Section 11.2 The Quadratic Formula
143.
42
8150xx
Let u =
2
x
42
8150
xx
Substitute
2
x
for u.
22
53
53
uu
xx
Apply the square root property.
53xx
The solution set is
5, 3 .
145.
3
33
2
18 1 2
12 12 4
xx
xxx
146.
3 152 0 6
147. a.
2
8210
(2 1)(4 1) 0
xx
xx
210 or 410
21 41
xx
xx
148. a.
2
2
9610
(3 1) 0
310
31
xx
x
x
x
149. a.
2
22
2
2
2
42
3
42
3
34 2
3420
xx
xx
xx
xx
xx
b.
22
444(3)(2)16248bac
1.
2
2950
2 9 5
xx
abc
2
4
2
4
911
4
bb ac
xa
Evaluate the expression to obtain two solutions.
911 1 911
Chapter 11 Quadratic Equations and Functions
2.
2
2
261
2610
2 6 1
xx
xx
ab c
2
4
2
bb ac
xa
3.
2
2
356
3650
3 6 5
xx
xx
abc
2
2
4
2
66435
bb ac
xa
4. a.
22
464(1)(9)0bac
Since the discriminant is zero, there is one real
rational solution.
b.
22
4 ( 7) 4(2)( 4) 81bac
Since the discriminant is positive and a perfect
2
2
53410
20 5 12 3 0
20 7 3 0
xx
xxx
xx
Thus, one equation is
2
20 7 3 0.xx
Other
equations can be obtained by multiplying both
sides of this equation by any nonzero real
number.
Section 11.2 The Quadratic Formula
c. Because the solution set is
7,7 ,ii
we have
7or 7
70 70
xi xi
xi xi
Use the zero-product principle in reverse.
6.
2
( ) 0.01 0.05 107PA A A
2
2
115 0.01 0.05 107
0 0.01 0.05 8
AA
AA
0.01 0.05 8abc
A woman’s normal systolic blood pressure is 115
mm at about 26 years of age.
This is represented by the point (26,115) on the blue
graph.
11.2 Concept and Vocabulary Check
1.
2
4
2
bb ac
a
6.
2
4bac
7. no
8. two
11.2 Exercise Set
1.
2
8120
1 8 12
xx
abc
2
884112
2.
2
884115 86460
21 2
84 82
22
x
82 82
or
22
10 6
22
53
xx
xx
xx
Chapter 11 Quadratic Equations and Functions
3.
2
2
27 5
2750
xx
xx
4.
2
2
58 3
5830
5 8 3
xx
xx
abc
The solution set is
1, .
5
5.
2
3200
1 3 20
xx
abc
6.
2
5100
1 5 10
xx
abc
7.
2
2
373
3730
3 7 3
xx
xx
ab c
39 32 341
88
Section 11.2 The Quadratic Formula
9.
2
2
621
6210
6 2 1
xx
xx
ab c
10.
2
2
245
2450
xx
xx
2 4 5abc
2
44425
22
x
11.
2
2
43 6
4360
4 3 6
xx
xx
ab c
920
xx
9 1 2abc
2
11492 1172
29 18
1711
171
18 18
x
Chapter 11 Quadratic Equations and Functions
14.
2
6130
1 6 13
xx
abc
15.
2
2
387
3870
3 8 7
xx
xx
ab c
16.
2
2
346
3460
xx
xx
3 4 6ab c
2
44436
23
x
17.
2
22 12
24 12
xx x
xxx
Evaluate the expression to obtain two solutions.
511 3 511
or 4
42 4
xx
42
4
The solution set is
3,1.
2
19.
2
830
1 8 3
xx
abc
which are conjugates.
Section 11.2 The Quadratic Formula
21.
2
680
1 6 8
xx
abc
22.
22
4241341216bac
23.
2
230
2 1 3
xx
abc
24.
2
2
4442316248bac
Since the discriminant is negative, there is no real
solution. There are two imaginary solutions that are
complex conjugates.
25.
2
260
2 6 0
xx
abc
26.
2
2
4543025025bac
27.
2
530
5 0 3
x
abc
Since the discriminant is negative, there is no real
solution. There are two imaginary solutions that are
complex conjugates of each other.
9 12 4
ab c
2
2
412494
bac
2
420250
xx
4 20 25ab c
2
2
4204425
400 400 0
bac
Since the discriminant is zero, there is one repeated
real rational solution.
