 1� ≥ 3x �52
150 Chapter 1 Equations, Inequalities, and Mathematical Modeling
34.
0 21 3
x
−1
16
x > 16
�12x < �2
36x � 4 < 48x � 6
9x � 1 < 34�16x � 2� 35.
−2−3−4−5−6
x
x ≥ �4
3.6x ≥ �14.4
3.6x � 11 ≥ �3.4 36.
15 16 17 18
x
14
x > 16
�1.3x < �20.8
15.6 � 1.3x < �5.2
37.
30 1 2−1
x
�1 < x < 3
�2 < 2x < 6
�1 < 2x � 3 < 9 38.
10−1−2−3−4−5−6
x
�6 < x ≤ 1
�3 ≤ �3x < 18
�8 ≤ �3x � 5 < 13
�8 ≤ ��3x � 5� < 13 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
40.
−2−4
−3 7
86420
x
�3 ≤ x < 7
0 ≤ x � 3 < 10
0 ≤x � 3
2< 5 41.
0 1−1
−
x
−43
41
� 34
< x < �14
�14
> x > �34
34
> x � 1 >14
42.
1 3 5 7 9 11
x
3 < x < 9
�9 < �x < �3
�3 < 6 � x < 3
�1 < 2 �x3
< 1
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 44.
There is no solution.
2 > x > 10
3 > 1.5x > 15
9 > 1.5x � 6 > 21
4.5 >1.5x � 6
2> 10.5 45.
−2−4−6 0 2 4 6
x
�6 < x < 6
�x� < 6
46.
−6 −4 −2 0 2 4 6
x
x < �4 or x > 4
�x� > 4 47.
−2−3 −1 0 1 2 3
x
x < �2 x > 2
x2
< �1 or x2
> 1
�x2� > 1 48.
−20 −10 0 10
x
20
x < �15 x > 15
x5
< �3 or x5
> 3
�x5� > 3
49.
No solution. The absolute value of a number cannot beless than a negative number.
�x � 5� < � 1 50. There is no solution because the absolute value of a num-ber cannot be less than a negative number.
Section 1.7 Linear Inequalities in One Variable 151
51.
201510 25 30
x
14 26
14 ≤ x ≤ 26
�6 ≤ x � 20 ≤ 6
�x � 20� ≤ 6 52.
All real numbers x
1 2 30−3 −2 −1
x
x ≤ 8
�x ≥ �8
x ≥ 8 �x � 8 ≥ 0
x � 8 ≥ 0 or ��x � 8� ≥ 0
�x � 8� ≥ 0
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 54.
−1−2 10 2 3
x
�2 < x < 3
3 > x > �2
�6 < �2x < 4
�5 < 1 � 2x < 5
�1 � 2x� < 5
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 56.
−1 0 1 2 3 4
x
0 < x < 3
3 > x > 0
�2 < �2x
3< 0
�1 < 1 �2x
3< 1
�1 �2x
3 � < 1
57.
x
3 654
4 < x < 5
5 > x > 4
�10 < �2x < �8
�1 < 9 � 2x < 1
�9 � 2x� < 1
�9 � 2x� � 2 < �1 58.
−35 −28 −21 −14 70−7
x
x < �28 x > 0
x � 14 < �14 or x � 14 > 14
�x � 14� > 14
�x � 14� � 3 > 17
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 60.
15 ≤ x ≤ 75
75 ≥ x ≥ 15
�7 ≤ �5x ≤ �1
�3 ≤ 4 � 5x ≤ 3
�4 � 5x� ≤ 32
755
1
1
0
x
3�4 � 5x� ≤ 9
152 Chapter 1 Equations, Inequalities, and Mathematical Modeling
61.
−10 10
−10
10
6x > 2
6x > 12 62.
−10 10
−10
10
x ≤ 2
3x ≤ 6
3x � 1 ≤ 5 63.
−10 10
−10
10
5 � 2x ≤ 2
5�2x ≥ �4
5 � 2x ≥ 1
64.
−10 10
−10
10
x < 2
2x < 4
3x � 3 < x � 7
3�x � 1� < x � 7 65.
10
−10
−10 24
�6 ≤ x ≤ 22
�14 ≤ x � 8 ≤ 14
�x � 8� ≤ 14 66.
−15 9
−10
10
x < �11 x > 2
2x < �22 2x > 4
2x � 9 < �13 or 2x � 9 > 13
�2x � 9� > 13
67.
or
−15 1
−10
10
x ≥ �12 x ≤ �
272
x � 7 ≥ 132 x � 7 ≤ �
132
�x � 7� ≥ 132
2�x � 7� ≥ 13 68.
−10 10
−10
10
�7 ≤ x ≤ 5
�6 ≤ x � 1 ≤ 6
�x � 1� ≤ 6
12�x � 1� ≤ 3
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 70.
(a)
(b)
x ≥ �32
23x ≥ �1
23x � 1 ≥ 0
y ≥ 0
x ≤ 6
23x ≤ 4
23x � 1 ≤ 5
y ≤ 5
−5 7
−2
6y �23x � 1
Section 1.7 Linear Inequalities in One Variable 153
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 72.
(a)
(b)
x ≥ 83
�3x ≤ �8
�3x � 8 ≤ 0
y ≤ 0
53 ≤ x ≤ 3
3 ≥ x ≥ 53
�9 ≤ �3x ≤ �5
�1 ≤ �3x � 8 ≤ 3
�1 ≤ y ≤ 3
−3
9−9
9y � �3x � 8
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� 74.
(a)
(b)
x ≤ �4 x ≥ 0
12x ≤ �2 12x ≥ 0
12x � 1 ≤ �1 or 12x � 1 ≥ 1
�12x � 1� ≥ 1
y ≥ 1
�10 ≤ x ≤ 6
�5 ≤ 12x ≤ 3
�4 ≤ 12x � 1 ≤ 4
�12x � 1� ≤ 4
y ≤ 4
−5 1
−1
3y � �12x � 1�
75.
�5, ��
x ≥ 5
x � 5 ≥ 0 76.
�10, ��
x ≥ 10
x � 10 ≥ 0
�x � 10 77.
��3, ��
x ≥ �3
x � 3 ≥ 0 78.
���, 3�
3 ≥ x
3 � x ≥ 0
�3 � x
79.
���, 72� x ≤ 7
2
�2x ≥ �7
7 � 2x ≥ 0 80.
��52, ��
x ≥ �52
6x ≥ �15
6x � 15 ≥ 0
4�6x � 15 81.
All real numbers within8 units of 10.
�x � 10� < 8 82.
All real numbers morethan 4 units from 8
�x � 8� > 4
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� 84. The graph shows all real numbersmore than 3 units from 0.
�x� > 3
�x � 0� > 3
85. The graph shows all real numbersat least 3 units from 7.
�x � 7� ≥ 3
154 Chapter 1 Equations, Inequalities, and Mathematical Modeling
86. The graph shows all real numbersno more than 4 units from
�x � 1� ≤ 4
�1.87. All real numbers within 10 units
of 12
�x � 12� < 10
88. All real numbers at least 5 unitsfrom 8
�x � 8� ≥ 5
89. All real numbers more than 4 units from
�x � 3� > 4
�x � ��3�� > 4
�3 90. All real numbers no more than 7 units from
�x � 6� ≤ 7
�6
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. 92.
You must make more than 42,857 copies to justify buyingthe copier.
42,857 < x
3000 < 0.07x
3000 � 0.03x ≤ 0.1x
93.
r > 3.125%
r > 0.03125
2r > 0.0625
1 � 2r > 1.0625
1000�1 � r�2�� > 1062.50 94.
The rate must be more than 5%.
0.05 < r
75 < 1500r
825 < 750 � 1500r
825 < 750�1 � 2r�
825 < 750�1 � r�2��
95.
120.95x ≥ 36 units
120.95x > 35.7995
120.95x > 750
115.95x > 95x � 750
115.95R > C 96.
Because the number of units x must be an integer, theproduct will return a profit when at least 16,394 units are sold.
x > 16,393.44262
9.15 > 150,000
24.55x > 15.4x � 150,000
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. 98. The goal is to lose pounds. At poundsper week, it will take 24 weeks.
� 24
� 12 2
36 � 112 � 36
23
112164 � 128 � 36
(e) An athlete’s weight is not a particularly good indicator of the athlete’s maximum bench pressweight. Other factors, such as muscle tone and exercise habits, influence maximum bench pressweight.
