Section 11.5 Polynomial and Rational Inequalities
61.
2
23612
xx
Express the inequality without the absolute value symbol.
22
22
2 36 12 or 2 36 12
2240 2480
xx xx
xx xx
Solve the related quadratic equations.
The boundary points are
8, 6, 4 and 6
.
2
Test Interval Test Number Test Conclusion
9
, 8 , 8 belongs to the
9293612
solution set.
27 12, True
The solution set is
,8 6,4 6,
.
62.
2
618
xx
Express the inequality without the absolute value symbol.
22
61 8or 618
xx xx
Chapter 11 Quadratic Equations and Functions
The boundary points are
7, 3, and 1
.
2
Interval Test Value Test Conclusion
8
, 7 , 7 belongs to the
86818
solution set.
17 8, True
63.
33
32
xx
Express the inequality so that one side is zero.
33
0
32
xx
Find the values of x that make the denominator zero.
30 20
32
xx
xx
The boundary points are
3
and 2.
Interval Test Value Test Conclusion
433
, 3 , 3 does not belong
False
43 42 to the solution set.
Section 11.5 Polynomial and Rational Inequalities
64.
12
11
xx
Express the inequality so that one side is zero.
Find the values of x that make the numerator and denominator zero.
30 10 10
311
xx x
xx x
The boundary points are
3
,
1
, and 1.
Interval Test Value Test Conclusion
412
, 3 , 3 belongs to the
41 41 solution set.
12
, True
The solution set is
,3 1,1
.
Chapter 11 Quadratic Equations and Functions
65.
2
2
20
43
xx
xx
Find the values of x that make the numerator and denominator zero.
The boundary points are 1, 1, 2 and 3.
2
2
Interval Test Value Test Conclusion
2
, 1 , 1 belongs to the
222
0solution set.
2423
40, True
15
2
2
2.5
2,3 2,3 does not belong
2.5 2.5 2 0to the solution set.
2.5 4 2.5 3
70, False
3
Section 11.5 Polynomial and Rational Inequalities
66.
2
2
32
0
23
xx
xx
Find the values of x that make the numerator and denominator zero.
The boundary points are 1, 1, 2 and 3.
2
2
Interval Test Value Test Conclusion
2
, 1 , 1 belongs to the
2322
0solution set.
2223
12 0, True
5
2
2
2.5
2,3 2,3 does not belong
2.5 3 2.5 2 0to the solution set.
2.5 2 2.5 3
30, False
7
Chapter 11 Quadratic Equations and Functions
67.
32
32
211 76
211760
xxx
xxx
The graph of
32
21176
fx x x x
appears to cross the x-axis at
6
,
1
, and 1. Verify this numerically by
2
68.
32
32
211 76
211760
xxx
xxx
69.
13
42 42
13
0
4242
xx
xx
70.
13
42 42
xx
Section 11.5 Polynomial and Rational Inequalities
71.
2
16 48 160
st t t
To find when the height exceeds the height of the building, solve the inequality
2
16 48 160 160.
tt
Solve the related quadratic equation.
2
16 48 160 160
tt
The boundary points are 0 and 3.
Interval Test Value Test Conclusion
0,3
1
2
16 1 48 1 160 160
192 160, true
0,3
belongs to the solution set.
72.
2
16 8 87 87
tt
2
2
16 8 87 87
16 8 0
tt
tt
Interval Test Value Test Conclusion
1
0, 2
1
4
2
11
16 8 87 87 true
44
1
0, 2
belongs to the solution set.
Chapter 11 Quadratic Equations and Functions
73.
2
( ) 0.0875 0.4 66.6fx x x
2
( ) 0.0875 1.9 11.6gx x x
d.
2
2
0.0875 0.4 66.6 540
0.0875 0.4 473.4 0
xx
xx
Solve the related quadratic equation.
2
0.0875 0.4 473.4 0xx
Since the function’s domain is
30,x
we must test the following intervals.
2
Interval Test Value Test Conclusion
50
30,76 30,76 does not belong
0.0875(50) 0.4(50) 66.6 540
to the solution set.
265.35 540, False
74.
2
( ) 0.0875 0.4 66.6fx x x
2
( ) 0.0875 1.9 11.6gx x x
a.
2
(55) 0.0875(55) 0.4(55) 66.6 309 feetf
2
(55) 0.0875(55) 1.9(55) 11.6 381 feetg
Section 11.5 Polynomial and Rational Inequalities
2
2
4
2
(1.9) (1.9) 4(0.0875)(528.4)
2(0.0875)
89 or 68
bb ac
xa
x
x
Since the function’s domain is
30,x
we must test the following intervals.
Interval Test Value Test Conclusion
75.
500,000 400x
Cx x
To find when the cost of producing each wheelchair does not exceed $425, solve the inequality
500,000 400 425.
x
x
Express the inequality so that one side is zero.
500,000 400 425 0
x
The boundary points are 0 and 20,000.
Interval Test Value Test Conclusion
0,20000
1
500,000 400 1 425
1
500,400 425, false
0,20000
does not belong to the solution
set.
Chapter 11 Quadratic Equations and Functions
76.
500,000 400 410
500,000 400 410 0
x
x
x
Interval Test Value Test Conclusion
0,50000
1
500,000 400 1 410 false
1
0,50000
does not belong to the solution
set.
77. Let x = the length of the rectangle.
Since
Perimeter 2 length 2 width
, we know
50 2 2 width
50 2 2 width
x
x
Solve the related equation
2
2
25 114
025114
0196
xx
xx
xx
Section 11.5 Polynomial and Rational Inequalities
The boundary points are 6 and 19.
2
Interval Test Value Test Conclusion
1
0,6 0,6 belongs to the
25 1 1 114
solution set.
