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4-5 Completing the Square
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
2. SOLUTION: Factor the perfect square trinomial.
Use the Square Root Property.
The solution set is {0.39, 7.61}.
4. SOLUTION: Factor the perfect square trinomial.
Use the Square Root Property.
The solution set is {–2.77, –0.56}.
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Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect
square.
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The solution set is {0.39, 7.61}.
4-5 Completing
the Square
4. SOLUTION: Factor the perfect square trinomial.
Use the Square Root Property.
The solution set is {–2.77, –0.56}.
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect
square.
6. SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it.
So:
c = 25
Substitute 25 for c in the trinomial.
Solve each equation by completing the square.
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Substitute 25 for c in the trinomial.
4-5 Completing the Square
Solve each equation by completing the square.
8. SOLUTION: Use the Square Root Property.
The solution set is {–4, 2}.
10. SOLUTION: Use the Square Root Property.
The solution set is { –0.69, 2.19}.
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12. Page 3
The solution set is {–4, 2}.
4-5 Completing
the Square
10. SOLUTION: Use the Square Root Property.
The solution set is { –0.69, 2.19}.
12. SOLUTION: Use the Square Root Property.
The solution set is
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
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14. Page 4
The solution set is
4-5 Completing the Square
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
14. SOLUTION: Use the Square Root Property.
The solution set is { –5.16, 1.16}.
16. SOLUTION: Use the Square Root Property.
The solution set is {–8.24, 0.24}.
18. SOLUTION: Use the Square Root Property.
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The solution set
{–8.24, 0.24}.
4-5 Completing
theisSquare
18. SOLUTION: Use the Square Root Property.
The solution set is {–8.24, –3.76}.
20. SOLUTION: Use the Square Root Property.
The solution set is {0.5, 4.5}.
22. SOLUTION: Use the Square Root Property.
The solution set is {–17, –15}.
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24. Page 6
The solution set
{0.5, 4.5}.
4-5 Completing
theisSquare
22. SOLUTION: Use the Square Root Property.
The solution set is {–17, –15}.
24. SOLUTION: Use the Square Root Property.
The solution set is {–5.5, –1.5}.
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect
square.
26. SOLUTION: To find the value of c, divide the coefficient of x by 2, and square the result.
Substitute 16 for c in the trinomial.
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28. Page 7
The solution set
{–5.5, –1.5}.
4-5 Completing
theisSquare
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect
square.
26. SOLUTION: To find the value of c, divide the coefficient of x by 2, and square the result.
Substitute 16 for c in the trinomial.
28. SOLUTION: To find the value of c, divide the coefficient of x by 2, and square it.
Substitute
in the trinomial.
Solve each equation by completing the square.
30. SOLUTION: Use the Square Root Property.
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;
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4-5 Completing the Square
Solve each equation by completing the square.
30. SOLUTION: Use the Square Root Property.
;
The solution set is
.
32. SOLUTION: Use the Square Root Property.
The solution set is {– 4, – 2}
.
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Page 9
The solution set is {– 4, – 2}
4-5 Completing
the Square
.
34. SOLUTION: Use the Square Root Property.
The solution set is
.
36. SOLUTION: The solution set is
.
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Page 10
The solution set is
4-5 Completing the Square
.
38. SOLUTION: The solution set is
.
40. SOLUTION: The solution set is
.
42. SOLUTION: The solution set is
.
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SOLUTION: Page 11
The solution set
4-5 Completing
theisSquare
.
44. SOLUTION: ]
The solution set is {–0.71, 3.11}.
46. SOLUTION: The solution set is {–1.39, 1.59}.
48. CCSS MODELING An architect’s blueprints call for a dining room measuring 13 feet by 13 feet. The customer
would like the dining room to be a square, but with an area of 250 square feet. How much will this add to the
dimensions of the room?
SOLUTION: The area of a square is given by
, where s is the side length.
Therefore:
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The solution set
{–1.39, 1.59}.