140140 260
300
Section 1.7 Linear Inequalities in One Variable 155
99. (a)
075 150
5
y � 0.067x � 5.638 (b) From the graph we see that when Algebraically we have:
IQ scores are not a good predictor of GPAs. Other factorsinclude study habits, class attendance, and attitude.
x ≥ 129
8.638 ≤ 0.067x
3 ≤ 0.067x � 5.638
x ≥ 129.y ≥ 3
100. (a) and (b)
(c) One estimate is pounds.
(d)
x ≥ 181.5385 � 181.54 pounds
1.3x ≥ 236
1.3x � 36 ≥ 200
x ≥ 181
x 165 184 150 210 196 240
y 170 185 200 255 205 295
179 203 159 237 219 2761.3x � 36
x 202 170 185 190 230 160
y 190 175 195 185 250 155
227 185 205 211 263 1721.3x � 36
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 ≤ 12 102. (a)
The number of eggs produced was between 70 and80 billion from late 1991 until late 1997.
(b) when
So the annual egg production will exceed 95 billionin 2007.
t ≥ 16.95.E � 1.64t � 67.2 ≥ 95
1.7 ≤ t ≤ 7.8
2.8 ≤ 1.64t ≤ 12.8
70 ≤ 1.64t � 67.2 ≤ 80
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� ≤ 116 104.
The interval containing the possible side lengths s in centimeters of the square is so the inter-val containing the possible areas in square centimeters is
or �573.6025, 597.8025�.�23.952, 24.452�,
�23.95, 24.45�,
23.95 ≤ s ≤ 24.45
24.2 � 0.25 ≤ s ≤ 24.2 � 0.25
156 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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� ≤ 110 106. 1 oz lb, so oz lb
so you could have been under- or overcharged as much as $0.47.14.99 � 1
32 � 0.4684375,
�132
12�
116
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 � < 1 108.
6968676665 70 71 72h
71.265.8
65.8 inches ≤ h ≤ 71.2 inches
�2.7 ≤ h � 68.5 ≤ 2.7
�1 ≤h � 68.5
2.7≤ 1
�h � 68.5
2.7 � ≤ 1
109.
The minimum relative humidity is 20 and the maximumis 80.
20 ≤ h ≤ 80
�30 ≤ h � 50 ≤ 30
�h � 50� ≤ 30 110. (a) Estimate from the graph: when the plate thickness is2 millimeters, the frequency is approximately 330vibrations per second.
(b) Estimate from the graph: when the frequency is 600,the plate thickness is approximately 3.6 millimeters.
(c) Estimate from the graph: when the frequency isbetween 200 and 400 vibrations per second, the platethickness is between 1.2 and 2.4 millimeter.
(d) Estimate from the graph: when the plate thickness is less than 3 millimeters, the frequency is less than500 vibrations per second.
114. must be greater than or equal to zero.
Let then and This is true when One set of values is
(Note: This solution is not unique. Any positive multiple of these values will also work, such as or a � 3, b � c � 15.�b � c � 10a � 2,
c � 5.b � 5,a � 1,b � c � 5.b � c � 10.b � c � 0a � 1,
b � c ≤ ax ≤ b � c
�c ≤ ax � b ≤ c
�ax � b� ≤ c ⇒ c
111. False. If c is negative, thenac ≥ bc.
112. False. If thenand �x ≥ �8.10 ≥ �x
�10 ≤ x ≤ 8, 113. Matches (b).
or
x ≥ a � 2
x � a ≥ 2
x ≤ a � 2
x � a ≤ �2
�x � a� ≥ 2
115. and
Midpoint: �4 � 12
, 2 � 12
2 � �32
, 7d � ��1 � ��4��2 � �12 � 2�2 � �52 � 102 � �125 � 5�5
�1, 12���4, 2�
Section 1.8 Other Types of Inequalities 157
116.
Midpoint: �1 � 102
, �2 � 3
2 � � �112
, 12�
d � ��1 � 10�2 � ��2 � 3�2 � ���9�2 � ��5�2 � �81 � 25 � �106
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�
118.
Midpoint: �0 � 62
, �3 � 9
2 � � ��62
, 62� � ��3, 3�
d � ��0 � ��6��2 � ��3 � 9�2 � �62 � ��12�2 � �36 � 144 � �180 � 6�5
119.
x � 10
6x � 60
6x � 24 � 36
�12 � 6x � 12 � 36
�6�2 � x� � 12 � 36 120.
x �132
�2x � �13
4x � 28 � 9 � 6x � 6
4�x � 7� � 9 � �6��x � 1�
121.
or
x � �12 x �
17
2x � �1 7x � 1
2x � 1 � 0 7x � 1 � 0
�7x � 1��2x � 1� � 0
14x2 � 5x � 1 � 0 122.
x2 � 4 � 0 ⇒ x � ±2
x � 5 � 0 ⇒ x � �5
�x � 5��x2 � 4� � 0
x2�x � 5� � 4�x � 5� � 0
x3 � 5x2 � 4x � 20 � 0
123. ��3, 10� 125. Answers will vary.124. If then the point couldbe located in either Quadrant I or II.
�x, y�y > 0,
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
158 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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
2.
(a)
Yes, is a solution.
x � 5
8 ≥ 0
�5�2 � �5� � 12 ≥?
0
x � 5
x2 � x � 12 ≥ 0
(b)
No, is not a solution.
x � 0
�12 ≥� 0
�0�2 � 0 � 12 ≥?
0
x � 0 (c)
Yes, is a solution.x � �4
8 ≥ 0
16 � 4 � 12 ≥?
0
��4�2 � ��4� � 12 ≥?
0
x � �4 (d)
Yes, is a solution.x � �3
0 ≥ 0
9 � 3 � 12 ≥?
0
��3�2 � ��3� � 12 ≥?
0
x � �3
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
5
17≥� 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
4.
(a)
No, is nota solution.
x � �2
12
8<� 1
3��2�2
��2�2 � 4<?
1
x � �2
3x2
x2 � 4< 1
(b)
Yes, is a solution.
x � �1
3
5< 1
3��1�2
��1�2 � 4<?
1
x � �1 (c)
Yes, is a solution.
x � 0
0 < 1
3�0�2
�0�2 � 4<?
1
x � 0 (d)
No, is not a solution.
x � 3
27
13<� 1
3�3�2
�3�2 � 4<?
1
x � 3
5.
Critical numbers: x � �32
, x � 2
x � 2 � 0 ⇒ x � 2
2x � 3 � 0 ⇒ x � �32
2x2 � x � 6 � �2x � 3��x � 2� 6.
The critical numbers are 0 and 25
9.
9x � 25 � 0 ⇒ x �25
9
x2 � 0 ⇒ x � 0
x2�9x � 25� � 0
9x3 � 25x2 � 0 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
Section 1.8 Other Types of Inequalities 159
8.
The critical numbers are �2, �1, 1, 4.
x � 1 � 0 ⇒ x � 1
x � 2 � 0 ⇒ x � �2
�x � 2��x � 1� � 0
x � 1 � 0 ⇒ x � �1
x � 4 � 0 ⇒ x � 4
�x � 4��x � 1� � 0
��x � 4��x � 1��x � 2��x � 1�
�x2 � x � 2x � 4
�x � 2��x � 1�
x
x � 2�
2
x � 1�
x�x � 1� � 2�x � 2��x � 2��x � 1�
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
10.
Critical numbers:
Test intervals:
Solution interval:
x
−8 −6 −4 −2 0 2 4 6 8
��6, 6�
�6, �� ⇒ �x � 6��x � 6� > 0
��6, 6� ⇒ �x � 6��x � 6� < 0
���, �6� ⇒ �x � 6��x � 6� > 0
x � �6, x � 6
�x � 6��x � 6� < 0
x2 � 36 < 0
x2 < 36 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
12.
Critical numbers:
Test intervals:
Solution intervals:
31 2 54
x
���, 2� � �4, ��
�4, �� ⇒ �x � 2��x � 4� > 0
�2, 4� ⇒ �x � 2��x � 4� < 0
���, 2� ⇒ �x � 2��x � 4� > 0
x � 2, x � 4
�x � 2��x � 4� ≥ 0
x2 � 6x � 8 ≥ 0
�x � 3�2 ≥ 1
160 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 14.
