Chapter 11
Quadratic Equations and Functions
Exercise Set 11.1
2. a) 5x2=35
x2=7
6. 3x2−7=0
x≈−1.449 or x ≈3.449
10. (x+3)
2=9
12. (x−9)2=34
14. (x+1)
2=−9
x+1= ±3i
x=−1±3i
18. y+3
42
=17
16
(p−4)2=1
26. x2−18x=10
x2−18x+81 =10+81
28. x2−x=3
296 Chapter 11: Quadratic Equations and Functions
30. y2+9y=8
y2+9y+81
4=8+81
4
x−2
32
=10
9
34. r2+2
5r=4
5
36. x2−8x+15=0
x2−8x+16 =−15+16
(x−4)2=1
38. x2+18x+74=0
40. x2+10x−4=0
x2+10x+25 =4+25
42. a) x2−7x−2=0
x2−7x+49
44. a) 2x2−3x+9= 0
x2−3
2x=−9
2
b) The equation 2x2−3x+ 9 = 0 has no real-number
solutions, so f(x)=2x2−3x+9 has no x-intercepts.
46. x2+3
x+3
42
=41
16
48. 2x2+3x−1=0
x2+3
2x−1
2=0
y
f(x) ⴝ 5
ⴚ
2x2
Exercise Set 11.1 297
50. 3x2+4x−3=0
x2+4
52. x2−x=−1
2±i√3
2
54. x2−6x=−13
56. 48 = 48T2
58. 640=16t2
60. 1550 = 16t2
62. 2684 = 16t2
64.
70. t3−8
t2−5t+6 =(t−2)(t2+2t+4)
(t−2)(t−3)
72.
x
=6
=6
1
b=±√300 = ±10√3
4(r – 7),
or 4r – 28
B
298 Chapter 11: Quadratic Equations and Functions
80. Let rrepresent the speed of the fishing boat. Then r−7
Exercise Set 11.2
2. x2−6x−4=0
4. 3u2=18u−6
6. x2+x+2=0
8. x2+13=6x
10. h2+4=6h
12. 1+ 5
x2=2
x
14. a) 4x+x(x−3) = 0
b) (−1,0), (0,0)
16. a) 7x2+8x=−2
18. a) 49x2−14x+1=0
3x2+3x−7x2−14x=6
22. 11(x−2)+(x−5)=(x+ 2)(x−6)
Exercise Set 11.2 299
24. 15x=2x2+16
28. 3
x+x
3=5
2
30. 1
x+1
x+4 =1
7
x=−(−10) ±(−10)2−4·1·(−28)
2y2−(y2−y−6)=12
34. (x+3)
2+(x−1)2=0
x2+6x+9+x2−2x+1=0
36. x3+27 =0
38. x2+4x−7=0
x=−4±2√11
2=−2±√11
x≈−5.317 or x ≈1.317
x≈0.268 or x ≈3.732
42. 3x2−3x−2=0
44. 2y2+2y−3=0
46. √x+1+2= √3x+1
4√x+1=2x−4
2√x+1= x−2
300 Chapter 11: Quadratic Equations and Functions
48. 3
√4x−7=2
50. 100x2+1=20x
52. 3
x−2=5
6x
56. 5
x=−(−44) ±(−44)2−4·7·140
60. (1 + √3)x2−(3+2
√3)x+3=0
√3±√9
√3±3
x2+14x+13=0
RC2. (e)
4. Let w= the width of the flag, in inches. Then 2w−3=
6. Let h= the height of the sail, in meters. Then h−9 = the
8. Let h= the height of the sail, in feet, and h−8 = the
10. Let x= the width of the border, in feet, as shown in
the drawing in the text. Solve (8 −2x)(6 −2x)=1
12. Let w= the width of the flag, in inches. Then w+12=
14. We use the labels on the drawing in the text. Solve
Exercise Set 11.3 301
16. Let x= the width, in cm, and 2x= the length. Solve 328 =
18. Let x= the width of the frame, in cm. Solve
20. Let x= the length of the longer leg, in meters, and x−10
= the length of the shorter leg. Solve x2+(x−10)2=22
2.
22. Let rrepresent the speed and tthe time for the first part
of the trip.
Canoe trip Distance Speed Time
5, or 12 km/h, and the speed on
24. Let rrepresent the speed and tthe time.
26. Let rrepresent the speed and trepresent the time of the
Super-prop 2800 r t
t−3= 2000
r+50.
28. Let rrepresent the speed and tthe time of the trip to
Richmond.
