9-4 Solving Quadratic Equations by Completing the Square
So, c must be
to make the trinomial a perfect square.
Find the value of c that makes each trinomial a
perfect square.
2
4. x − 7x + c
SOLUTION: In this trinomial, b = –7.
2
1. x − 18x + c
SOLUTION: In this trinomial, b = –18.
So, c must be
to make the trinomial a perfect square.
So, c must be 81 to make the trinomial a perfect
square.
Solve each equation by completing the square.
Round to the nearest tenth if necessary.
2
5. x + 4x = 6
2
2. x + 22x + c
SOLUTION: In this trinomial, b = 22.
SOLUTION: The solutions are about –5.2 and 1.2.
2
6. x − 8x = −9
So, c must be 121 to make the trinomial a perfect
square.
SOLUTION: 2
3. x + 9x + c
SOLUTION: In this trinomial, b = 9.
The solutions are about 1.4 and 6.6.
2
7. 4x + 9x − 1 = 0
SOLUTION: So, c must be
to make the trinomial a perfect square.
2
4. x − 7x + c
SOLUTION: In this trinomial, b = –7.
eSolutions Manual - Powered by Cognero
The solutions are about –2.4 and 0.1.
2
8. −2x + 10x + 22 = 4
Page 1
9-4 Solving Quadratic Equations by Completing the Square
The solutions are about 1.4 and 6.6.
2
7. 4x + 9x − 1 = 0
SOLUTION: The solutions are about –2.4 and 0.1.
The solutions are about –1.4 and 6.4.
9. CCSS MODELING Collin is building a deck on the
back of his family’s house. He has enough lumber
for the deck to be 144 square feet. The length should
be 10 feet more than its width. What should the
dimensions of the deck be?
SOLUTION: Let x = the width of the deck and x + 10 = the length
of the deck.
2
8. −2x + 10x + 22 = 4
SOLUTION: The dimensions of the deck can not be negative. So,
x = –5 + 13 or 8. The width of the deck is 8 feet and
the length of the deck is 8 + 10 or 18 feet.
Find the value of c that makes each trinomial a
perfect square.
2
10. x + 26x + c
The solutions are about –1.4 and 6.4.
SOLUTION: In this trinomial, b = 26.
9. CCSS MODELING Collin is building a deck on the
back of his family’s house. He has enough lumber
for the deck to be 144 square feet. The length should
be 10 feet more than its width. What should the
dimensions of the deck be?
So, c must be 169 to make the trinomial a perfect
square.
2
11. x − 24x + c
SOLUTION: Let x = the width of the deck and x + 10 = the length
of the deck.
eSolutions Manual - Powered by Cognero
The dimensions of the deck can not be negative. So,
x = –5 + 13 or 8. The width of the deck is 8 feet and
SOLUTION: In this trinomial, b = –24.
Page 2
So, c must be 144 to make the trinomial a perfect
square.
So, c must be 169 to make the trinomial a perfect
9-4 Solving
square. Quadratic Equations by Completing the Square
2
So, c must be
to make the trinomial a perfect square.
2
11. x − 24x + c
14. x + 5x + c
SOLUTION: In this trinomial, b = –24.
SOLUTION: In this trinomial, b = 5.
So, c must be 144 to make the trinomial a perfect
square.
So, c must be
2
15. x − 13x + c
SOLUTION: In this trinomial, b = –19.
to make the trinomial a perfect square.
SOLUTION: In this trinomial, b = –13.
So, c must be
2
2
16. x − 22x + c
SOLUTION: In this trinomial, b = 17.
to make the trinomial a perfect square.
2
SOLUTION: In this trinomial, b = –22.
So, c must be 121 to make the trinomial a perfect
square.
2
14. x + 5x + c
17. x − 15x + c
SOLUTION: In this trinomial, b = 5.
SOLUTION: In this trinomial, b = –15.
eSolutions Manual - Powered by Cognero
So, c must be
to make the trinomial a perfect square.
13. x + 17x + c
So, c must be
square.