24 114, True
78.
22
2 2 180
21802
90
lwP
lw
lw
lw
We want to restrict the area to 800 square feet. That is,
Assuming the width is the shorter side, we ignore the larger solution.
2
Interval Test Value Test Conclusion
0,10 5 90 5 5 800 true 0,10 is part of the solution set
79. – 81. Answers will vary.
82.
2
3100xx
Chapter 11 Quadratic Equations and Functions
83.
2
2530xx
84.
40
1
x
x
The graph crosses the x-axis at
1
and 4. The graph is below the x-axis when
14x
. Thus, the solution set is
1, 4 .
85.
22
3
x
x
86.
12
14
xx
The solution set is
4, 1 2, .
88. a.
2
( ) 0.1125 0.1 55.9fx x x
b.
2
2
0.1125 0.1 55.9 455
0.1125 0.1 399.1 0
xx
xx
Solve the related quadratic equation.
59 or 60
x
Since the function’s domain must be
0,x
we must test the following intervals.
2
Interval Test Value Test Conclusion
50
0,60 0,60 does not belong
0.1125(50) 0.1(50) 55.9 455
to the solution set.
332.15 455, False
Section 11.5 Polynomial and Rational Inequalities
89. a.
2
( ) 0.1375 0.7 37.8fx x x
b.
2
0.1375 0.7 37.8 446
xx
Since the function’s domain must be
0,x
we must test the following intervals.
2
Interval Test Value Test Conclusion
10
0,52 0,52 does not belong
0.1375(10) 0.7(10) 37.8 446
to the solution set.
58.55 446, False
90. makes sense
91. does not make sense; Explanations will vary. Sample explanation: Polynomials are defined for all values.
92. makes sense
96. false; Changes to make the statement true will vary. A sample change is: The inequalities have different solution sets.
The value, 1, is included in the domain of the first inequality, but not included in the domain of the second inequality.
97. true
98. Answers will vary. An example is as follows.
Chapter 11 Quadratic Equations and Functions
99. Answers will vary. An example is
30.
4
x
x
101.
2
20x
Since the left hand side of the inequality is a square,
102. There is no value of x for which the left hand side
will be less than –1. The solution set is
.
103.
2
10
2
x
104. a. The x–axis is the divider between y–values that
are positive and y–values that are negative.
Since the entire graph falls above the x–axis, we
know that all of the corresponding y–values are
c. First, consider
2
4870xx
from part a.
2
4870xx
24
8 64 112
8
1
888 2
x
i
The graph of
2
4870xx
is a parabola. Since
the values of x are complex numbers, we know
there are no x–intercepts. This means that the entire
graph lies above the x–axis and the value of
2
105. The radicand must be greater than or equal to zero:
2
27 3 0x
The inequality is true for values between 3 and –3.
This means that the radicand is positive for values
Section 11.5 Polynomial and Rational Inequalities
107.
2
22
26 9
816 34
xx
xx xx
108.
44
16
xy
109.
() 2
x
fx
3
() 2 (, )
11
32 3,
88
x
xfx xy
110.
() 2
x
fx
2
11
22 2,
44
x
1
0
44
11
12 11 1,1
22
0212 0,2
Chapter 11 Quadratic Equations and Functions
Chapter 11 Review Exercises
1.
2
2 3 125
x
2.
2
2
31500
3150
x
x
3.
2
2
320
32
x
x
4.
2
418
418
x
x
5.
2
736
736
x
x
Since b = 20, add
22
2
20 10 100
22
b
.
2
2
93
342
xx x
2
93
xx
The solution set is
3, 9 .
Chapter 11 Review Exercises
9.
2
2
710
7 1
xx
xx
Apply the square root property.
753
24
753753
22 2
x
x
The solution set is
753
.
2
10.
2
2
2340
320
2
xx
xx
2
2
39 9
2
216 16
3329
341
4
xx
x
x
The solution set is
341
.
4
11.
2
2
1
2916 2500 1
2916 1
2500
t
AP r
r
r
Apply the square root property.
2916
Apply the square root property.
2
196
196
14
t
t
t
The solutions are –14 and 14. Disregard –14, because
we cannot have a negative time measurement. The
fetus will weigh 588 grams after 14 weeks.
Chapter 11 Quadratic Equations and Functions
13.
Use the Pythagorean Theorem.
The solutions are
60 5
meters. Disregard
60 5
meters, because we can’t have a negative
length measurement. Therefore, the building is
60 5
meters, or approximately 134.2 meters high.
15.
()
()
()
()
22
222
24 53
24 2 2 4
44 8 42222.83
d=−−− +−
=−+ + = +
=+= =⋅= ≈
18.
2
2
24
240
1 2 4
xx
xx
ab c
2
22414
21
x
19.
2
2190
1 2 19
xx
ab c
2
2
262 2132 132
22
ii
i
The solution set is
132.i
x
2x
300
Chapter 11 Review Exercises
20.
2
2
234
2430
xx
xx
2
21.
2
4130
1 4 13
xx
ab c
22.
2
2
923
9320
xx
xx
9 3 2abc
Find the discriminant.
23.
2
2
243
2430
xx
xx
The solution set is
3
25.
2
22
23 2 24
xx xx
21
3 9 ( 40)
2
349 37 5 or 2
22
Chapter 11 Quadratic Equations and Functions
27.
2
2
16 0
16
x
x
28.
2
2
380
38
x
x
Apply the square root principle.
29.
2
320xx
Use the quadratic formula.
3 1 2ab c
30.
51
2
14
51
41 412
x
x
x
xx
2
884111
2
845
2
825 24 5 45
22
x
or
35
31 53
310 530.
xx
xx
22
2
2
990
9981 0
81 1 0
81 0
xixi
xixixi
x
x