Exercise Set 11.7 315
b) The solutions of x2
8+x
4−3
8= 1 are the first coor-
dinates of the points of intersection of the graph of
f(x)=x2
8+x
4−3
8and y= 2. From the graph we
Exercise Set 11.7
2. 2l+2w= 100,
Dimensions: 25 yd by 25 yd
4. We make a drawing and label it.
A= (60 −2w)w=−2w2+60w=
6. A=x(180 −3x)=−3x2+ 180x=−3(x−30)2+ 2700
8. We make an overhead drawing and label it.
l+w=8,
V=3lw
10. V(x)=x2−6x+13=(x−3)2+4
12. Find the total profit:
To find the maximum value of the total profit and the value
of xat which it occurs we complete the square:
14. x+y=45,
P=xy,
16. x−y=6,
Second coordinate of vertex: P=(−3)2+6(−3) = −9
18. Let x,y, and Prepresent the two numbers and their prod-
uct, respectively.
316 Chapter 11: Quadratic Equations and Functions
The minimum function value of −81
26. Polynomial, neither quadratic nor linear
a=3,b=−1, c=2
f(x)=3x2−x+2
30. f(x)=ax2+bx +c
−30 = a(−3)2+b(−3) + c,
130 = a(80)2+b(80) + c,
200 = a(100)2+b(100) + c
34. D(x)=ax2+bx +c
0=a·02+b·0+c,
36. a) P(d)=ad2+bd +c
64 ·142+5
16 ·14 + 13
2≈$13.94
38. (2mn2−n2−3m2n)−(m2n+2mn2−n2)
Exercise Set 11.8
RC2. To solve x
x+1 >0, we look for intervals for which
✲✛
A
B
C
4. 9−x2≤0
Exercise Set 11.8 317
6. (x−7)(x+3)≤0
C: Test 8,y=(8−7)(8 + 3) = 11 >0
8. x2+x−2<0
(x+ 2)(x−1) <0
C: Test 2,y=2
2+2−2=4>0
10. x2+6x+9<0
(x+3)
2<0
12. x2−12 >4x
x2−4x−12 >0
✲✛
A
B
C
D
{x|−1<x<0or x > 1},or(−1,0) ∪(1,∞)
16. (x−1)(x+ 8)(x−2) <0
A: Test −9,y=(−9−1)(−9 + 8)(−9−2) = −110 <0
18. (x−2)(x−3)(x+1)<0
✲✛
−123
A: Test −2,y=(−2−2)(−2−3)(−2+1)=−20 <0
20. 1
x+4 >0
1
x+4 =0
Find the numbers for which the rational expression is un-
defined.
x+4=0
B: Test 0,1
x+4 >0
22. x−2
x+5 <0
Solve the related equation.
318 Chapter 11: Quadratic Equations and Functions
Find the numbers for which the rational expression is un-
defined.
x+5 <0
0−2
0+5 ?0
24. 5−2x
4x+3 ≤0
Solve the related equation.
B: Test 0,5−2x
4x+3 ≤0
26. x+1
2x−3<1
Solve the related equation.
3
24
A: Test 0,x+1
Exercise Set 11.8 319
28. (x+ 4)(x−1)
x+3 >0
A
B
C
D
2+4
2−1
?0
2
C: Test 0,(x+ 4)(x−1)
x+3 >0
30. x
x−2≥0
Solve the related equation.
✲✛
02
C: Test 3,x
x−2≥0
32. x−5
x<1
Solve the related equation.
x−5
x=1
320 Chapter 11: Quadratic Equations and Functions
Solve the related equation.
x+2
A: Test −8,x+2
(x−2)(x+7) >0
C: Test 0,x+2
(x−2)(x+7) >0
−1
7FALSE
36. 1
x≤2
2
A: Test −1,1
x≤2
1
1?2
38. x2−11x+30
x2−8x−9≥0
Exercise Set 11.8 321
(−2−6)(−2−5)
(−2−9)(−2+1) ?0
(0 −6)(0 −5)
(0 −9)(0 + 1) ?0
C: Test 11
2,(x−6)(x−5)
2−611
2−5
11
?0
(7 −9)(7 + 1) ?0
1·2
40. 25
4a2=√25
√4a2=5
2a
46. 3√10+8
√20 −5√80
48. a) {x|x<0},or(−∞,0)
50. x2+2x>4
52. x4+3x2≤0
54. a) −x2+ 812x−9600 >0
322 Chapter 11: Quadratic Equations and Functions
✲✛
A
B
C
of xdo not make sense in this problem, the solution set is
Chapter 11 Vocabulary Reinforcement
the discriminant.