2
12. x − 19x + c
So, c must be
to make the trinomial a perfect to make the trinomial a perfect Page 3
So, c must be
to make the trinomial a perfect 9-4 Solving
Quadratic
by Completing
So, c must
be 121 toEquations
make the trinomial
a perfectthe Square
square.
2
2
17. x − 15x + c
20. x − 2x − 14 = 0
SOLUTION: In this trinomial, b = –15.
So, c must be
The solutions are –8 and 2.
to make the trinomial a perfect square.
2
18. x + 24x + c
SOLUTION: The solutions are about –2.9 and 4.9.
2
21. x − 8x − 1 = 8
SOLUTION: SOLUTION: In this trinomial, b = 24.
The solutions are –1 and 9.
2
22. x + 3x + 21 = 22
So, c must be 144 to make the trinomial a perfect
square.
SOLUTION: Solve each equation by completing the square.
Round to the nearest tenth if necessary.
2
19. x + 6x − 16 = 0
SOLUTION: The solutions are about –3.3 and 0.3.
2
The solutions are –8 and 2.
23. x − 11x + 3 = 5
SOLUTION: 2
20. x − 2x − 14 = 0
SOLUTION: The solutions are about –0.2 and 11.2.
eSolutions
Manual - Powered by Cognero
The solutions are about –2.9 and 4.9.
2
Page 4
2
24. 5x − 10x = 23
9-4 Solving
Quadratic Equations by Completing the Square
The solutions are about –3.3 and 0.3.
2
23. x − 11x + 3 = 5
SOLUTION: The square root of a negative number has no real
roots. So, there is no solution.
2
26. 3x + 12x + 81 = 15
SOLUTION: The solutions are about –0.2 and 11.2.
2
24. 5x − 10x = 23
SOLUTION: The solutions are about –1.4 and 3.4.
2
25. 2x − 2x + 7 = 5
SOLUTION: The square root of a negative number has no real
roots. So, there is no solution.
2
27. 4x + 6x = 12
SOLUTION: The solutions are about –2.6 and 1.1.
2
28. 4x + 5 = 10x
SOLUTION: The square root of a negative number has no real
roots. So, there is no solution.
2
26. 3x + 12x + 81 = 15
SOLUTION: The solutions are about 0.7 and 1.8.
2
29. −2x + 10x = −14
SOLUTION: eSolutions Manual - Powered by Cognero
Page 5
9-4 Solving
Quadratic Equations by Completing the Square
The solutions are about 0.7 and 1.8.
2
29. −2x + 10x = −14
SOLUTION: The stock is worth $60 on the 30th and 40th days
after purchase.
GEOMETRY Find the value of x for each
figure. Round to the nearest tenth if necessary.
2
32. area = 45 in
SOLUTION: The solutions are about –1.1 and 6.1.
2
30. −3x − 12 = 14x
SOLUTION: The height cannot be negative. So, x ≈ 6.3.
33. area = 110 ft
2
SOLUTION: The solutions are about –3.5 and –1.1.
31. STOCK The price p in dollars for a particular stock
can be modeled by the quadratic equation p = 3.5t −
2
0.05t , where t represents the number of days after
the stock is purchased. When is the stock worth $60?
SOLUTION: Let p = 60.
The dimensions of the rectangle cannot be negative.
So, x ≈ 5.3.
34. NUMBER THEORY The product of two
consecutive even integers is 224. Find the integers.
The stock is worth $60 on the 30th and 40th days
after purchase.
for each
figure. Round to the nearest tenth if necessary.
GEOMETRY Find the value of x
eSolutions
Manual - Powered by Cognero
2
32. area = 45 in
SOLUTION: Let x = the first integer and x + 2 = the second
integer.
Page 6
9-4 Solving
Quadratic
Equations
Completing
the Square
The dimensions
of the
rectangleby
cannot
be negative.
So, x ≈ 5.3.
34. NUMBER THEORY The product of two
consecutive even integers is 224. Find the integers.
The integers must be negative. So, they are –23 and
–21.
36. GEOMETRY Find the area of the triangle.
SOLUTION: Let x = the first integer and x + 2 = the second
integer.