Critical numbers:
Test intervals:
Solution interval:
−2
−1
2 40 6
7
8
x
��1, 7�
�7, �� ⇒ �x � 1��x � 7� > 0
��1, 7� ⇒ �x � 1��x � 7� < 0
���, �1� ⇒ �x � 1��x � 7� > 0
x � �1, x � 7
�x � 1��x � 7� < 0
x2 � 6x � 7 < 0
x2 � 6x � 9 < 16
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 16.
Critical numbers:
Test intervals:
Solution intervals:
−1−2−3−4 0 1 2
x
���, �3� � �1, ��
�1, �� ⇒ �x � 3��x � 1� > 0
��3, 1� ⇒ �x � 3��x � 1� < 0
���, �3� ⇒ �x � 3��x � 1� > 0
x � �3, x � 1
�x � 3��x � 1� > 0
x2 � 2x � 3 > 0
x2 � 2x > 3
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 18.
Critical numbers:
Test intervals:
Solution intervals:
0−2−4 2 4 6 8
x2 − 5 2 + 5
���, 2 � �5� � �2 � �5, ���2 � �5, �� ⇒ x2 � 4x � 1 > 0
�2 � �5, 2 � �5 � ⇒ x2 � 4x � 1 < 0
���, 2 � �5 � ⇒ x2 � 4x � 1 > 0
x � 2 � �5, x � 2 � �5
x �4 ± �16 � 4
2� 2 ± �5
x2 � 4x � 1 > 0
Section 1.8 Other Types of Inequalities 161
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
20.
Critical numbers:
Test intervals:
Solution interval:
x
543210−1−2
32 2
39− 32 2
39+
���, 32
��39
2 � � 32
��39
2, ��
�32
��39
2, �� ⇒ �2x2 � 6x � 15 < 0
�32
��39
2,
32
��39
2 � ⇒ �2x2 � 6x � 15 > 0
���, 32
��39
2 � ⇒ �2x2 � 6x � 15 < 0
x �32
��39
2, x �
32
��39
2
�6 ± 2�39
4�
32
±�39
2
x ����6� ± ���6�2 � 4�2���15�
2�2� �6 ± �156
4
2x2 � 6x � 15 ≥ 0
�2x2 � 6x � 15 ≤ 0 21.
Critical numbers:
Test intervals:
Test: Is ?
Interval x-Value Value of Conclusion
Negative
Positive
Negative
Positive
Solution set:
0 1 2 43
x
−1
��1, 1� � �3, ��
�5��3��1� � 15x � 4�3, ��
�3��1���1� � �3x � 2�1, 3�
�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
22.
Critical numbers:
Test intervals: Solution interval:
�2, �� ⇒ x3 � 2x2 � 4x � 8 > 00 1 2 3 4
x��2, 2� ⇒ x3 � 2x2 � 4x � 8 < 0
���, 2����, �2� ⇒ x3 � 2x2 � 4x � 8 < 0
x � �2, x � 2
�x � 2��x2 � 4� ≤ 0
x2�x � 2� � 4�x � 2� ≤ 0
x3 � 2x2 � 4x � 8 ≤ 0
162 Chapter 1 Equations, Inequalities, and Mathematical Modeling
23.
Critical numbers:
Test intervals:
Test: Is
Interval x-Value Value of Conclusion
Negative
Positive
Negative
Positive
Solution set: ��3, 2� � �3, ��
x
−4 −3 −2 −1 0 1 2 3 4 5
�2��7��1� � 14x � 4�3, ��
�0.5��5.5���0.5� � �1.375x � 2.5�2, 3�
��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
24.
Critical numbers:
Test intervals:
Solution interval: ��132 , �2�, �2, ��
�2, �� ⇒ 2x3 � 13x2 � 8x � 52 > 0
��2, 2� ⇒ 2x3 � 13x2 � 8x � 52 < 0
��132 , �2� ⇒ 2x3 � 13x2 � 8x � 52 > 0
4
x
−2−4−6−8 20
132
−���, �132 � ⇒ 2x3 � 13x2 � 8x � 52 < 0
x � �132 , x � �2, x � 2
�2x � 13��x2 � 4� ≥ 0
x2�2x � 13� � 4�2x � 13� ≥ 0
2x3 � 13x2 � 8x � 52 ≥ 0
2x3 � 13x2 � 8x � 46 ≥ 6
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�, �1
2, ��
x �12
�2x � 1�2 ≤ 0
4x2 � 4x � 1 ≤ 0 26.
The critical numbers are imaginary:
So the set of real numbers is the solution set.
1 2 30−3 −2 −1
x
�32
±i�23
2
x2 � 3x � 8 > 0
Section 1.8 Other Types of Inequalities 163
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 28.
Critical numbers:
Test intervals:
Solution interval: �3, ��
�3, �� ⇒ 4x2�x � 3� > 0
�0, 3� ⇒ 4x2�x � 3� < 0
���, 0� ⇒ 4x2�x � 3� < 0
x � 0, x � 3
4x2�x � 3� > 0
4x3 � 12x2 > 0
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 30.
Critical numbers:
Test intervals:
Solution intervals: ���, 0� � �2, ��
�2, �� ⇒ x3�2 � x� < 0
�0, 2� ⇒ x3�2 � x� > 0
���, 0� ⇒ x3�2 � x� < 0
x � 0, x � 2
x3�2 � x� ≤ 0
2x3 � x4 ≤ 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 32.
Critical numbers:
Test intervals:
Solution intervals: ���, 0� � �0, 3� or ���, 3�
�3, �� ⇒ x4�x � 3� > 0
�0, 3� ⇒ x4�x � 3� < 0
���, 0� ⇒ x4�x � 3� < 0
x � 0, x � 3
x4�x � 3� ≤ 0
33.
−5 7
−2
6
y � �x2 � 2x � 3 (a)
(b) y ≥ 3 when 0 ≤ x ≤ 2.
y ≤ 0 when x ≤ �1 or x ≥ 3.
34.
−10
−4
14
12
y �1
2x2 � 2x � 1 (a)
when 2 � �2 ≤ x ≤ 2 � �2.y ≤ 0
�4 ± �8
2� 2 ± �2
x ����4� ± ���4�2 � 4�1��2�
2�1�
x2 � 4x � 2 ≤ 0
12
x2 � 2x � 1 ≤ 0
y ≤ 0 (b)
y ≥ 7 when x ≤ �2, x ≥ 6.
�x � 6��x � 2� ≥ 0
x2 � 4x � 12 ≥ 0
1
2x2 � 2x � 1 ≥ 7
y ≥ 7
164 Chapter 1 Equations, Inequalities, and Mathematical Modeling
35.
−12 12
−8
8
y �18x3 �
12x (a)
(b) y ≤ 6 when x ≤ 4.
y ≥ 0 when �2 ≤ x ≤ 0, 2 ≤ x < �.
36.
−12 12
−24
48
y � x3 � x2 � 16x � 16 (a)
when 1 ≤ x ≤ 4.�� < x ≤ �4,y ≤ 0
�x � 1��x2 � 16� ≤ 0
x2�x � 1� � 16�x � 1� ≤ 0
x3 � x2 � 16x � 16 ≤ 0
y ≤ 0 (b)
5 ≤ x < �.y ≥ 36 when x � �2,
�x � 2��x � 5��x � 2� ≥ 0
x3 � x2 � 16x � 20 ≥ 0
x3 � x2 � 16x � 16 ≥ 36
y ≥ 36
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 38.
Critical numbers:
Test intervals:
Solution interval:
10
14
−1
x
���, 0� � �1
4, ��
�1
4, �� ⇒ 1 � 4x
x< 0
�0, 1
4� ⇒ 1 � 4x
x> 0
���, 0� ⇒ 1 � 4x
x< 0
x � 0, x �1
4
1 � 4x
x< 0
1
x� 4 < 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 40.
Critical numbers:
Test intervals:
Solution interval:
−1−2 0 1 2 3
x
��2, 3�
�3, �� ⇒ 6 � 2x
x � 2< 0
��2, 3� ⇒ 6 � 2x
x � 2 > 0
���, �2� ⇒ 6 � 2x
x � 2< 0
x � �2, x � 3
6 � 2x
x � 2≥ 0
x � 12 � 3�x � 2�
x � 2≥ 0
x � 12
x � 2� 3 ≥ 0
Section 1.8 Other Types of Inequalities 165
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:
3 6 9
5
12 15 18
x
�5, 15�
15 � x
x � 5> 0?