4-5 Completing
theisSquare
48. CCSS MODELING An architect’s blueprints call for a dining room measuring 13 feet by 13 feet. The customer
would like the dining room to be a square, but with an area of 250 square feet. How much will this add to the
dimensions of the room?
SOLUTION: The area of a square is given by
, where s is the side length.
Therefore:
Solve for x.
Therefore, about 2.81 ft should be added to the dimensions of the room.
Solve each equation. Round to the nearest hundredth if necessary.
50. SOLUTION: Write the equation in standard form and solve using the quadratic formula.
The quadratic formula is given by:
The solution set is {–2.77, –0.56}.
52. SOLUTION: Write the equation in standard form and solve using the quadratic formula.
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The solution set is {–2.77, –0.56}.
4-5 Completing
the Square
52. SOLUTION: Write the equation in standard form and solve using the quadratic formula.
The quadratic formula is given by:
The solution set is {–1.44, 0.24}.
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect
square.
54. SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it.
Substitute the value of c in the trinomial.
56. SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it.
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Substitute the value of c in the trinomial.
4-5 Completing
the Square
56. SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it.
Substitute the value of c in the trinomial.
2
58. ERROR ANALYSIS Alonso and Aida are solving x + 8x – 20 = 0 by completing the square. Is either of them
correct? Explain your reasoning.
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by add
Cognero
Alonso;
Aida
did not
16 to
each side; she added it only to the left side.
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4-5 Completing the Square
2
58. ERROR ANALYSIS Alonso and Aida are solving x + 8x – 20 = 0 by completing the square. Is either of them
correct? Explain your reasoning.
SOLUTION: Alonso; Aida did not add 16 to each side; she added it only to the left side.
60. CCSS ARGUMENTS Without solving, determine how many unique solutions there are for each equation. Are they
rational, real, or complex? Justify your reasoning.
a. (x + 2)2 = 16 2
b. (x - 2) = 16 c. -(x - 2)2 = 16
2
d. 36 - (x - 2) = 16 e . 16(x + 2)2 = 0 2
2
f. (x + 4) = (x + 6)
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SOLUTION: a. 2; rational; 16 is a perfect square so x + 2 and x are rational.
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SOLUTION: Alonso; Aida did not add 16 to each side; she added it only to the left side.
4-5 Completing
the Square
60. CCSS ARGUMENTS Without solving, determine how many unique solutions there are for each equation. Are they
rational, real, or complex? Justify your reasoning.
a. (x + 2)2 = 16 2
b. (x - 2) = 16 c. -(x - 2)2 = 16
2
d. 36 - (x - 2) = 16 e . 16(x + 2)2 = 0 2
2
f. (x + 4) = (x + 6)
SOLUTION: a. 2; rational; 16 is a perfect square so x + 2 and x are rational.
b. 2; rational; 16 is a perfect square so x – 2 and x are rational.
c. 2; complex; If the opposite of square is positive, the square is negative. The square root of a negative number is
complex.
d. 2; real; The square must equal 20. Since that is positive but not a perfect square, the solutions will be real but not
rational.
e . 1; rational; The expression must be equal to 0 and only -2 makes the expression equal to 0.
f. 1; rational; The expressions (x + 4) and (x + 6) must either be equal or opposites. No value makes them equal, -5
makes them opposites. The only solution is -5.
62. WRITING IN MATH Explain what it means to complete the square. Include a description of the steps you would
take.
SOLUTION: Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once
in this form, the equation can be solved by using the Square Root Property.
64. GEOMETRY Find the area of the shaded region.
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eSolutions
2
F 14 m
Page 17
SOLUTION: Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once
in this form, the
4-5 Completing
theequation
Square can be solved by using the Square Root Property.
64. GEOMETRY Find the area of the shaded region.
F 14 m2
2
G 18 m
H 42 m2
2
J 60 m
SOLUTION: The area of the outer rectangle is given by:
The area of the inner rectangle is given by:
The area of the shaded region is
.
The correct choice is H.
