Mid-Chapter Check Point
11.
2
2
(2 3) 4
23 4
xx
xx
12.
2
2
323
xx
Multiply both sides of the equation by the least common denominator 6.
2
2
66
xx
13.
2
22
42
68
22
(4)(2) 4 2
xx
xx
xx
xx
xx x x
Multiply both sides of the equation by the least common denominator (4)(2)xx
.
Chapter 11 Quadratic Equations and Functions
14.
2
2
10 3 0
10 3
xx
xx
Since
10b
, add
2
2
10 525
2
.
15. distance:
2
2
22
22 2 2
44
d
16. distance:
22
22
10 5 14 8
56
d
17.
2
34fx x
Since
1a
is positive, the parabola opens upward.
The vertex of the parabola is
,3,4hk
.
Replace
fx
with 0 to find x–intercepts.
2
03 4 5y
Since
1a
is negative, the parabola opens
downward. The vertex of the parabola is
,2,5hk
. Replace
gx
with 0 to find x–
intercepts.
2
025
x
Mid-Chapter Check Point
19.
2
45hx x x
Since
1a
is negative, the parabola opens
downward. The x–coordinate of the vertex of the
parabola is
44
2
221 2
b
a
and the y–
485 9
The vertex is
2, 9 .
Replace
hx
with 0 to find
the x–intercepts.
2
2
045
045
051
xx
xx
xx
20.
2
361fx x x
Since a = 3, the parabola opens upward. The x–
coordinate of the vertex of the parabola is
31 61 1
361 2
The vertex
1, 1
. Replace
fx
with 0 to find the
x-intercepts.
2
03 6 1xx
6
624
6
626 3 6 0.18 or 1.82
63
Set
0x
and solve for y to obtain the y–intercept:
2
Chapter 11 Quadratic Equations and Functions
22.
2
10 4 15 15
10 40 15 15
xx x
xxx
23. Because the solution set is
13
,,
24
we have
13
or
24
xx
24. Because the solution set is
23,23
we have
23 or 23
xx
25.
2
150 4425Px x x
Since
1a
is negative, the function opens down
and has a maximum at
26. Let x = one of the numbers.
Let
18 x
= the other number.
The vertex is
9,81
. The maximum product is
81. This occurs when the two number are
9
and
18 ( 9) 9
.
27. Let x = the measure of the height.
Let
40 2x
= the measure of the base.
Since
1a
is negative, the function opens down and
has a maximum at
20 20 10
221 2
b
xa
.
2
10 10 20 10
A
Section 11.4 Equations Quadratic in Form
11.4 Check Points
1. Let
2
.ux
42
560
xx
2. Let .ux
2
280
280
xx
xx
16
x
Disregard –2 because the square root of x cannot be
a negative number. We must check 16, because
both sides of the equation were raised to an even
power.
Check:
280
xx
3. Let
2
4.ux
222
2
(4)(4)60
60
xx
uu
2
11
2
210
210
(1)(21)0
xx
uu
uu
1
1
11
1
1
1
2
(1)
1
12
2
x
x
xx
x
The solution set is
1, 2 .
Chapter 11 Quadratic Equations and Functions
5. Let
1
3
.ux
21
33
2
31140
xx
3
Replace
1
3
with .ux
1or 4
uu
The solution set is
1,64 .
27
11.4 Concept and Vocabulary Check
1.
2
x
;
2
13 36 0uu
2.
1
2
or xx
;
2
280uu
11.4 Exercise Set
1. Let
2
.ux
42
41
uu
Replace
2
with .ux
22
or
41
xx
2
13 36 0
990
ut
uu
90or 40
94
uu
uu
22
or
94
32
xx
xx
The solution set is
3, 2, 2,3
.
Section 11.4 Equations Quadratic in Form
4. Let
2
.ux
42
2
22
9200
xx
5. Let
2
.ux
42
42
28
280
xx
xx
6. Let
2
.ux
42
42
2
22
2
45
450
450
450
xx
xx
xx
uu
7. Let .ux
2
20
20
xx
xx
Disregard –2 because the square root of x cannot be
a negative number. We must check 1, because both
sides of the equation were raised to an even power.
Check:
1120
30 or 20
32
uu
uu
3x or 2
4
x
x
Chapter 11 Quadratic Equations and Functions
9. Let
1
2
.ux
70or 30
73
uu
uu
Replace
1
2
with .ux
Check:
1
2
49 4 49 21 0
49 4 7 21 0
10. Let
1
2
.
ux
1
2
680
xx
40 or 20
42
uu
uu
2
2
13 40 0
13 40 0
13 40 0
580
xx
xx
uu
uu
Both solutions must be checked since both sides of
the equation were raised to an even power.
25
25 13 25 40 0
x
64 13 64 40 0
64 13 8 40 0
64 104 40 0
00
Both solutions check. The solution set is
25,64 .
Section 11.4 Equations Quadratic in Form
12. Let
.ux
27 300
xx
250 or 60
25 6
5
2
uu
uu
u
13. Let 5.ux
2
2
54 5210
4210
730
xx
uu
uu
14. Let 3.ux
2
37 3180
xx
15. Let
2
1.ux
2
22
112
210
xx
uu
Apply the zero product principle.
20or 10
21
uu
uu
0
3
x
x
The solution set is
3,0, 3 .