���, 5�, �5, 15�, �15, ��
x � 5, x � 15
15 � x
x � 5> 0
3x � 5 � 4�x � 5�x � 5
> 0
3x � 5
x � 5� 4 > 0
3x � 5
x � 5> 4 42.
Critical numbers:
Test intervals:
Solution intervals:
−2 −1
−
0 21
12
x
���, �1
2� � �1, ��
�1, �� ⇒ 1 � x
1 � 2x< 0
�� 1
2, 1� ⇒ 1 � x
1 � 2x> 0
���, �1
2� ⇒ 1 � x
1 � 2x< 0
x � �1
2, x � 1
1 � x
1 � 2x< 0
5 � 7x � 4�1 � 2x�
1 � 2x< 0
5 � 7x
1 � 2x< 4
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:
2
−1−2−3−4−5
−
0
3
x
��5, �32� � ��1, ��
7�x � 1��x � 5��2x � 3�
> 0?
��3
2, �1�, ��1, ��
���, �5�, ��5, �3
2�,
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 � 344.
Critical numbers:
Test intervals:
Solution intervals:
−14
−5
−2
−10−15 0 5
6
10
x
��14, �2� � �6, ��
�6, �� ⇒ 2x � 28
�x � 6��x � 2�> 0
��2, 6� ⇒ 2x � 28
�x � 6��x � 2�< 0
��14, �2� ⇒ 2x � 28
�x � 6��x � 2�> 0
���, �14� ⇒ 2x � 28
�x � 6��x � 2�< 0
x � �14, x � �2, x � 6
2x � 28
�x � 6��x � 2�> 0
5�x � 2� � 3�x � 6�
�x � 6��x � 2�> 0
5
x � 6>
3
x � 2
166 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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:
−2−4
−33
4 6 820
4x
��34, 3� � �6, ��
30 � 5x
�x � 3��4x � 3�≤ 0?
���, �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 � 346.
Critical numbers:
Test intervals:
Solution intervals:
−1−2−3−4 10
x
���, �3� � �0, ��
�0, �� ⇒ 3
x�x � 3�> 0
��3, 0� ⇒ 3
x�x � 3�< 0
���, �3� ⇒ 3
x�x � 3�> 0
x � �3, x � 0
3
x�x � 3�≥ 0
1�x � 3� � 1�x�
x�x � 3�≥ 0
1
x≥
1
x � 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 48.
Critical numbers:
Test intervals:
Solution intervals:
−1−2−3 0 1 2 3
x
��3, 0� � �2, ��
�2, �� ⇒ �x � 3��x � 2�x
> 0
�0, 2� ⇒ �x � 3��x � 2�x
< 0
��3, 0� ⇒ �x � 3��x � 2�x
> 0
���, �3� ⇒ �x � 3��x � 2�x
< 0
x � �3, x � 0, x � 2
�x � 3��x � 2�
x≥ 0
x2 � x � 6
x≥ 0
Section 1.8 Other Types of Inequalities 167
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 50.
Critical numbers:
Test intervals:
Solution intervals:
0 2 64
x
−2−4
1
���, �4�, ��2, 1�, �6, ��
�6, �� ⇒ ��x � 6��x � 2��x � 1��x � 4� < 0
�1, 6� ⇒ ��x � 6��x � 2��x � 1��x � 4� > 0
��2, 1� ⇒ ��x � 6��x � 2��x � 1��x � 4� < 0
��4, �2� ⇒ ��x � 6��x � 2��x � 1��x � 4� > 0
���, �4� ⇒ ��x � 6��x � 2��x � 1��x � 4� < 0
x � �4, x � �2, x � 1, x � 6
��x � 6��x � 2��x � 1��x � 4� ≤ 0
�x2 � 4x � 12�x � 1��x � 4� ≤ 0
3x�x � 4� � x�x � 1� � 3�x � 4��x � 1�
�x � 1��x � 4� ≤ 0
3x
x � 1≤
xx � 4
� 3
51.
−6 12
8
−4
y �3x
x � 2
(a)
(b) y ≥ 6 when 2 < x ≤ 4.
y ≤ 0 when 0 ≤ x < 2.
52.
−15 15
−6
14
y �2�x � 2�
x � 1(a)
y ≤ 0 when �1 < x ≤ 2.
2�x � 2�x � 1
≤ 0
y ≤ 0 (b)
y ≥ 8 when �2 ≤ x < �1.
�6�x � 2�
x � 1≥ 0
�6x � 12
x � 1≥ 0
2�x � 2� � 8�x � 1�
x � 1≥ 0
2�x � 2�
x � 1≥ 8
y ≥ 8
168 Chapter 1 Equations, Inequalities, and Mathematical Modeling
53.
−6 6
−2
6
y �2x2
x2 � 4(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
54.
(a)
y ≥ 1 when 1 ≤ x ≤ 4.
��x � 4��x � 1�
x2 � 4≥ 0
5x � �x2 � 4�
x2 � 4≥ 0
5x
x2 � 4≥ 1
y ≥ 1
y �5x
x2 � 4
(b)
y ≤ 0 when �� < x ≤ 0.
5x
x2 � 4≤ 0 −6 6
−4
4 y ≤ 0
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 56.
Critical numbers:
Test intervals:
Domain: ���, �2� � �2, ��
�2, �� ⇒ �x � 2��x � 2� > 0
��2, 2� ⇒ �x � 2��x � 2� < 0
���, �2� ⇒ �x � 2��x � 2� > 0
x � �2, x � 2
�x � 2��x � 2� ≥ 0
x2 � 4 ≥ 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 58.
Critical numbers:
Test intervals:
Domain: ��4, 4�
�4, �� ⇒ 9�4 � x��4 � x� < 0
��4, 4� ⇒ 9�4 � x��4 � x� > 0
���, �4� ⇒ 9�4 � x��4 � x� < 0
x � �4, x � 4
9�4 � x��4 � x� ≥ 0
144 � 9x2 ≥ 0
59.
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: ��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
Section 1.8 Other Types of Inequalities 169
60.
Critical numbers:
Test intervals:
Domain: ��3, 0� � �3, ��
�3, �� ⇒ x
�x � 3��x � 3�> 0
�0, 3� ⇒ x
�x � 3��x � 3�< 0
��3, 0� ⇒ x
�x � 3��x � 3�> 0
���, �3� ⇒ x
�x � 3��x � 3�< 0
x � �3, x � 0, x � 3
x
�x � 3��x � 3�≥ 0
x
x2 � 9≥ 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
62.
Critical numbers:
Test intervals:
Solution set: ��1.13, 1.13�
���, �1.13�, ��1.13, 1.13�, �1.13, ��
±1.13
�1.3x2 � 1.66 > 0
�1.3x2 � 3.78 > 2.12 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
64.
Critical numbers:
Test intervals:
Solution set: ��4.42, 0.42�
���, �4.42�, ��4.42, 0.42�, �0.42, ��
�4.42, 0.42
1.2x2 � 4.8x � 2.2 < 0
1.2x2 � 4.8x � 3.1 < 5.3 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
66.
23.46 � 17.98x
3.1x � 3.7> 0
2 � 5.8�3.1x � 3.7�
3.1x � 3.7> 0
2
3.1x � 3.7> 5.8
Critical numbers:
Test intervals:
Solution interval: �1.19, 1.30�
�1.30, �� ⇒ 23.46 � 17.98x
3.1x � 3.7< 0
�1.19, 1.30� ⇒ 23.46 � 17.98x
3.1x � 3.7> 0
���, 1.19� ⇒ 23.46 � 17.98x
3.1x � 3.7< 0
x � 1.19, x � 1.30
170 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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 68.
(a)
It will be back on the ground in 8 seconds.
(b)
Critical numbers:
Test intervals:
Solution set:and 4 � 2�2 seconds < t ≤ 8 seconds0 seconds ≤ t < 4 � 2�2 seconds
�4 � 2�2, �����, 4 � 2�2�, �4 � 2�2, 4 � 2�2�,
4 � 2�2, 4 � 2�2
�16t2 � 128t � 128 < 0
�16t2 � 128t < 128
t � 8 � 0 ⇒ t � 8
�16t � 0 ⇒ t � 0
�16t�t � 8� � 0
�16t2 � 128t � 0
s � �16t2 � v0t � s0 � �16t 2 � 128t
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 70.