SOLUTION: The integers are 14 and 16 or –16 and –14.
35. CCSS PRECISION The product of two
consecutive negative odd integers is 483. Find the
integers.
SOLUTION: Let x = the first integer and x + 2 = the second
integer.
The dimensions of the triangle cannot be negative.
So, x = 18. The base of the triangle is 18 meters and
the height is 18 + 6 or 24 meters.
The area of the triangle is 216 square meters.
The integers must be negative. So, they are –23 and
–21.
36. GEOMETRY Find the area of the triangle.
Solve each equation by completing the square.
Round to the nearest tenth if necessary.
2
37. 0.2x − 0.2x − 0.4 = 0
SOLUTION: SOLUTION: The solutions are –1 and 2.
2
38. 0.5x = 2x − 0.3
SOLUTION: The dimensions of the triangle cannot be negative.
So, x = 18. The base of the triangle is 18 meters and
the height is 18 + 6 or 24 meters.
eSolutions Manual - Powered by Cognero
Page 7
9-4 Solving
Quadratic
by Completing the Square
The solutions
are –1Equations
and 2.
The solutions are –0.5 and 2.5.
2
38. 0.5x = 2x − 0.3
41. SOLUTION: SOLUTION: The solutions are about 0.2 and 3.8.
2
The solutions are about –8.2 and 0.2.
39. 2x −
SOLUTION: 42. SOLUTION: The solutions are about 0.2 and 0.9.
The solutions are about –5.1 and 0.1.
40. SOLUTION: 43. ASTRONOMY The height of an object t seconds
after it is dropped is given by the equation
, where h 0 is the initial height and g
is the acceleration due to gravity. The acceleration
2
The solutions are –0.5 and 2.5.
41. SOLUTION: eSolutions Manual - Powered by Cognero
due to gravity near the surface of Mars is 3.73 m/s ,
2
while on Earth it is 9.8 m/s . Suppose an object is
dropped from an initial height of 120 meters above
the surface of each planet.
a. On which planet would the object reach the
ground first?
b. How long would it take the object to reach the
ground on each planet? Round each answer to the
nearest tenth.
c. Do the times that it takes the object to reach the
ground seem reasonable? Explain your reasoning.
Page 8
SOLUTION: a. The object on Earth will reach the ground first
because it is falling at a faster rate.
So, t ≈ 4.9 seconds.
9-4 Solving
Quadratic Equations by Completing the Square
The solutions are about –5.1 and 0.1.
43. ASTRONOMY The height of an object t seconds
after it is dropped is given by the equation
, where h 0 is the initial height and g
c. Sample answer: Yes; the acceleration due to
gravity is much greater on Earth than on Mars, so the
time to reach the ground should be much less.
2
44. Find all values of c that make x + cx + 100 a perfect
square trinomial.
SOLUTION: is the acceleration due to gravity. The acceleration
2
due to gravity near the surface of Mars is 3.73 m/s ,
2
while on Earth it is 9.8 m/s . Suppose an object is
dropped from an initial height of 120 meters above
the surface of each planet.
a. On which planet would the object reach the
ground first?
b. How long would it take the object to reach the
ground on each planet? Round each answer to the
nearest tenth.
c. Do the times that it takes the object to reach the
ground seem reasonable? Explain your reasoning.
SOLUTION: a. The object on Earth will reach the ground first
because it is falling at a faster rate.
So, c can be –20 or 20 to make the trinomial a
perfect square.
2
45. Find all values of c that make x + cx + 225 a perfect
square trinomial.
SOLUTION: b. Mars:
So, c can be –30 or 30 to make the trinomial a
perfect square.
So, t ≈ 8.0 seconds.
Earth:
46. PAINTING Before she begins painting a picture,
Donna stretches her canvas over a wood frame. The
frame has a length of 60 inches and a width of 4
inches. She has enough canvas to cover 480 square
inches. Donna decides to increase the dimensions of
the frame. If the increase in the length is 10 times the
increase in the width, what will the dimensions of the
frame be?