16. Let
2
2.ux
2
22
226
xx
50
0
5
xx
x
x
The solution set is
5,0, 5 .
Chapter 11 Quadratic Equations and Functions
Replace
2
with 3 .uxx
First, consider u = 10.
2
2
310
3100
520
xx
xx
xx
The solution set is
5, 2,
1, 2 .
18. Let
2
2.ux x
2
22
2
211 2240
xx xx
First, consider u = 8.
2
28xx
2
280xx
420xx
19. Let
1
.ux
21
2
11
2
20 0
20 0
xx
xx
11
54
xx
The solution set is
11
,.
45
20. Let
1
.ux
60
320
uu
uu
30or 20
32
uu
uu
Section 11.4 Equations Quadratic in Form
21. Let
1
.ux
21
2
11
2730
2730
xx
xx
Replace
1
with .ux
11
1or 3
2
11 13
xx
22. Let
1
.ux
21
2
11
2
20 9 1 0
20 9 1 0
20 9 1 0
xx
xx
uu
23. Let
1
.ux
21
21
2
11
43
430
430
xx
xx
xx
22
22 7
427 27
22
Replace
1
with .ux
Rationalize the denominator.
2
2
127 27
2727
27
2727 27
47 3 3
x
Chapter 11 Quadratic Equations and Functions
24. Let
1
.ux
21
21
64
640
xx
xx
35 1
1
35
135
x
x
x
x
25. Let
1
3
.ux
21
33
60xx
27 8
xx
The solution set is
8, 27 .
26. Let
1
3
.ux
Section 11.4 Equations Quadratic in Form
27. Let
1
5
.ux
Replace
1
5
with .ux
28. Let
1
5
.ux
21
55
2
11
55
20
20
xx
xx
29. Let
1
4
.ux
2
u
1
16
Since both sides of the equations were raised to an
even power, the solutions must be checked.
1
First, check .
16
x
11
24
11
21
Chapter 11 Quadratic Equations and Functions
30. Let
1
4
.ux
11
24
2
11
44
253
2530
xx
xx
31. Let 8.uxx
2
88
5140
xx
xx
2
780
810
xx
xx
80 or 10
81
xx
xx
Next, consider u= 2 .
2
6270
930
xx
uu
uu
90 or 30
93
uu
uu
Section 11.4 Equations Quadratic in Form
10 3
10 3
xx
xx x
x
33.
42
42
54
54
fx x x
yx x
Set y = 0 to find the x–intercept(s).
42
054xx
34.
42
42
42
13 36
13 36
01336
fx x x
yx x
xx
94
uu
Substitute
2
for .xu
22
or
94
32
xx
xx
The intercepts are
2 and 3
. The corresponding
Substitute
1
6
for .xu
11
66
66
11
66
66
31
or
31
729 1
xx
xx
xx
Chapter 11 Quadratic Equations and Functions
Since both sides of the equations were raised to an
even power, the solutions must be checked.
First check x = 729.
Next check x = 1.
11
36
12130
36.
21
21
21
6
6
06
fx x x
yx x
xx
Let
1
.ux
37.
2
2
29 220
29 220
fx x x
yx x
2
2
29 2200
9200
xx
uu
25or 24
32
xx
xx
The intercepts are 2 and 3. The corresponding
graph is graph f.
2
Section 11.4 Equations Quadratic in Form
39. Let
2
32ux x
2
22
2
16
32 10 32 16
10 16
fx
xx xx
uu
Apply the zero product principle.
50 or 20
52
xx
xx
Next, consider
2u
.
2
322
xx
40. Let
2
22ux x
.
6
fx
First, consider
1u
.
2
2
221
230
(3)(1)0
xx
xx
xx
2
2
2
2
11
31512
352
3520
fx
xx
uu
uu
First, consider
1
3
u
.
Chapter 11 Quadratic Equations and Functions
Next, consider
2u
.
112
x
42. Let
3
.ux
21
33
2
232
fx
xx
Replace
1
3
with .ux
11
33
1
2or
2
xx
43. Let
4
x
ux
.
94
uu
Replace u with
4
x
x
.
First, consider
9u
.
81 324
80 324
324 81
80 20
xx
x
x
Next, consider
4u
.
case, both check, so the solutions are
20
and
15
.
Section 11.4 Equations Quadratic in Form
44. Let
2
x
ux
.
10 11
22
fx gx
xx
xx
Replace u with
2
x
x
.
First, consider
10u
.
10
2
x
x
45. Let
1
4.ux
21
12
34 164 12
fx gx
xx
First, consider
2
3
u
.
1
2
43
12
43
x
x
16
4
641
6241
625
x
x
x
x
2
2
33
656
6560
32230
xx
uu
uu
uu
Chapter 11 Quadratic Equations and Functions
For
2
3
u
: For
3
2
u
:
22
x
23
x
The solutions are
9
and
4
.
47.
2
2
0.04 40 3 40 104
60 0.04 40 3 40 104
Px x x
xx
2
3340.0444
20.04
397.04
0.08
31.9631.4
55 or 20
0.08 0.08
u
48.
2
2
0.04 40 3 40 104
50 0.04 40 3 40 104
Px x x
xx
2
3340.0454
20.04
398.6430.36
0.08 0.08
u
house is very important. The function models the
data well.
49. – 51. Answers will vary.
52.
63
780xx
Using a graphing utility we find the solution set is
1, 2
.