By the Quadratic Formula we have:
Critical numbers:
Test:
Solution set:
45.97 feet ≤ L ≤ 174.03 feet
110 � 10�41 ≤ L ≤ 110 � 10�41
Is �L2 � 220L � 8000 ≥ 0?
L � 110 ± 10�41
�L2 � 220L � 8000 ≥ 0
L�220 � L� ≥ 8000
LW ≥ 8000
2L � 2W � 440 ⇒ W � 220 � L
71. and
Critical numbers: (These wereobtained by using the Quadratic Formula.)
Test intervals:
By testing values in each test interval in the inequality, wesee that the solution set is or
The 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� 72. What is the price per unit?
When
When
Solution interval: $30.00 ≤ p ≤ $32.00
R � $3,000,000 ⇒ 3,000,000100,000
� $30 per unit
x � 100,000:
R � $2,880,000 ⇒ 2,880,00090,000
� $32 per unit
x � 90,000:
Section 1.8 Other Types of Inequalities 171
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 37 and 42. Thesevalues 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
74. (a)
(b)
The minimum depth is 3.83 inches.
3.83 ≤ d
14.67 ≤ d2
2472.1 ≤ 168.5d2
2000 ≤ 168.5d2 � 472.1
Depth of the beam
Max
imum
saf
e lo
ad
d
L
4 6 8 10 12
5,000
10,000
15,000
20,000
25,000
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
2d 4 6 8 10 12
Load 2223.9 5593.9 10,312 16,378 23,792
76. (a)
So the number of master’s degrees earned by womenexceeded 220,000 in 1995.
(c)
So the number of master’s degrees earned by womenwill exceed 320,000 in 2006.
� 320 ⇒ t � 16.2
N � �0.03t 2 � 9.6t � 172
� 220 ⇒ t � 5
N � �0.03t 2 � 9.6t � 172 (b) and (d)
t
N
Mas
ter's
deg
rees
ear
ned
(in
thou
sand
s)
Year (0 ↔ 1990)
2 6 10 14 18
160
200
240
280
320
N = 220
N = 320
172 Chapter 1 Equations, Inequalities, and Mathematical Modeling
77. True
The test intervals are and�4, ��.
���, �3�, ��3, 1�, �1, 4�,
x3 � 2x2 � 11x � 12 � �x � 3��x � 1��x � 4�
78. True
The y-values are greater than zero for all values of x.
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 80.
To have at least one real solution,
This inequality is true for all real values of b. Thus, theinterval for b such that the equation has at least one realsolution is ���, ��.
b2 � 16 ≥ 0.
b2 � 4�1���4� ≥ 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 82.
To have at least one real solution,
This occurs when or Thus,the interval for b such that the equation has at least one real solution is ���, �2�10� � �2�10, ��.
b ≥ 2�10.b ≤ �2�10
b2 � 40 ≥ 0.
b2 � 4�2��5� ≥ 0
2x2 � bx � 5 � 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 > 084. (a)
(b)
(c) The real zeros of the polynomial
ax
b
+ − +
− − +
− + +
x � a, x � b
85. 4x2 � 20x � 25 � �2x � 5�2 86.
� �x � 7��x � 1�
�x � 3�2 � 16 � ��x � 3� � 4���x � 3� � 4�
87.
� �x � 2��x � 2��x � 3�
x2�x � 3) � 4�x � 3� � �x2 � 4��x � 3� 88.
� 2x�x � 3��x2 � 3x � 9�
2x4 � 54x � 2x�x3 � 27�
89.
� 2x2 � x
� �2x � 1��x�
Area � �length��width� 90.
� 32b2 � b
� 12�b��3b � 2�
Area �12�base��height�
Review Exercises for Chapter 1 173
1.
–3 –2 –1 1 2 3
–5
–4
–3
–2
–1
1
x
y
y � 3x � 5 2.
x−2
−2−4−6 2 4
4
6
8
10
y
y � �12x � 2
x 0 1 2
y 1�2�5�8�11
�1�2 x �4 �2 0 2 4
y 4 3 2 1 0
3.
–3 –2 –1 1 2 54
–3
–2
4
5
x
y
y � x2 � 3x 4.
x
−3
5431
1
−2
−4
−9
−3−4−5 −1
y
y � 2x2 � x � 9
x 0 1 2 3 4
y 4 0 0 4�2�2
�1 x �2 �1 0 1 2 3
y 1 �6 �9 �8 �3 6
5.
Line with x-intercept andy-intercept
–5 –4 –3 –1 1 2 3
–2
1
3
4
5
6
x
y
�0, 3���3
2, 0�
y � 2x � 3
y � 2x � 3 � 0 6.
Line with x-intercept andy-intercept
–5 –4 –3 1 2 3
–6
–5
–4
–3
1
2
x
y
�0, �3���2, 0�
y � �32x � 3
2y � �3x � 6
3x � 2y � 6 � 0 7.
Domain:
–2 –1 1 2 3 4 5 6
–2
1
3
4
5
6
x
y
���, 5�
y � �5 � x
x 5 4 1
y 0 1 2 3
�4x 0 1
y 1 3 5�1
�1�2
x 0 1
y 0 �92�3�
32
32
�1�2�3
Review Exercises for Chapter 1
174 Chapter 1 Equations, Inequalities, and Mathematical Modeling
9.
is a parabola.
–3 –2 –1 1 2 3
–5
–4
–3
–2
1
x
y
y � �2x2
y � 2x2 � 0 10. is a parabola.
–2 –1 1 2 3 5 6
–4
–3
–2
x
y
y � x2 � 4x
x 0
y 0 �8�2
±2±1
x �1 0 1 2 3 4
y 5 0 �3 �4 �3 0
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
12.
For
For
or
The x-intercepts are and the y-intercept is �0, �2�.��4, 0�;�2, 0�
y � �2y � �0 � 1� � 3
y � �x � 1� � 3
x � 1 < 0, 0 � ��x � 1� � 3, or �4 � x.
x � 1 > 0, 0 � x � 1 � 3, or 2 � x.
0 � �x � 1� � 3
y � �x � 1� � 3
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
14.
Intercepts:
No symmetry
�65, 0�, �0, �6�
x6
1
2 3 4 5−1−2−1
−2
−3
−4
−5
−6
−7
yy � 5x � 6
8.
−3 −2 −1 1 2 3 4 5
2
−2
3
4
5
6
x
y
y � �x � 2, domain: ��2, ��
x �2 0 2 7
y 0 2 3�2
Review Exercises for Chapter 1 175
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 16.
Intercepts:
y-axis symmetry
x
2
42−4−6 6 8
−6
−12
−4
−2
y
�±�10, 0�, �0, �10�
y � x2 � 10
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 18.
Intercepts:
No symmetry
x
2
42 6−4−6
−8
−10
y
� 3��6, 0�, �0, �6�
y � �6 � x3
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
20.
Intercepts:
y-axis symmetry
�0, 9�
y
x−3−6−9 3 6 9
−3
3
6
9
12
15
y � �x� � 9 21.
Center:
Radius: 3
�0, 0�
–4 –2 –1 1 2 4
–4
–2
–1
1
2
4
x(0, 0)
yx2 � y2 � 9
176 Chapter 1 Equations, Inequalities, and Mathematical Modeling
22.
Center:
Radius: 2
�0, 0�
x
3
3−3
1
1−1−1
−3
(0, 0)
yx2 � y2 � 4 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
24.
Center:
Radius: 9
�0, 8�
x4
8
10
4
−2−4−6−8
2
6
(0, 8)
6 8
10
12
14
16
18
yx2 � �y � 8�2 � 81 25.
Center:
Radius: 6
�12, �1�
�x �12�2
� �y � ��1��2 � 62
x2
8
8−2
−8
2
4
−4
−2−4−6 4
( (12, −1
y�x �12�2
� �y � 1�2 � 36
26.