So, t ≈ 4.9 seconds.
c. Sample answer: Yes; the acceleration due to
gravity is much greater on Earth than on Mars, so the
time to reach the ground should be much less.
2
SOLUTION: Let x = the increase in the width and let 10x = the
increase in the length. So, x + 4 = the new width and
10x + 60 = the new length.
44. Find all values of c that make x + cx + 100 a perfect
square trinomial.
SOLUTION: eSolutions Manual - Powered by Cognero
Page 9
9-4 Solving
by Completing
So, c canQuadratic
be –30 or Equations
30 to make the
trinomial a the Square
perfect square.
46. PAINTING Before she begins painting a picture,
Donna stretches her canvas over a wood frame. The
frame has a length of 60 inches and a width of 4
inches. She has enough canvas to cover 480 square
inches. Donna decides to increase the dimensions of
the frame. If the increase in the length is 10 times the
increase in the width, what will the dimensions of the
frame be?
Complete the last column of the table with the
number of roots of each equation.
c. VERBAL Compare the number of roots of each
2
equation to the result in the b − 4ac column. Is there
a relationship between these values? If so, describe
it.
2
d. ANALYTICAL Predict how many solutions 2x
− 9x + 15 = 0 will have. Verify your prediction by solving the equation.
SOLUTION: a.
Trinomial
SOLUTION: Let x = the increase in the width and let 10x = the
increase in the length. So, x + 4 = the new width and
10x + 60 = the new length.
b2 − 4ac
2
2
x − 8x + 16
(–8) – 4(1)(16) =
0
2
(–11) – 4(2)(3) =
97
2
3x + 6x + 9
(6) – 4(3)(9) =
−72
2
2
2
x − 2x + 7
(–2) – 4(1)(7) =
−24
2
x + 10x + 25
(10) – 4(1)(25) = 0
2
47. MULTIPLE REPRESENTATIONS In this
problem, you will investigate a property of quadratic
equations.
a. TABULAR Copy the table shown and complete
the second column.
Trinomial
Number of
b2 − 4ac
Roots
2
0
1
x − 8x + 16
2
2x − 11x + 3
2
3x + 6x + 9
2
x − 2x + 7
2
x + 10x + 25
2
x + 3x − 12
b. ALGEBRAIC Set each trinomial equal to zero,
and solve the equation by completing the square.
Complete the last column of the table with the
number of roots of each equation.
c. VERBAL Compare the number of roots of each
2
equation to the result in the b − 4ac column. Is there
a relationship between these values? If so, describe
it.
2
d. ANALYTICAL
eSolutions
Manual - Powered byPredict
Cognerohow many solutions 2x
− 9x + 15 = 0 will have. Verify your prediction by solving the equation.
2
2
x + 3x − 12
The dimensions cannot be negative, so x = 2. The
width is 2 + 4 or 6 inches and the length is 10(2) + 60
or 80 inches.
2
2x − 11x + 3
(3) – 4(1)(–12) =
57
Number
of
Roots
1
b.
2
The trinomial x − 8x + 16 has 1 root.
2
The trinomial 2x − 11x + 3 has 2 roots.
The square root of a negative number has no real
2
roots. So, the trinomial 3x + 6x + 9 has 0 roots.
Page 10
2
The square
root of aEquations
negative number
has no realthe Square
9-4 Solving
Quadratic
by Completing
2
roots. So, the trinomial 3x + 6x + 9 has 0 roots.
The equation 2x − 9x + 15 = 0 has 0 real solutions
2
because b − 4ac is negative.
The square root of a negative number has no real
2
roots. So, the trinomial x − 2x + 7.
The equation cannot be solved because the square
root of a negative number has no real roots.
2
48. CCSS PERSEVERANCE Given y = ax + bx + c
with a ≠ 0, derive the equation for the axis of symmetry by completing the square and rewriting the
2
equation in the form y = a(x − h) + k.
2
The trinomial x + 10x + 25 has 1 root.
SOLUTION: 2
The trinomial x + 3x − 12 has 2 roots.
