Find the value of c that makes each trinomial a perfect square. 1. x

Transcription

9-4 Solving Quadratic Equations by Completing the Square
Find the value of c that makes each trinomial a perfect square.
2
1. x − 18x + c
SOLUTION: In this trinomial, b = –18.
So, c must be 81 to make the trinomial a perfect square.
2
2. x + 22x + c
SOLUTION: In this trinomial, b = 22.
2
3. x + 9x + c
So, c must be
to make the trinomial a perfect square.
2
4. x − 7x + c
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So, cManual
must -be
Solve each equation by completing the square. Round to the nearest tenth if necessary.
Page 1
9-4 Solving
Quadratic
Equations by Completing the Square
So, c must
be
2
4. x − 7x + c
So, c must be
2
5. x + 4x = 6
SOLUTION: The solutions are about –5.2 and 1.2.
2
6. x − 8x = −9
SOLUTION: The solutions are about 1.4 and 6.6.
2
7. 4x + 9x − 1 = 0
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SOLUTION: Page 2
The solutions are about 1.4 and 6.6.
2
7. 4x + 9x − 1 = 0
2
8. −2x + 10x + 22 = 4
SOLUTION: eSolutions Manual - Powered by Cognero
Page 3
The solutions are about –2.4 and 0.1.
2
8. −2x + 10x + 22 = 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.
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Page 4
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.
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.
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
2
11. x − 24x + c
2
12. x − 19x + c
Page 5
9-4 Solving
Quadratic
by Completing
Square
So, c must
be 169 toEquations
make the trinomial
a perfectthe
square.
2
11. x − 24x + c
2
12. x − 19x + c
So, c must be
2
13. x + 17x + c
So, c must be
2
14. x + 5x + c
So, c must be
2
15. x − 13x + c
Page 6
9-4 Solving
Quadratic
So, c must
be
2
14. x + 5x + c
So, c must be
2
15. x − 13x + c
So, c must be
2
16. x − 22x + c
2
17. x − 15x + c
So, c must be
2
18. x + 24x + c
Page 7
2
17. x − 15x + c
So, c must be
2
18. x + 24x + c
2
19. x + 6x − 16 = 0
SOLUTION: The solutions are –8 and 2.
2
20. x − 2x − 14 = 0
Page 8
9-4 Solving
Quadratic
by Completing the Square
The solutions
are –8Equations
and 2.
2
20. x − 2x − 14 = 0
2
21. x − 8x − 1 = 8
2
22. x + 3x + 21 = 22
Page 9
9-4 Solving
Quadratic Equations by Completing the Square
The solutions are –1 and 9.
2
22. x + 3x + 21 = 22
2
23. x − 11x + 3 = 5
2
24. 5x − 10x = 23
Page 10
9-4 Solving
2
24. 5x − 10x = 23
2
25. 2x − 2x + 7 = 5
SOLUTION: The square root of a negative number has no real roots. So, there is no solution.
2
26. 3x + 12x + 81 = 15
Page 11
9-4 Solving
The square root of a negative number has no real roots. So, there is no solution.
2
26. 3x + 12x + 81 = 15
SOLUTION: The square root of a negative number has no real roots. So, there is no solution.
2
27. 4x + 6x = 12
2
28. 4x + 5 = 10x
Page 12
9-4 Solving
2
28. 4x + 5 = 10x
2
29. −2x + 10x = −14
Page 13
9-4 Solving
2
29. −2x + 10x = −14
2
30. −3x − 12 = 14x
Page 14
9-4 Solving
2
30. −3x − 12 = 14x
SOLUTION: The solutions are about –3.5 and –1.1.
2
31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t , wh
the number of days after the stock is purchased. When is the stock worth $60?
SOLUTION: Let p = 60.
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9-4 Solving
The solutions are about –3.5 and –1.1.
2
31. STOCK The price p in dollars for a particular stock can be modeled by the quadratic equation p = 3.5t − 0.05t , wh
the number of days after the stock is purchased. When is the stock worth $60?
SOLUTION: Let p = 60.
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
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9-4 Solving
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 height cannot be negative. So, x ≈ 6.3.
33. area = 110 ft
2
Page 17
9-4 Solving
Quadratic
by x ≈ 6.3.
Completing the Square
The height
cannot beEquations
negative. So,
33. area = 110 ft
2
SOLUTION: 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.
SOLUTION: Let x = the first integer and x + 2 = the second integer.
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.
Page 18
9-4 Solving
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.
The integers must be negative. So, they are –23 and –21.
36. GEOMETRY Find the area of the triangle.
Page 19
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.
9-4 Solving
The integers must be negative. So, they are –23 and –21.
36. GEOMETRY Find the area of the triangle.
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.
The area of the triangle is 216 square meters.
2
37. 0.2x − 0.2x − 0.4 = 0
Page 20
The area of the triangle is 216 square meters.
2
37. 0.2x − 0.2x − 0.4 = 0
2
38. 0.5x = 2x − 0.3
2
Page 21
9-4 Solving
Quadratic
The solutions
are –1Equations
and 2.
2
38. 0.5x = 2x − 0.3
2
39. 2x −
Page 22
9-4 Solving
2
39. 2x −
40. SOLUTION: eSolutions Manual - Powered by Cognero