2
66. If 5 – 3i is a solution for x + ax + b = 0, where a and b are real numbers, what is the value of b?
A 10
B 14
C 34
D 40
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SOLUTION: Substitute 5 – 3i for x in the equation.
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The correct choice
is H.
4-5 Completing
the Square
2
66. If 5 – 3i is a solution for x + ax + b = 0, where a and b are real numbers, what is the value of b?
A 10
B 14
C 34
D 40
SOLUTION: Substitute 5 – 3i for x in the equation.
Equate the real and the imaginary parts.
And:
Substitute a = –10.
The correct choice is C.
Simplify.
68. SOLUTION: Write a quadratic equation in standard form with the given root(s).
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SOLUTION: 4-5 Completing
the Square
Write a quadratic equation in standard form with the given root(s).
70. SOLUTION: 72. SOLUTION: 74. SHOPPING Main St. Media sells all DVDs for one price and all books for another price. Alex bought 4 DVDs and
6 books for $170, while Matt bought 3 DVDs and 8 books for $180. What is the cost of a DVD and the cost of a
book?
SOLUTION: Let x and y are the cost of a DVD and a book.
Multiply the first and the second equation by 3 and –4 respectively then add.
Substitute 15 for y in the first equation and solve for x.
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4-5 Completing the Square
74. SHOPPING Main St. Media sells all DVDs for one price and all books for another price. Alex bought 4 DVDs and
6 books for $170, while Matt bought 3 DVDs and 8 books for $180. What is the cost of a DVD and the cost of a
book?
SOLUTION: Let x and y are the cost of a DVD and a book.
Multiply the first and the second equation by 3 and –4 respectively then add.
Substitute 15 for y in the first equation and solve for x.
The cost of a DVD is $20.
The cost of a book is $15.
Graph each inequality.
76. SOLUTION: The boundary of the graph is the graph of 2x – 3y = 6. Since the inequality symbol is <, the line will be dashed.
Test the point (0, 0).
Shade the region that contains (0, 0).
The graph of the inequality is:
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The cost of a DVD is $20.
The cost of athe
book
is $15.
4-5 Completing
Square
Graph each inequality.
76. SOLUTION: The boundary of the graph is the graph of 2x – 3y = 6. Since the inequality symbol is <, the line will be dashed.
Test the point (0, 0).
Shade the region that contains (0, 0).
The graph of the inequality is:
Write the piecewise function shown in each graph.
78. SOLUTION: The left portion is the graph of the constant function f (x) = –4. There is a dot at –2, so the constant function is
defined for x ≤ –2.
The center portion is the graph of f (x) = –2x + 3. There are circles at –2 and 1, so the function is defined for –2 < x
< 1.
The right portion is the graph of f (x) = x – 5. There is a dot at 2, so the function is defined for x ≥ 2.
The piece-wise function is:
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4-5 Completing
the Square
Write the piecewise function shown in each graph.
78. SOLUTION: The left portion is the graph of the constant function f (x) = –4. There is a dot at –2, so the constant function is
defined for x ≤ –2.
The center portion is the graph of f (x) = –2x + 3. There are circles at –2 and 1, so the function is defined for –2 < x
< 1.
The right portion is the graph of f (x) = x – 5. There is a dot at 2, so the function is defined for x ≥ 2.
The piece-wise function is:
80. SOLUTION: The left portion is the graph of f (x) = x + 12. There is dot at –6, so the function is defined for x ≤ –6.
The center portion is the graph of the constant function f (x) = 8. There are circles at –6 and 2, so the constant
function is defined for –6 < x < 2.
The right portion is the graph of f (x) = –2.5x + 15. There is a dot at 2, so the function is defined for x ≥ 2.
The piece-wise function is:
2
Evaluate b – 4ac for the given values of a, b, and c.
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82. a = –2, b = –7, c = 3
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The piece-wise function is:
4-5 Completing the Square
2
Evaluate b – 4ac for the given values of a, b, and c.
82. a = –2, b = –7, c = 3
SOLUTION: Substitute a = –2, b = –7, and c = 3.
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