Trinomial
b2 − 4ac
0
Number of
Roots
1
97
2
−72
0
−24
0
0
1
57
2
2
x − 8x + 16
2
2x − 11x + 3
2
3x + 6x + 9
2
x − 2x + 7
2
x + 10x + 25
2
x + 3x − 12
c. If b 2 − 4ac is negative, the equation has no real
2
solutions. If b − 4ac is zero, the equation has one
2
solution. If b − 4ac is positive, the equation has 2
solutions.
d.
Let and , then the
2
equation becomes y = (x - h) + k.
2
For an equation in the form y = ax + bx + c, the .
equation for the axis of symmetry is 2
So, for an equation in the form y = (x - h) + k, the
equation for the axis of symmetry becomes x = h.
49. REASONING Determine the number of solutions
2
x + bx = c has if
. Explain.
SOLUTION: None; Sample answer: If you add
to each side
of the equation and each side of the inequality, you
2
The equation 2x − 9x + 15 = 0 has 0 real solutions
2
because b − 4ac is negative.
get
. Since
the left side of the last equation is a perfect square, it
cannot equal the negative number
eSolutions Manual - Powered by Cognero
. So,
Page 11
there are no real solutions.
50. WHICH ONE DOESN’T BELONG? Identify the
2
For an equation in the form y = ax + bx + c, the .
equation for the axis of symmetry is 2
(x - h) + k, the
9-4 Solving
Quadratic Equations by= Completing
the Square
So, for an equation in the form y
equation for the axis of symmetry becomes x = h.
49. REASONING Determine the number of solutions
2
x + bx = c has if
. Explain.
is not a perfect square. So, it does not belong.
51. OPEN ENDED Write a quadratic equation for
which the only solution is 4.
SOLUTION: Find a quadratic that has two roots of 4.
SOLUTION: None; Sample answer: If you add
The trinomial
to each side
of the equation and each side of the inequality, you
get
Verify that the solution is 4.
. Since
the left side of the last equation is a perfect square, it
cannot equal the negative number
. So,
2
there are no real solutions.
50. WHICH ONE DOESN’T BELONG? Identify the
expression that does not belong with the other three.
Explain your reasoning.
So, the quadratic equation x − 8x + 16 = 0 has a
solution of 4.
52. WRITING IN MATH Compare and contrast the
2
following strategies for solving x − 5x − 7 = 0:
completing the square, graphing, and factoring.
SOLUTION: Sample answer: Because the leading coefficient is 1,
solving the equation by completing the square is
simpler and yields exact answers for the solutions. SOLUTION: The first 3 trinomials are perfect squares.
The trinomial
To solve the equation by graphing, use a graphing
2
calculator to graph the related function y = x - 5x 7. Select the zero option from the 2nd [CALC]
menu to determine the roots. is not a perfect square. So, it does not belong.
51. OPEN ENDED Write a quadratic equation for
which the only solution is 4.
SOLUTION: Find a quadratic that has two roots of 4.
eSolutions Manual - Powered by Cognero
Page 12
for this equation the solutions would have to be given
as estimations.
2
So, the quadratic
equation
x − 8x
16 = 0 has athe Square
9-4 Solving
Quadratic
Equations
by+Completing
solution of 4.
52. WRITING IN MATH Compare and contrast the
2
following strategies for solving x − 5x − 7 = 0:
completing the square, graphing, and factoring.
SOLUTION: Sample answer: Because the leading coefficient is 1,
solving the equation by completing the square is
simpler and yields exact answers for the solutions. There are no factors of –7 that have a sum of –5, so
solving by factoring is not possible.
53. The length of a rectangle is 3 times its width. The
area of the rectangle is 75 square feet. Find the
length of the rectangle in feet.
A 25
B 15
C 10
D5
SOLUTION: Let x = the width of the rectangle and let 3x = the
length of the rectangle.
To solve the equation by graphing, use a graphing
2
calculator to graph the related function y = x - 5x 7. Select the zero option from the 2nd [CALC]
menu to determine the roots. The length of the rectangle is 3(5) or 15 feet. Choice
B is the correct answer.