31.
2
344
xx
The solution set is
2,2
3
.
Chapter 11 Quadratic Equations and Functions
32.
2
2
21
210
21 10
xx
xx
xx
33.
2
21xx
Since
2b
, add
22
2
211.
22
b
34.
2
2
231
2310
xx
xx
2 3 1abc
35.
2
2
39
390
3 1 9
xx
xx
ab c
1 107
6
1 107
66
i
i
63656
4
620
4
Section 11.2 The Quadratic Formula
37.
2
2
25 12
22552
2370
xx
xxx
xx
38.
23 41xx
2
2
283121
211121
xxx
xx
39.
2
34 16x
Apply the square root property.
3 4 16 or 3 4 16
xx
40.
2
27 25x
or
2 7 25 2 7 25
xx
2
31240xx
Apply the quadratic formula.
3 12 4ab c
63
The solution set is
626
.
3
Chapter 11 Quadratic Equations and Functions
42.
2
10
36
xx
2
2610
2 6 1
xx
ab c
43.
2
32 10
x
Apply the square root property.
44.
2
41 15
x
4 1 15 4 1 15
or
xx
45.
111
23
xx
The LCD is
32
xx
.
11 1
32 32
xx xx
4404210
210
22
2
2
4124 3
12 8 3
0512
1 5 12
xxxx
xx x
xx
ab c
2
Section 11.2 The Quadratic Formula
47.
2
26 25 112
246125512
xx x
xxx x
48.
2
2
2
72324
714328
714 25
71250
xx x
xx x
xxx
xx
49.
2
2
2
2
10 2(2 1)
10 4 2
410 2
4120
xx
xx
xx
xx
1 4 12ab c
2
444112
x
50.
2
(6) 12
612
xx
xx
i
The solution set is
422.i
51. Because the solution set is
3, 5 ,
we have
3or 5
xx
20 60
xx
2
2
260
62120
4120
xx
xxx
xx
53. Because the solution set is
21
,,
34
we have
Chapter 11 Quadratic Equations and Functions
54.
51
or
63
xx
55. Because the solution set is
2, 2 ,
we have
2or 2
20 20.
xx
xx
56.
3or 3
30 30
xx
xx
2
330
30
xx
x
57. Because the solution set is
25,
25
we have
Use the zero-product principle in reverse.
25 25 0
xx
59. Because the solution set is
6,6 ,ii
we have
6or 6
xi x i
2
36 0
x
60.
8or 8
80 80
xi x i
xi xi
880
xixi
61. Because the solution set is
1, 1 ,ii
we have
11
or
10 10.
xi xi
xi xi
Use the zero-product principle in reverse.
2
2
110
11110
xixi
xx ix i i i
xxxi
xxi
2
10
i
20 20
xi xi
Section 11.2 The Quadratic Formula
63. Because the solution set is
12,12
, we
have
2 1 0xx
64.
13or 13
130 130
xx
xx
67. a. The equation has two non-integer solutions
32
, so the graph crosses the x-axis at
32
and
32
.
2
22416
21
x
2
2
2120
2210
xx
xx
Apply the quadratic formula.
2 2 1ab c
2
.
71.
22
115
2
32 4
115
x
xx x
Chapter 11 Quadratic Equations and Functions
Apply the quadratic formula.
72.
2
11
23 56
11
23(2)(3)
xx
xxxx
xx
xx xx
Apply the quadratic formula.
2 6 2ab c
.
2
(6) (6) 42 2
22
63616620
44
645625
44
x
73.
2
23220xx
325 35
22 22
Evaluate the expression to obtain two solutions.
35 35
xx
The solution set is
2
22, 2
.
74.
2
36730xx
Apply the quadratic formula.
3 6 7 3abc
2
664373
23
x
Section 11.2 The Quadratic Formula
75.
2
22
22
23
23or 23
230 230
xx
xx xx
xx xx
Apply the quadratic formula to solve
2
Apply the zero product principle to solve
2
230.xx
(3)(1)0xx
3 0 or 1 0
3 or 1
xx
xx
The solution set is
3, 1, 1 2i
.
2
334(1)(2)
2(1)
39(8) 317
22
x
77.