Trip Distance Speed Time
30. Let rrepresent the speed of the boat in still water, and let
Trip Distance Speed Time
302 Chapter 11: Quadratic Equations and Functions
32. A=4πr2
34. N=kQ1Q2
s2
36. V=1
3s2h
38. a2+b2+c2=d2
40. s=v0t+gt2
2
42. A=πr2+πrs
0=πr2+πrs −A
LC
L=1
48. m=m0
1−v2
m21−v2
c2=(m0)2
(x−1)(x2+x+1)
=x3+x2+2x+2
(x−1)(x2+x+1)
LCD = (2x+i)(x−i)(x+i)
4(x−i)(x+i)−(2x+i)(x+i) = 2(2x+i)(x−i)
56. Let t= the number of hours it takes Chad to make 100
large pizza crusts. Then t−1.2 = Ron’s time. Solve:
Exercise Set 11.4 303
58. From Example 7, we know
Exercise Set 11.4
2. b2−4ac =12
2−4·1·36 = 0
There is one real solution.
6. b2−4ac =0
2−4·1·(−3)=12
There are two real solutions.
There are two nonreal solutions.
2−4·4·0=49
18. x=−11 or x =9
20. x=−ior x=i
22. x=−3or x =−3
24. x=−1
26. x=c
2or x =d
2
x−c
28. x=√2or x =3
√2
30. x=8ior x=−8i
32. x4−7x2+12=0
All four numbers check.
34. 2x−9√x+4=0
Both numbers check.
304 Chapter 11: Quadratic Equations and Functions
36. (x2+5x)2+2(x2+5x)−24 = 0
Let u=x2+5x.
x=−5±52−4·1·(−4)
38. 3x−2−x−1−14 = 0
3or x−1=−2
1
40. (2 + √x)2−3(2 + √x)−10 = 0
Let u=2+√x.
42. (2t2+t)2−4(2t2+t)+3=0
Let u=2t2+t.
2t2+t=3 or 2t2+t=1
44. t4−10t2+9=0
Let u=t2.
46. m−2+9m−1−10 = 0
u=−10 or u =1
48. 6x4−17x2+5=0
u=1
3or u =5
2
x2=1
3or x2=5
2
(u+ 4)(u−2) = 0
u=−4or u =2
52. x+3
−x+3
y
5x ⴚ 2y
ⴝ
8
Exercise Set 11.4 305
u=3 or u =−2
54. 16x−1
x−82
+8
x−1
x−8+1=0
x=12
5
56. y2−1
y2
−4y2−1
y−12 = 0
y2−1=6yor y
2−1=−2y
y=−2±22−4·1·(−1)
2·1
58. Solve: 3x+10
√x−8=0
60. Solve: (x2−x)2−8(x2−x)+12=0
Let u=x2−x.
62. The x-intercepts occur where f(x) = 0. Thus, we must
have x2/5+x1/5−6=0.
u2+u−6 = 0 Substituting ufor x1/5
(u+ 3)(u−2) = 0
64. Let xand yrepresent the amounts of A and B, respectively,
that should be used.
70.
y
x
f(x) ⴝ
ⴚ
x
ⴚ
3
306 Chapter 11: Quadratic Equations and Functions
72.
74. a) 28.423
76. a) kx2−17x+33=0
b) 2x2−17x+33=0
78. kx2−4x+(2k−1) = 0
x=2±√4+k−2k2
k
The product of the solutions is 3, so we have:
k=−1.
u=12 or u =−2
x
x−3=12 or x
x−3=−2
x
x−3= 144 or No solution
x= 144x−432
432 = 143x
432
143 =x
This number checks.
Let u=a3/2.
u2−26u−27 = 0
a=27
2/3No solution
u2+7u−8=0
Chapter 11 Mid-Chapter Review
5. 5x2+3x=4
Chapter 11 Mid-Chapter Review 307
6. 5x2+3x=4
x=−3
10 ±√89
10
7. x2+1= −4x
8. 2x2+5x−3=0
2x2+5x=3
x+5
=49
9. x2+10x−6=0
10. x2−x=5
11. x2−10x+25=0
a=1,b=−10, c=25
We compute the discriminant.
a=1,b=0,c=−11
We compute the discriminant.
13. y2=1
3y−4
7
308 Chapter 11: Quadratic Equations and Functions
14. x2+5x+9=0
=−9
15. x2−4=2x
x2−2x−4=0
We compute the discriminant.
b2−4ac =(−2)2−4·1·(−4)
a=1,b=−8, c=0
We compute the discriminant.
b2−4ac =(−8)2−4·1·0
17. The solutions are −1 and 10.
x=−1or x =10
18. The solutions are −13 and 13.
19. The solutions are −√5 and 3√5.
√5
x+4i=0 or x −4i=0
21. There is only one solution, −6. It must be a double solu-
x+6=0 or x +6= 0
(x+ 6)(x+6)=0
22. The solutions are −4
3and 2
7.