Center:
Radius: 10
��4, 32� �x � ��4��2 � � y �
32�2
� 100
x9
3
3−3−6−9−3
−15
6
9
15
( (32
−4,
−6
y �x � 4�2 � � y �32�2
� 100 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�
28. Endpoints of a diameter:
Center:
Radius:
Standard form:
�x � 1�2 � �y �132 �
2
�854
�x � 1�2 � �y � ��132 ��
2
� ��854 �
2
r ���1 � ��2��2 � ��132
� ��3��2
��9 �494
��854
��2 � 42
, �3 � ��10�
2 � � �1, �132 �
��2, �3� and �4, �10�
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
Review Exercises for Chapter 1 177
30. (a)
(b) The number of stores was 1300 in 2003.z � 9.94;
t
1800
Num
ber
of T
arge
t sto
res
Year (4 ↔ 1994)
4 6 8 10 12
800
1000
1200
1400
1600
N 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
32.
Conditional equation
x � 4
3x � 12
3x � 6 � 2x � 2x � 6
3�x � 2� � 2x � 2�x � 3� 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
34.
Conditional equation
3x2 � x � 19 � 0
6x2 � 2x � 38 � 0
3x2 � 12x � 24 � �3x2 � 10x � 14
3x2 � 12x � 24 � �10x � 20 � 3x2 � 6
3�x2 � 4x � 8� � �10�x � 2� � 3x2 � 6 35.
x � 20
3x � 2x � 10 � 10
3x � 2�x � 5� � 10
36.
x � �92
2x � �9
4x � 14 � 2x � 5
4x � 2�7 � x� � 5 37.
x � �12
10x � �5
4x � 9 � �6x � 4
4x � 12 � 3 � 8 � 6x � 4
4�x � 3� � 3 � 2�4 � 3x� � 4 38.
x � �173
�32x �
172
12x �32 � 2x � 2 � 5
12�x � 3� � 2�x � 1� � 5
39.
x � �5
�4x � 20
x � 15 � 5x � 5
5�x5
� 3� � �x � 1�5
x5
� 3 �2x2
� 1 40.
x � 18
�x � �18
8x � 6 � 3x � 12x � 24
2�4x � 3� � 3x � 12x � 24
4x � 3
6�
x4
� x � 2 41.
x � 9
8x � 72
18x � 72 � 10x
18�x � 4� � 10x
18x
�10
x � 4
42.
x �113
�3x � �11
10x � 15 � 13x � 26
5�2x � 3� � 13�x � 2�
5
x � 2�
132x � 3
43.
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
178 Chapter 1 Equations, Inequalities, and Mathematical Modeling
44.
The x-intercept is and the y-intercept is �0, 6�.�65, 0�
65 � x
y � 6 �6 � �5x
y � �5�0� � 6 0 � �5x � 6
y � �5x � 6 y � �5x � 6 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�
46.
The x-intercept is and the y-intercept is �0, 4�.��17, 0�
�17
� x
�4 � 28x
y � 4 0 � 28x � 4
y � 4�7�0� � 1� 0 � 4�7x � 1�
y � 4�7x � 1� y � 4�7x � 1� 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 �2312
�43
y � �12
x �23
48.
The x-intercept is and the y-intercept is �0, �14�.�1
3, 0�
13
� x
y � �14
43
�14
�43
�34
x
y �34
�0� �14
0 �34
x �14
y �34
x �14
y �34
x �14
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
50.
The x-intercept is and the y-intercept is �0, 0.8�.�0.6, 0�
y � 0.8 x � 0.6
1.5y � 1.2 2x � 1.2
1.5y � 2�0� � 1.2 � 0 1.5�0� � 2x � 1.2 � 0
1.5y � 2x � 1.2 � 0 1.5y � 2x � 1.2 � 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
52.
For F �95�100 �
1609 � � 212�FC � 100�:
F �95�C �
1609 �
59F � C �160
9
C �59F �
1609
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
Review Exercises for Chapter 1 179
54. Model:
Labels: Original price
Discount rate
Sale price
Equation:
The percent discount is 20%.
x � 0.2
1 � x � 0.8
425�1 � x� � 340
� 340
� x
� 340 � 85 � 425
�Original price��1 � discount rate� � �sale price� 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 �
56. Let the number of members in the group.
Cost per member
If two more members join the group, the cost per member will be
Since this new cost is $4000 less than the original cost:
There are four members presently in the group.
x � 4 � 0 ⇒ x � 4
x � 6 � 0 ⇒ x � �6, extraneous
0 � �x � 6��x � 4�
0 � x2 � 2x � 24
12x � 24 � x2 � 2x � 12x
12�x � 2� � x�x � 2� � 12x Divide both sides by 4000.
48,000�x � 2� � 4000x�x � 2� � 48,000x
48,000
x� 4000 �
48,000
x � 2
48,000
x � 2.
�48,000
x
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:
extraneous or
There are currently nine investors.
x � 9x � �12,
�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
.90,000
x.
x �
180 Chapter 1 Equations, Inequalities, and Mathematical Modeling
58.
Using the positive value for r, we have miles per hour. The average speed on the trip home was miles per hour.r � 8 � 56
r � 48
0 � �r � 48��r � 56�
0 � r2 � 8r � 2688
336r � 2688 � 336r � r2 � 8r
6�r � 8��56� � 6r�56� � r�r � 8�
56
r�
56
r � 8�
1
6 Convert minutes to portion of an hour.
Time to work � time from work � 10 minutes
Time �distance
rate
Rate Time Distance
To work r 56
From work 5656
r � 8r � 8
56r
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 �
60. Model:
Labels: Amount invested at amount invested at
Interest from interest from total interest $305
Equation:
The amount invested at was $2500 and the amount invested at was 6000 � 2500 � $3500.512%41
2%
x � 2500
�0.01x � �25
0.045x � 330 � 0.055x � 305
0.045x � 0.055�6000 � x� � 305
�512% � �6000 � x��0.055��1�,41
2% � x�0.045��1�,
512% � 6000 � x41
2% � x,
�Interest from 412%� � �interest from 51
2%� � �total interest�
61.
3V
�r 2� h
3V � �r 2 h
V �1
3�r 2h 62.
m �2Ev2
mv2 � 2E
E �12
mv2
Review Exercises for Chapter 1 181
63.
or 40 minutes t �23 hour
15t � 10
55t � 40t � 10
64.
h �81�
9�� 9 feet
81� � � �3�2h
V � �r2hrate 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 66.
x � 4 � 0 ⇒ x � 4
2x � 7 � 0 ⇒ x � �72
�2x � 7��x � 4� � 0
2x2 � x � 28 � 0 67.
±�2 � x
2 � x2
6 � 3x2
68.
4x � 5 � 0 ⇒ x � �54
4x � 5 � 0 ⇒ x �54
�4x � 5��4x � 5� � 0
16x2 � 25 � 0
16x2 � 25 69.
x � �4 ± 3�2
x � 4 � ±�18
�x � 4�2 � 18 70.
x � 8 ± �15
x � 8 � ±�15
�x � 8�2 � 15
71.
x � 6 ± �6
x � 6 � ±�6
�x � 6�2 � 6
x2 � 12x � 36 � �30 � 36
x2 � 12x � �30
x2 � 12x � 30 � 0 72.
� �3 ± 2�3
��6 ± 4�3
2
��6 ± �48
2
��6 ± �62 � 4�1���3�
2�1�
x ��b ± �b2 � 4ac
2a
x2 � 6x � 3 � 0 73.
��5 ± �241
4
x ��5 ± �52 � 4�2���27�
2�2�
2x2 � 5x � 27 � 0
�2x2 � 5x � 27 � 0
74.
�3 ± �249
6�
12
±�249
6
����3� ± ���3�2 � 4�3���20�
2�3�
x ��b ± �b2 � 4ac
2a
�20 � 3x � 3x2 � 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�
182 Chapter 1 Equations, Inequalities, and Mathematical Modeling
76. (a)
(b)
(c)
The ball will hit the ground in about two seconds.
2.052 or �0.1767
��30 ± �1271.2
�32
t ��30 ± �302 � 4��16��5.8�
2��16�
�16t 2 � 30t � 5.8 � 0
h�1� � �16 � 12 � 30 � 1 � 5.8 � 19.8 feet
h�t� � �16t 2 � 30t � 5.8
77. 6 � ��4 � 6 � 2i 78. 3 � ��25 � 3 � 5i 79. i2 � 3i � �1 � 3i
80. �5i � i2 � �1 � 5i 81. �7 � 5i� � ��4 � 2i� � �7 � 4� � �5i � 2i� � 3 � 7i
82. � �2��2
2i� � ��2 i��2
2�
�2
2i� � ��2
2�
�2
2i� �
�2
2�
�2
2i �
�2
2�
�2
2i
83. 5i�13 � 8i� � 65i � 40i2 � 40 � 65i 84.