54. PROBABILITY At a festival, winners of a game
draw a token for a prize. There is one token for each
prize. The prizes include 9 movie passes, 8 stuffed
animals, 5 hats, 10 jump ropes, and 4 glow necklaces.
What is the probability that the first person to draw a
token will win a movie pass?
F
G
H
J
SOLUTION: There are 9 + 8 + 5 + 10 + 4 or 36 possible
outcomes.
P(movie pass) =
The roots are given as decimal approximations. So,
for this equation the solutions would have to be given
as estimations.
answer.
There are no factors of –7 that have a sum of –5, so
solving by factoring is not possible.
53. The length of a rectangle is 3 times its width. The
area of the rectangle is 75 square feet. Find the
length of the rectangle in feet.
A 25
eSolutions Manual - Powered by Cognero
B 15
C 10
D5
or . Choice J is the correct
55. GRIDDED RESPONSE The population of a town
can be modeled by P = 22,000 + 125t, where P
represents the population and t represents the
number of years from 2000. How many years after
2000 will the population be 26,000?
SOLUTION: Page 13
There are 9 + 8 + 5 + 10 + 4 or 36 possible
outcomes.
P(movie pass) =
or . Choice J is the correct
9-4 Solving Quadratic Equations by Completing the Square
answer.
55. GRIDDED RESPONSE The population of a town
can be modeled by P = 22,000 + 125t, where P
represents the population and t represents the
number of years from 2000. How many years after
2000 will the population be 26,000?
Percy delivered 40 pizzas. Choice C is the correct
answer.
Describe how the graph of each function is
2
related to the graph of f (x) = x .
57. g(x) = −12 + x
2
SOLUTION: 2
SOLUTION: The graph of f (x) = x + c represents a translation up
or down of the parent graph. Since c = –12, the
translation is down. So, the graph is shifted down 12
units from the parent function. In 32 years the population of the town will be 26,000.
56. Percy delivers pizzas for Pizza King. He is paid $6
an hour plus $2.50 for each pizza he delivers. Percy
earned $280 last week. If he worked a total of 30
hours, how many pizzas did he deliver?
A 250 pizzas
B 184 pizzas
C 40 pizzas
D 34 pizzas
SOLUTION: Let p = the number of pizzas Percy delivered.
2
58. h(x) = (x + 2)
SOLUTION: 2
The graph of f (x) = (x – c) represents a translation
left or right from the parent graph. Since c = –2, the
translation is to the left by 2 units. Percy delivered 40 pizzas. Choice C is the correct
answer.
Describe how the graph of each function is
2
related to the graph of f (x) = x .
57. g(x) = −12 + x
2
2
SOLUTION: 59. g(x) = 2x + 5
2
The graph of f (x) = x + c represents a translation up
or down of the parent graph. Since c = –12, the
translation is down. So, the graph is shifted down 12
units from the parent function. SOLUTION: 2
The function can be written f (x) = ax + c, where a
= 2 and c = 5. Since 2 > 0 and > 1, the graph of y
2
2
= 2x + 5 is the graph of y = x vertically stretched
and shifted up 5 units.
eSolutions Manual - Powered by Cognero
2
58. h(x) = (x + 2)
Page 14
9-4 Solving Quadratic Equations by Completing the Square
2
59. g(x) = 2x + 5
61. g(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
= 2 and c = 5. Since 2 > 0 and > 1, the graph of y
2
2
= 2x + 5 is the graph of y = x vertically stretched
and shifted up 5 units.
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = 6. Since
of y = 6 +
> 0 and 2
> 1, the graph 2
x is the graph of y = x vertically
stretched and shifted up 6 units.
60. SOLUTION: 2
The function can be written f (x) = a(x – b) , where
a=
and b = 6. Since
< 1, the graph > 0 and
2
of
is the graph of y = x vertically
compressed and shifted right 6 units.