2
2
2
0.013 1.19 28.24
3 0.013 1.19 28.24
0 0.013 1.19 25.24
fx x x
xx
xx
Apply the quadratic formula.
Thus, 33 year olds and 58 year olds are expected to
be in 3 fatal crashes per 100 million miles driven.
The function models the actual data well.
78.
2
2
2
0.013 1.19 28.24
10 0.013 1.19 28.24
0 0.013 1.19 18.24
fx x x
xx
xx
1.87383 0.50617
0.026 0.026
72.1 19
xx
xx
Drivers of approximately age 19 and age 72 are
expected to be involved in 10 fatal crashes per 100
Chapter 11 Quadratic Equations and Functions
79. Use the quadratic formula to solve
2
0.01 0.7 6.1 0.fx x x
0.01, 0.7, 6.1abc
80. Using a graphing calculator, we find that the height
of the shot put is approximately 0 feet when the
distance is 55.3 feet. Graph (a) shows the shot’s
path.
81. Let x = the width of the rectangle.
Let x + 4 = the length of the rectangle.
2
44418
21
416 32
x
82. Let x = the width of the rectangle.
Let
23x
= the length of the rectangle.
Alw
4
x
or
1.6
4
x
Disregard –3.1 because we can’t have a negative
length measurement. The solution is 1.6 and the
rectangle’s dimensions are roughly 1.6 inches by
2 1.6 3 6.2
inches.
2
16 48 0
xx
Apply the quadratic formula.
1 16 48ab c
2
16 16 4 1 48
Section 11.2 The Quadratic Formula
84. Let x = the length of the one leg.
Let x – 2 = the length of the other leg.
2
22
26
xx
2
224116
21
x
85.
2
2
20 2 13
20 2 13
02 20 13
xx
xx
xx
A gutter with depth 9.3 or 0.7 inches will have a
cross-sectional area of 13 square inches.
86.
22
82
44
xx
2
2
2
2
82
2166432
216320
x
x
xx
xx
87. Let x = the time for the first person to mow the yard
alone.
Let
1x
= the time for the second person to mow
the yard alone.
Fractional Time Fractional
2
2
2
44
111
1
414
444
074
xx xx
xx
xxxx
xxxx
xx
Chapter 11 Quadratic Equations and Functions
Apply the quadratic formula.
1 7 4ab c
2
77414
21
x
88. Let t = hours needed to fill the pool with inlet.
Let
2t
= hours needed to empty pool with outlet.
Then
1
t
is the inlet rate and
1
2t
is the outlet rate.
The difference between the inlet and outlet rates is
We can solve this equation for t.
11 1
82 82
28
828 2
tt tt
tt
tttt
89. – 96. Answers will vary.
97.
20 2fx x x
98. does not make sense; Explanations will vary.
Sample explanation: The factoring and the square
root methods are both likely to be quicker ways to
solve this equation.
101. does not make sense; Explanations will vary.
Sample explanation: There are two imaginary
solutions. They are not classified as irrational.
102. true
Section 11.2 The Quadratic Formula
106.
2
0
2
0
0
16
016
16, ,
stvt
tvts
abvcs
107. The dimensions of the pool are 12 meters by 8
meters. With the tile, the dimensions will be 12 +
2x meters by 8 + 2x meters. If we take the area of
the pool with the tile and subtract the area of the
pool without the tile, we are left with the area of the
tile only.
Evaluate the expression to obtain two solutions.
10 14.8 10 14.8
or
22
4.8 24.8
22
xx
xx
108. The shaded area is equal to the total area of the
triangle minus the area of the rectangle.
1
Abhlw
2
884110
21
86440
2
8 104
x
43
x
x
The solution set is
1
3, .
4
Chapter 11 Quadratic Equations and Functions
110.
22
25 31
25 31
25 31
xx
xx
xx
The solution set is
3, 7 .
111.
2
553535
3333
xx
xxxx
112.
22
() () 2
39 11
24 6
xfxxgxx
113.
22
() () ( 2)
39 1
24 0
xfxxgx x
2
2
2
2( 3) 8
(3)4
(3) 4
32
32
x
x
x
x
x
32 or 32
51
xx
xx
The x-intercepts are 1 and 5.
Since
1a
is negative, the parabola opens
downward. The vertex of the parabola is
,1,4hk
. Replace
fx
with 0 to find x–
intercepts.
2
014
x
31
xx