� 17 � 28i
� 5 � 28i � 12
�1 � 6i��5 � 2i� � 5 � 2i � 30i � 12i2
85.
� �4 � 46i
�10 � 8i��2 � 3i� � 20 � 30i � 16i � 24i2 86.
� 9 � 20i
� 20i � 9i2
� i�20 � 9i�
i�6 � i��3 � 2i� � i�18 � 12i � 3i � 2i2�
87.
�2317
�1017
i
�23 � 10i
17
�24 � 10i � i2
16 � 1
6 � i4 � i
�6 � i4 � i
�4 � i4 � i
88.
�1726
�7i26
�17 � 7i
26
�15 � 3i � 10i � 2i2
25 � i2
3 � 2i5 � i
�3 � 2i5 � i
�5 � i5 � 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 90.
�9 � 83i
85�
985
�83i85
�18 � 81i � 2i � 9i2
4 � 81i2
��9 � i
�2 � 9i�
��2 � 9i���2 � 9i�
�1 � 4i � 10 � 5i2 � 8i � i � 4i2
1
2 � i�
51 � 4i
��1 � 4i� � 5�2 � i�
�2 � i��1 � 4i�
Review Exercises for Chapter 1 183
91.
� ±�1
3 i � ±
�3
3i
x � ±��1
3
x2 � �1
3
3x2 � �1
3x2 � 1 � 0 92.
x � ±12
i
x2 � �14
8x2 � �2
2 � 8x2 � 0 93.
x � 1 ± 3i
x � 1 � ±��9
�x � 1�2 � �9
x2 � 2x � 1 � �10 � 1
x2 � 2x � 10 � 0
94.
��3 ± 3i�71
12� �
14
±�71
4i
��3 ± ��639
12
��3 ± �32 � 4�6��27�
2�6�
x ��b ± �b2 � 4ac
2a
6x2 � 3x � 27 � 0 95.
x � 0 or 5 � x �12
5
x3 � 0 or 5x � 12 � 0
x3�5x � 12� � 0
5x4 � 12x3 � 0 96.
4x � 6 � 0 ⇒ x �3
2
x2 � 0 ⇒ x � 0
x2�4x � 6� � 0
4x3 � 6x2 � 0
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 98.
9x2 � 4 � 0 ⇒ x � ± 23
x � 0
x � 3 � 0 ⇒ x � �3
�x � 3��x��9x2 � 4� � 0
�x � 3��9x3 � 4x� � 0
9x3�x � 3� � 4x�x � 3� � 0
9x 4 � 27x3 � 4x2 � 12x � 0
99.
x � 5
x � 4 � 9
��x � 4 �2� �3�2
�x � 4 � 3 100.
x � 66
x � 2 � 64
�x � 2 � 8
�x � 2 � 8 � 0
101.
extraneous or extraneous
No solution
x � 11,x � 3,
�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
184 Chapter 1 Equations, Inequalities, and Mathematical Modeling
102.
x �38 � 5�3
24, extraneous
x �38 � 5�3
24
�1824 ± �172,800
1152�
1824 ± 240�3
1152 x �
���1824� ± ���1824�2 � 4�576��1369�2�576�
576x2 � 1824x � 1369 � 0
576x2 � 1680x � 1225 � 144�x � 1�
24x � 35 � 12�x � 1
25x � 36 � 12�x � 1 � x � 1
5�x � 6 � �x � 1
5�x � �x � 1 � 6
103.
x � 126 or x � �124
x � 1 ± 125
x � 1 � ±�253
�x � 1�2 � 253
�x � 1�23 � 25
�x � 1�23 � 25 � 0 104.
x � 79
x � 2 � 81
x � 2 � 2743
�x � 2�34 � 27
105.
or
x � �4
5x2 � 20x � 1 � 0 �x � 4�12 � 0
�x � 4�12�5x2 � 20x � 1� � 0
�x � 4�12�1 � 5x�x � 4�� � 0
�x � 4�12 � 5x�x � 4�32 � 0
x � �2 ±�95
5
x ��20 ± 2�95
10
x ��20 ± �400 � 20
10
x ��20 ± �400 � 20
10
106.
3x � 2 � 0 ⇒ x � �2
3
3x � 2 � 0 ⇒ x �2
3
x � 2 � 0 ⇒ x � �2
x � 2 � 0 ⇒ x � 2
�x � 2�13�x � 2�13�3x � 2��3x � 2� � 0
�x2 � 4�13�9x2 � 4� � 0
�x2 � 4�13�8x2 � x2 � 4� � 0
8x2�x2 � 4�13 � �x2 � 4�43 � 0 107.
±�10 � x
10 � x2
5x � 10 � x2 � 2x � 3x
5�x � 2� � 1�x��x � 2� � 3x
5x
� 1 �3
x � 2
Review Exercises for Chapter 1 185
108.
0 � x � 3 ⇒ x � 3
0 � 3x � 10 ⇒ x � �10
3
0 � �3x � 10��x � 3�
0 � 3x2 � x � 30
14x � 30 � 3x2 � 15x
6�x � 5� � 8x�3x�x � 5�
x�x � 5�6
x� x�x � 5�
8
x � 5� 3x�x � 5� 109.
x �43
9x � 12
15x � 5x � 10 � x � 2
3�5x� � 5�x � 2� � x � 2
3
x � 2�
1x
�15x
110.
x � �25
�3x � 75
17x � 25 � 20x � 100
12x � 5x � 25 � 20x � 100
x�x � 5�12
x � 5�
x�x � 5�5x
�x�x � 5�20
x111.
x � �51 or 5 � x � 15
x � 5 � �10 or x � 5 � 10
�x � 5� � 10
112.
x � 2 x � �5
2x � 4 2x � �10
2x � 3 � 7 or 2x � 3 � �7
�2x � 3� � 7 113.
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
114.
x � 2 � 0 ⇒ x � �2, extraneous
x � 3 � 0 ⇒ x � 3
�x � 3��x � 2� � 0
x2 � x � 6 � 0
x2 � 6 � x
�x2 � 6� � x
x � 3 � 0 ⇒ x � �3, extraneous
x � 2 � 0 ⇒ x � 2
�x � 3��x � 2� � 0
x2 � x � 6 � 0
or ��x2 � 6� � x
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 � 2 116. (a)
(b) There were 800 daily evening papers about 1997.
(c)
t 27.98 ⇒ 1998
t �148.04�23
t 32 ��681�4.6
148.04
�681 � �4.6t 32
800 � 1481 � 4.6t 32
0 300
1500
186 Chapter 1 Equations, Inequalities, and Mathematical Modeling
117. Interval:
Inequality:
The interval is bounded.
�7 < x ≤ 2
��7, 2� 118. Interval:
Inequality:
The interval is unbounded.
4 < x < �
�4, �� 119. Interval:
Inequality:
The interval is unbounded.
x ≤ �10
���, �10�
120. Interval:
Inequality:
The interval is bounded.
�2 ≤ x ≤ 2
��2, 2� 121.
���, 12�
x ≤ 12
2x ≤ 24
9x � 8 ≤ 7x � 16 122.
��2, ��
x > �2
9x > �18
92x > �9
152 x � 4 > 3x � 5
123.
�3215, ��
x ≥ 3215
�152 x ≤ �16
20 � 8x ≤ 4 �12x
4�5 � 2x� ≤ 12�8 � x� 124.
��53, ��
x > �53
3x > �5
9 � 3x > 4 � 6x
12�3 � x� > 13�2 � 3x� 125.
��23, 17�
�23 < x ≤ 17
�2 < 3x ≤ 51
�19 < 3x � 17 ≤ 34
126.
��2, 10� �2 ≤ x < 10
�4 ≤ 2x < 20
�9 ≤ 2x � 5 < 15
�3 ≤2x � 5
3< 5 127.
��4, 4�
�4 ≤ x ≤ 4
�x� ≤ 4 128.