Page 23
9-4 Solving
40. SOLUTION: The solutions are –0.5 and 2.5.
Page 24
9-4 Solving
The solutions are –0.5 and 2.5.
41. SOLUTION: The solutions are about –8.2 and 0.2.
Page 25
9-4 Solving
42. SOLUTION: 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 is the acceleration due to gravity. The acceleration due to gravity near the surface
2
2
of Mars is 3.73 m/s , 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.
b. Mars:
So, t ≈ 8.0 seconds.
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Earth:
Page 26
9-4 Solving
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 due to gravity near the surface
2
2
of Mars is 3.73 m/s , 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.
b. Mars:
Earth:
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: So, c can be –20 or 20 to make the trinomial a perfect square.
2
Page 27
c. Sample
9-4 Solving
Quadratic
Equations
by Completing
the Square
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
2
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?
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.
Page 28
So, c can be –30 or 30 to make the trinomial a 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?
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.
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.
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.
2
Trinomial
Number of
b − 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.
2
c. VERBAL Compare the number of roots of each equation to the result in the b − 4ac column. Is there a
relationship between these values? If so, describe it.
Page 29
d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have. Verify your prediction by solving the equation.
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.
2
c. VERBAL Compare the number of roots of each equation to the result in the b − 4ac column. Is there a
relationship between these values? If so, describe it.
d. ANALYTICAL Predict how many solutions 2x2 − 9x + 15 = 0 will have. Verify your prediction by solving the equation.
SOLUTION: a.
Trinomial
2
x − 8x + 16
2
2x − 11x + 3
2
3x + 6x + 9
2
x − 2x + 7
2
x + 10x + 25
2
x + 3x − 12
2
b − 4ac
2
(–8) – 4(1)(16) = 0
2
(–11) – 4(2)(3) = 97
2
(6) – 4(3)(9) = −72
2
(–2) – 4(1)(7) = −24
2
(10) – 4(1)(25) = 0
2
(3) – 4(1)(–12) = 57
Number
of
Roots
1
b.
2
The trinomial x − 8x + 16 has 1 root.
2
+ 3 has
The Manual
trinomial
2x − by
11x
eSolutions
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2 roots.
Page 30
2
The trinomial 2x − 11x + 3 has 2 roots.
2
The square root of a negative number has no real roots. So, the trinomial 3x + 6x + 9 has 0 roots.
2
The square root of a negative number has no real roots. So, the trinomial x − 2x + 7.
2
The trinomial x + 10x + 25 has 1 root.
Page 31
9-4 Solving Quadratic
2
The trinomial x + 10x + 25 has 1 root.
2
The trinomial x + 3x − 12 has 2 roots.
Trinomial
2
0
Number of
Roots
1
97
2
−72
0
−24
0
0
1
57
2
b − 4ac
2
x − 8x + 16
2
2x − 11x + 3
2
3x + 6x + 9
2
x − 2x + 7
2
x + 10x + 25
2
x + 3x − 12
2
2
c. If b − 4ac is negative, the equation has no real solutions. If b − 4ac is zero, the equation has one solution. If b
− 4ac is positive, the equation has 2 solutions.
d.
2
2
2
The equation 2x − 9x + 15 = 0 has 0 real solutions because b − 4ac is negative.
Page 32
9-4 Solving Quadratic
2 Equations by Completing the Square 2
The equation 2x − 9x + 15 = 0 has 0 real solutions because b − 4ac is negative.
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 2
completing the square and rewriting the equation in the form y = a(x − h) + k.
SOLUTION: Let 2
and , then the 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.
2
49. REASONING Determine the number of solutions 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 get
Page 33
. Since the left side of the last equation is a perfect square, it cannot
Let 2
and , then the 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.
2
49. REASONING Determine the number of solutions 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 get
. Since the left side of the last equation is a perfect square, it cannot
equal the negative number
. So, 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.
SOLUTION: The first 3 trinomials are perfect squares.
The trinomial
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.
Verify that the solution is 4.
Page 34
9-4 Solving
Quadratic Equations
The trinomial
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.
Verify that the solution is 4.
2
So, the quadratic equation x − 8x + 16 = 0 has a solution of 4.
2
52. WRITING IN MATH Compare and contrast the 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. 2
To solve the equation by graphing, use a graphing calculator to graph the related function y = x - 5x - 7. Select the