5. The equation m6−m3−12 = 0 is quadratic in form.
Chapter 11 Concept Reinforcement
3. False; for f(x)=−3(x+2)
2−5=−3[x−(−2)]2−5, we
Chapter 11 Study Guide
2. x2−12x+31=0
x2−12x=−31
3. x2−10x=−23
x=−b±√b2−4ac
2a=−(−10) ±(−10)2−4·1·23
2·1
4. a) x2−3x=7
x2−3x−7=0
b) 2x2−5x+5=0
5. x=−2
(u−6)(u+1)=0
u−6=0 or u +1=0
u=6 or u =−1
Chapter 11 Summary and Review: Study Guide 323
7. f(x)=−x2−2x−3
f(x)=−(x2+2x+3)
The coefficient of x2is negative, so the graph opens down.
Thus, −2 is the maximum value of the function.
8. f(x)=x2−6x+4
f(0)=0
2−6·0+4 =4, so the y-intercept is (0,4). To
9. x2+40>14x
10. x+7
x−5≥3
Now find the numbers for which the rational expression is
not defined.
A: Test 0,x+7
x−5≥3
The number 6 is a solution of the inequality, so interval B
is in the solution set.
324 Chapter 11: Quadratic Equations and Functions
Chapter 11 Review Exercises
1. a) 2x2−7=0
b) The real-number solutions of the equation
2. 14x2+5x=0
x(14x+5)=0
3. x2−12x+27=0
4. x2−7x+13=0
a=1,b=−7, c=13
5. 4x2+6x=1
4x2+6x−1=0
2a
x=−6±62−4·4(−1)
2·4=−6±√36 + 16
8
x=3 or x =5
x=−b±√b2−4ac
2a
We can use a calculator to approximate the solutions:
8. x
x−2+4
x−6=0,LCM is (x−2)(x−6)
(x−2)(x−6)x
9. x
4−4
x=2,LCM is 4x
Chapter 11 Summary and Review: Review Exercises 325
a=1,b=−8, c=−16
10. 15 = 8
x+2 −6
x−2,LCM is (x+2)(x−2)
15x2−60 = 8x−16 −6x−12
x=2±√1924
30 =2±2√481
30
11. x2+6x+2=0
12. V(T)=48T2
13. Familiarize. Let l= the length of the screen, in cm.
Solve. We solve the equation.
l(l−5) = 126
only 14. If l= 14, then l−5=14−5 = 9. If the length is
14 cm and the width is 9 cm, the width is 5 cm less than
14. Familiarize. Using the labels on the drawing in the text,
we let x= the width of the mat, in inches. Then the
Solve. We solve the equation.
(16 −2x)(12 −2x) = 140
x−1=0 or x −13 = 0
15. Familiarize. We first make a drawing, labeling it with the
known and unknown information. We can also organize
1st part 50 r t
326 Chapter 11: Quadratic Equations and Functions
50
t−10 = 80
3−t,LCD is t(3 −t)
Check. Since the time cannot be negative (If t= 15,
speed of the second part was 40 mph.
16. x2+3x−6=0
17. x2+2x+5=0
18. x=1
5or x =−3
5
19. Since −4 is the only solution, it must be a double solution.
(x+ 4)(x+4)=0
20. N=3π1
p
21. 2A=3B
22. x4−13x2+36=0
Let u=x2.
(5u+ 1)(3u−1) = 0
x=−5x−1
5or 3x·1
x=3x·1
3
24. (x2−4)2−(x2−4) −6=0
y
x ⫽ ⫺2
Maximum: 4
(⫺2, 4)
Chapter 11 Summary and Review: Review Exercises 327
25. x−13√x+36=0
Let u=√x.
26. f(x)=−1
2(x−1)2+3
a) Vertex: (1,3)
27. f(x)=x2−x+6=(x2−x)+6
We add 1
4−1
4inside the parentheses.
we know that 23
d) We plot a few points and draw the curve.
x f(x)
28. f(x)=−3x2−12x−8=−3(x2+4x)−8
We complete the square inside the parentheses. We take
half the x-coefficient and square it.
x f(x)
−4−8
29. f(x)=x2−9x+14
30. g(x)=x2−4x−3
31. Familiarize. Let xand yrepresent the numbers.
328 Chapter 11: Quadratic Equations and Functions
32. We look for a function of the form ax2+bx +c=0.
Substituting the data points, we get
33. (x+ 2)(x−1)(x−2) >0
The solutions of (x+ 2)(x−1)(x−2) = 0 are −2, 1, and
2. They divide the real-number line into four intervals as
shown:
A
B
C
D
34. (x+ 4)(x−1)
x+2 <0
−1(−6)
−3
−2TRUE
C: Test 0, (x+ 4)(x−1)
x+2 <0
(0 + 4)(0 −1)
0+2 ?0
4(−1)
2FALSE
Interval Dis not part of the solution set.