�1, 3�
1 < x < 3
�1 < x � 2 < 1
�x � 2� < 1
129.
or
or
���, �1� � �7, ��
x > 7 x < �1
x � 3 > 4 x � 3 < �4
�x � 3� > 4 130.
���, 0� � �3, ��
x ≤ 0 x ≥ 3
x �32 ≤ �
32 or x �
32 ≥ 3
2
�x �32� ≥ 3
2
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
132.
x > 36 units
33.33x > 1200
125.33x > 92x � 1200
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 < 0 134.
Critical numbers:
Test intervals:
Solution interval: ���, �1� � �3, ��
�3, �� ⇒ �x � 3��x � 1� > 0
��1, 3� ⇒ �x � 3��x � 1� < 0
���, �1� ⇒ �x � 3��x � 1� > 0
x � �1, x � 3
�x � 3��x � 1� ≥ 0
x2 � 2x � 3 ≥ 0
x2 � 2x ≥ 3
Review Exercises for Chapter 1 187
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 < 4 136.
Critical numbers:
Test intervals:
Solution interval: ���, �3� � �52, ��
�52, �� ⇒ �2x � 5��x � 3� > 0
��3, 52� ⇒ �2x � 5��x � 3� < 0
���, �3� ⇒ �2x � 5��x � 3� > 0
x �52, x � �3
�2x � 5��x � 3� ≥ 0
2x2 � x � 15 ≥ 0
2x2 � x ≥ 15
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 ≥ 0 138.
Critical numbers:
Test intervals:
Solution interval: ���, 0� � �0, 53��5
3, �� ⇒ 12x3 � 20x2 > 0
�0, 53� ⇒ 12x3 � 20x2 < 0
���, 0� ⇒ 12x3 � 20x2 < 0
x � 0, x �53
4x2�3x � 5� < 0
12x3 � 20x2 < 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 < 0 140.
Critical numbers:
Test intervals:
Solution interval: ���, 3� � �20, ��
�20, �� ⇒ 3x � 8x � 3
< 4
�3, 20� ⇒ 3x � 8x � 3
≥ 4
���, 3� ⇒ 3x � 8x � 3
< 4
x � 3, x � 20
x ≥ 20
3x � 8 ≤ 4x � 12
3x � 8x � 3
≤ 4
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 � 1142.
Critical numbers:
Test intervals:
Solution intervals: ���, 3� � �5, ��
�5, �� ⇒ x � 53 � x
< 0
�3, 5� ⇒ x � 53 � x
> 0
���, 3� ⇒ x � 53 � x
< 0
x � 5, x � 3
x � 53 � x
< 0
188 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≥ 0 144.
Critical numbers:
Test intervals:
Solution interval: ���, 0� � �2, ��
�2, �� ⇒ 1
x � 2�
1x
> 0
�0, 2� ⇒ 1x � 2
�1x
< 0
���, 0� ⇒ 1x � 2
�1x
> 0
x � 2, x � 0
1
x � 2�
1x
> 0
1
x � 2>
1x
145.
r > 4.9%
r > 0.0488
1 � r > 1.0488
�1 � r�2 > 1.1
5000�1 � r�2 > 5500 146.
t ≥ 9 days
�1000t ≤ �9000
10,000 � 2000t ≤ 1000 � 3000t
2000�5 � t� ≤ 1000�1 � 3t�
2000 ≤1000�1 � 3t�
5 � t
P �1000�1 � 3t�
5 � t
147. False
���18���2� � �36 � 6
��18��2 � ��18i���2i� � �36i2 � �6
148. False. The equation has no real solution. The solutions are
717650
±�3311i
650.
149. Rational equations, equations involving radicals, andabsolute value equations, may have “solutions” that are extraneous. So checking solutions, in the originalequations, is crucial to eliminate these extraneous values.
150. Sample answer: The first equivalent inequality written isincorrect. It should be This leads tothe solution or �2, ��.���, �30
11�11x � 4 ≤ �26.
Problem Solving for Chapter 1
1.
Time (in seconds)
Dis
tanc
e (i
n fe
et)
y
x
Problem Solving for Chapter 1 189
2. (a)
(b)
When
When
When n � 10: 12�10��11� � 55
n � 8: 12�8��9� � 36
n � 5: 12�5��6� � 15
1 � 2 � 3 � . . . � n �12n�n � 1�
1 � 2 � 3 � 4 � 5 � 6 � 7 � 8 � 9 � 10 � 55
1 � 2 � 3 � 4 � 5 � 6 � 7 � 8 � 36
1 � 2 � 3 � 4 � 5 � 15 (c)
or
Since n is a natural number, choose n � 20.
n � 20n � �21
�n � 21��n � 20� � 0
n2 � n � 420 � 0
n�n � 1� � 420
12n�n � 1� � 210
3. (a)
thus:
(b)
(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�
A � �a�20 � a�
a � b � 20 ⇒ b � 20 � a,
A � �ab (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�
a 4 7 10 13 16
A 64 91 100 91 64�����
4.
(a)
miles per hour.
(b)
miles per hour.
No, actually it can survive wind blowing at timesthe speed found in part (a).
(c) The wind speed in the formula is squared, so a smallincrease in wind speed could have potentially seriouseffects on a building.
�2
s � 125
s2 � 15625
0.00256s2 � 40
s � 88.4
s2 � 7812.5
0.00256s2 � 20
P � 0.00256s2 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
190 Chapter 1 Equations, Inequalities, and Mathematical Modeling
6. (a) If then and
(b)
There is no integer m such that equals a negativenumber. cannot be factored over the integers.x2 � 9
m2
m��m� � 9 ⇒ �m2 � 9 ⇒ m2 � �9
m � n � 0 ⇒ n � �m
m � n � 0.mn � 9x2 � 9 � �x � m��x � n� 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
8.Equation
(a)1
(b)
(c)
(d)x � 5 ± 3i
34105 � 3i, 5 � 3ix2 � 10x � 34 � 0
�2x � 3��2x � 3� � 0 �940�
32, 3
2
4x2 � 9 � 0
�2x � 1��x � 3� � 0 �32�
52
12, �3
2x2 � 5x � 3 � 0
�x � 2��x � 3� � 0�6�2, 3
x2 � x � 6 � 0
x1 � x2x1 � x2x1, x29.
x1 � x2 �ca
x1 � x2 � �ba
x2 �ba
x �ca
� 0
ax2 � bx � c � 0
10. (a) (i)
(ii) ��5 � 5�3i2 �3
� 125
��5 � 5�3i2 �3
� 125 (b) (i)
(ii) ��3 � 3�3i2 �3
� 27
��3 � 3�3i2 �3
� 27 (c) (i) The cube roots of 1 are:
(ii) The cube roots of 8 are:
(iii) The cube roots of 64 are:4, �2 ± 2�3i
2, �1 ± �3i
1, �1 ± �3i
2
12.
Since a and b are real numbers, is also a real number.a2 � b2
�a � bi��a � bi� � a2 � abi � abi � b2i2 � a2 � b2
11. (a)
�1 � i
2�
12
�12
i
�1
1 � i�
11 � i
�1 � i1 � i
zm �1z
(b)
�3 � i
10�
310
�1
10i
�1
3 � i�
13 � i
�3 � i3 � i
zm �1z
(c)
��2 � 8i
68� �
134
�2
17i
�1
�2 � 8i�
�2 � 8i�2 � 8i
�1
�2 � 8i
zm �1z
Problem Solving for Chapter 1 191
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
14.
This equation will have two solutions when or when
This equation will have one solution when or when
This equation will have no solutions when or when k > 2.1 �k2
< 0
k � 2.1 �k2
� 0
k < 2.1 �k2
> 0
��x � 1�2 � 1 �k2
x � 2�x � 1 � 1 �k2
x � 2�x � �k2
2x � 4�x � k � 0 Complete the square.
4�x � 2x � k Number of solutions (real) Some k-values
2
1 2 only
0 3, 4, 5
�1, 0, 1
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 16.
The greatest amount you could expect to get is 66 ounces and the least amount is 62 ounces.
4�15.5� � 62, 4�16.5� � 66
15.5 ≤ w ≤ 16.5
�12 ≤ w � 16 ≤ 1
2
w � 16 ≤ 12
Practice Test for Chapter 1
192 Chapter 1 Equations, Inequalities, and Mathematical Modeling
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.