62. h(x) = − 1 − x
2
SOLUTION: 2
The function can be written f (x) = –ax + c, where a
=
and c = –1. Since
graph of y = –1 –
2
< 0 and > 1, the x is the graph of y = x
2
vertically stretched, shifted down 1 unit and reflected
across the x-axis.
61. g(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = 6. Since
of y = 6 +
> 0 and 2
> 1, the graph 2
x is the graph of y = x vertically
stretched and shifted up 6 units.
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63. RIDES A popular amusement park ride whisks
riders to the top of a 250-foot tower and drops them.
2
A function for the height of a rider is h = −16t +
250, where h is the height and t is the time in
seconds. The ride stops the descent of the rider 40
feet above the ground. Write an equation that models
the drop of the rider. How long does it take to fall
from 250 feet to 40 feet?
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SOLUTION: The ride stops the descent of the rider at 40 feet
9-4 Solving Quadratic Equations by Completing the Square
63. RIDES A popular amusement park ride whisks
riders to the top of a 250-foot tower and drops them.
2
A function for the height of a rider is h = −16t +
250, where h is the height and t is the time in
seconds. The ride stops the descent of the rider 40
feet above the ground. Write an equation that models
the drop of the rider. How long does it take to fall
from 250 feet to 40 feet?
SOLUTION: The ride stops the descent of the rider at 40 feet
above ground, so h = 40. Then, the equation 40 =
66. SOLUTION: 67. 2
−16t + 250 models the drop of the rider.
SOLUTION: Time cannot be negative. So, it takes about 3.6
seconds to complete the ride.
Simplify. Assume that no denominator is equal
to zero.
68. 64. SOLUTION: SOLUTION: 3
−3
−6
69. b (m )(b )
65. SOLUTION: SOLUTION: Solve each open sentence.
70. |y − 2| > 7
66. SOLUTION: SOLUTION: Case 1 y – 2
is positive.
and
Case 2 y – 2 is
negative.
The solution set is {y|y > 9 or y < −5}.
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67. 71. |z + 5| < 3
SOLUTION: Page 16
74. |9 − 4m| < −1
SOLUTION: SOLUTION: 9-4 Solving Quadratic Equations by Completing the Square
cannot be negative. So,
cannot be
less than or equal to –1. The solution set is empty.
Solve each open sentence.
70. |y − 2| > 7
SOLUTION: Case 1 y – 2
is positive.
75. |5c − 2| ≤ 13
and
Case 2 y – 2 is
negative.
SOLUTION: Case 1 5c – 2
is positive.
and
Case 2 5c – 2 is
negative.
The solution set is {y|y > 9 or y < −5}.
The solution set is {c|−2.2 ≤ c ≤ 3}.
71. |z + 5| < 3
SOLUTION: Case 1 z +5 is
positive.
and
Case 2 z +5 is
negative.
Evaluate
for each set of values.
Round to the nearest tenth if necessary.
76. a = 2, b = −5, c = 2
SOLUTION: The solution set is {z|−8 < z < −2}.
72. |2b + 7| ≤ −6
SOLUTION: 77. a = 1, b = 12, c = 11
cannot be negative. So,
cannot be SOLUTION: less than or equal to –6. The solution set is empty.
73. |3 − 2y| ≥ 8
SOLUTION: Case 1 3 – 2y is
positive.
and
Case 2 3 – 2y is
negative.
78. a = −9, b = 10, c = −1
SOLUTION: The solution set is {y|y ≥ 5.5 or y ≤ −2.5}.
74. |9 − 4m| < −1
SOLUTION: cannot be negative. So,
cannot be
less than or equal to –1. The solution set is empty.
79. a = 1, b = 7, c = −3
SOLUTION: 75. |5c − 2| ≤ 13
SOLUTION: Case 1 5c – 2
is positive.
and
Case 2 5c – 2 is
negative.
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80. a = 2, b = −4, c = −6
9-4 Solving Quadratic Equations by Completing the Square
79. a = 1, b = 7, c = −3
SOLUTION: 80. a = 2, b = −4, c = −6
SOLUTION: 81. a = 3, b = 1, c = 2
SOLUTION: This value is not a real number.
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