23. Familiarize. Let r= the speed, in mph, and let t= the
time of the trip, in hr, at the speed r. We organize the
information in a table.
5 mph
faster
780 r+5 t−1
Solve. We substitute 780
r(r+5)
r−1=r(r+5)·780
r+5
780(r+5)−r(r+ 5) = 780r
Chapter 11 Mid-Chapter Review 309
24. R=as2
25. 3x2+x=4
3x2+x−4=0
26. x4−8x2+15=0
Let u=x2.
27. 4x2=15x−5
x=−b±√b2−4ac
2a
28. 7x2+2= −9x
7x2+9x+2=0
x(x+1) =0
31. 49x2+16=0
The solutions are ±4
7i.
x2=−4or x2=6
x=√−4or x =−√−4or x =√6or x =−√6
33. r2+5r=12
r2+5r−12 = 0
2=−5±√73
2
The solutions are −5±√73
2.
310 Chapter 11: Quadratic Equations and Functions
34. s2+12s+37=0
The solutions are −6±i.
x−5
2=√11
2or x −5
2=−√11
2
36. x+1
x=7
3,LCM is 3x
The solutions are 7±√13
6.
37. 4x+1=4x2
x=4±√16+16
38. (x−3)2+(x+5)
2=0
a=1,b=2,c=17
39. b2−16b+64=3
(b−8)2=3
40. (x−3)2=−10
The solutions are 3 ±√10i.
41. 1
x=8±√64 + 40
2=8±√104
2=8±√4·26
2
42. x−√x−6=0
Let u=√x.
x=9 or x =4
Vertex:
(0, 0)
x ⫽ 0
⫺4⫺224
⫺4
⫺2
⫺5⫺3⫺1135
⫺5
⫺3
⫺1
x
Vertex:
(⫺1, 0)
⫺4
⫺5
1
Vertex:
(0, 0)
24
2
4
135
1
3
y
Exercise Set 11.5 311
43. Given the solutions of a quadratic equation, it is possible
to find an equation equivalent to the original equation but
44. Given the quadratic equation ax2+bx +c=0,wefind
x=−b+√b2−4ac
2aor x=−b−√b2−4ac
2ausing the
45. Set the product
46. Write an equation of the form a(3x2+1)2+b(3x2+1)+c=0
Exercise Set 11.5
2. f(x)=5x2
The vertex is (0,0).
x f(x)
0 0 ←Vertex
−2 1
10.
⫺4
4
⫺5
5
y
x ⫽ 1
4
5
y
x ⫽ ⫺4
4
3
5
y
f(x) ⴝ 4(x ⴚ 2)2
⫺
4
⫺
2
⫺
5
⫺
3
y
f(x) ⴝ ⴚe(x ⴙ 3)
2
⫺4⫺224
⫺5⫺3⫺1135
⫺1
x
f(x) ⴝ ⴚⴚ(x ⴚ 1)2 ⴚ 3
1
2
x ⫽ 1
⫺4
4
⫺5
5
y
Maximum: ⫺3
f(x) ⴝ ⴚ2(x ⴚ 5)2 ⴚ 3
⫺5
y
x ⫽ 4
⫺4
⫺5
⫺3
y
12.
14. x f(x)
−4 0
−5−2
16.
18.
22.
24.
26.
Exercise Set 11.6
2. f(x)=x2+2x−5
⫺2
⫺1
x
y
4224
531135
1
˛
81
16
Maximum:
(√, ˛)
81
y
5
Exercise Set 11.6 313
4. f(x)=−x2+4x+1
f(x)=−(x2−4x+4−4)+1
6. f(x)=4x2+8x+1
Line of symmetry: x=−1
8. f(x)=−2x2+2x+1
10. f(x)=x2−3x
12. f(x)=−4x2−7x+2
y
x
—4
4
5
—5
f
(
x
) = — + – – –
8
4
8
314 Chapter 11: Quadratic Equations and Functions
14. f(x)=x2+2x+12
16. f(x)=−x2+5x+24
18. f(x)=3x2−6x+1
20. f(x)=2x2+4x−1
f(0) = 2 ·02+4·0−1=−1, so the y-intercept is (0,−1).
22. C=kt
24. y=k
26. y=kx
28. (−2,1), (0,1)
30.
or 5, so by symmetry the other x-intercept is (−1−5,0), or
(−6,0). Substituting the three ordered pairs (4,0), (−1,7),
and (−6,0) in the equation f(x)=ax2+bx +cyields a
7=a−b+c,
25x2−14
25x+168
25 ,or