zero option from the 2nd [CALC] menu to determine the roots. eSolutions Manual - Powered by Cognero
Page 35
2
So, the quadratic equation x − 8x + 16 = 0 has a solution of 4.
2
52. WRITING IN MATH Compare and contrast the 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. 2
To solve the equation by graphing, use a graphing calculator to graph the related function y = x - 5x - 7. Select the
zero option from the 2nd [CALC] menu to determine the roots. The roots are given as decimal approximations. So, for this equation the solutions would have to be given as
estimations.
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
eSolutions
Page 36
C 10Manual - Powered by Cognero
D5
The roots are given as decimal approximations. So, for this equation the solutions would have to be given as
estimations.
9-4 Solving
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.
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) =
or . Choice J is the correct 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?
SOLUTION: 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
Page 37
D 34 pizzas
SOLUTION: 9-4 Solving Quadratic Equations by Completing the Square
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.
Percy delivered 40 pizzas. Choice C is the correct answer.
Describe how the graph of each function is related to the graph of f (x) = x 2.
57. g(x) = −12 + x
2
SOLUTION: 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. 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. 2
59. g(x) = 2x + 5
SOLUTION: The Manual
function
can bebywritten
f (x)
eSolutions
- Powered
Cognero
2
2
= ax + c, where a = 2 and c = 5. Since 2 > 0 and
is the graph of y = x vertically stretched and shifted up 5 units.
2
+ 5 38
> 1, the graph of y = 2x Page
2
59. g(x) = 2x + 5
SOLUTION: 2
The function can be written f (x) = ax + c, where a = 2 and c = 5. Since 2 > 0 and
2
> 1, the graph of y = 2x + 5
2
is the graph of y = x vertically stretched and shifted up 5 units.
60. SOLUTION: 2
The function can be written f (x) = a(x – b) , where a =
and b = 6. Since
> 0 and
< 1, the graph of 2
is the graph of y = x vertically compressed and shifted right 6 units.
61. g(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
2
and c = 6. Since
> 0 and > 1, the graph of y = 6 +
2
x is the graph of y = x vertically stretched and shifted up 6 units.
2
= − 1 -− x by Cognero
62. h(x)Manual
eSolutions
Powered
SOLUTION: Page 39
62. h(x) = − 1 − x
2
SOLUTION: 2
The function can be written f (x) = –ax + c, where a =
= –1 –
2
and c = –1. Since
< 0 and > 1, the graph of y
2
x is the graph of y = x vertically stretched, shifted down 1 unit and reflected across the x-axis.
63. RIDES A popular amusement park ride whisks riders to the top of a 250-foot tower and drops them. A function for
2
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: 2
The ride stops the descent of the rider at 40 feet above ground, so h = 40. Then, the equation 40 = −16t + 250
models the drop of the rider.
Time cannot be negative. So, it takes about 3.6 seconds to complete the ride.
Simplify. Assume that no denominator is equal to zero.
64. SOLUTION: 65. SOLUTION: eSolutions Manual - Powered by Cognero
Page 40
65. SOLUTION: 66. SOLUTION: 67. SOLUTION: 68. SOLUTION: 3
−3
−6
69. b (m )(b )
SOLUTION: Solve each open sentence.
70. |y − 2| > 7
SOLUTION: Case 1 y – 2 is
Page 41
and
Case 2 y – 2 is
Solve each open sentence.
70. |y − 2| > 7
SOLUTION: Case 1 y – 2 is
positive.
and
Case 2 y – 2 is
negative.
The solution set is {y|y > 9 or y < −5}.
71. |z + 5| < 3
SOLUTION: Case 1 z +5 is
positive.
and
Case 2 z +5 is
negative.
The solution set is {z|−8 < z < −2}.
72. |2b + 7| ≤ −6
SOLUTION: cannot be less than or equal to –6. The solution set is empty.
cannot be negative. So,
73. |3 − 2y| ≥ 8
SOLUTION: Case 1 3 – 2y is
positive.
and
Case 2 3 – 2y is
negative.
The solution set is {y|y ≥ 5.5 or y ≤ −2.5}.
74. |9 − 4m| < −1
SOLUTION: cannot be less than or equal to –1. The solution set is empty.
75. |5c − 2| ≤ 13
SOLUTION: Case 1 5c – 2 is
positive.
and
Case 2 5c – 2 is
negative.
Page 42
74. |9 − 4m| < −1
cannot be less than or equal to –1. The solution set is empty.
75. |5c − 2| ≤ 13
SOLUTION: Case 1 5c – 2 is
positive.
and
Case 2 5c – 2 is
negative.
The solution set is {c|−2.2 ≤ c ≤ 3}.
Evaluate
76. a = 2, b = −5, c = 2
for each set of values. Round to the nearest tenth if necessary.
SOLUTION: 77. a = 1, b = 12, c = 11
SOLUTION: 78. a = −9, b = 10, c = −1
SOLUTION: 79. a = 1, b = 7, c = −3
80. a = 2, b = −4, c = −6
SOLUTION: Page 43
80. a = 2, b = −4, c = −6
SOLUTION: 81. a = 3, b = 1, c = 2
SOLUTION: This value is not a real number.
Page 44

Find the value of c that makes each trinomial a perfect square. 1. x

Transcription

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