Solving Quadratic Equations: Completing the Square Answers
14 verified answers
1
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Which are the solutions of x2 = –13x – 4?
A
0, 13B
0, –13C
StartFraction 13 minus StartRoot 153 EndRoot Over 2 EndFraction comma StartFraction 13 + StartRoot 153 EndRoot Over 2 EndFraction
D
StartFraction negative 13 minus StartRoot 153 EndRoot Over 2 EndFraction comma StartFraction negative 13 + StartRoot 153 EndRoot Over 2 EndFraction
1
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Which are the solutions of the quadratic equation?x2 = 9x + 6
A
StartFraction negative 9 minus StartRoot 105 EndRoot Over 2 EndFraction comma StartFraction negative 9 + StartRoot 105 EndRoot Over 2 EndFraction
B
StartFraction negative 9 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction negative 9 + StartRoot 57 EndRoot Over 2 EndFraction
C
StartFraction 9 minus StartRoot 105 EndRoot Over 2 EndFraction comma StartFraction 9 + StartRoot 105 EndRoot Over 2 EndFraction
D
StartFraction 9 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction 9 + StartRoot 57 EndRoot Over 2 EndFraction
2
What number should be added to both sides of the equation to complete the square?x2 + 3x = 6
A
Option
B
(StartFraction 3 Over 2 EndFraction) squared
C
3D
622
Which are the solutions of x2 = –13x – 4?
A
0, 13B
0, –13C
StartFraction 13 minus StartRoot 153 EndRoot Over 2 EndFraction comma StartFraction 13 + StartRoot 153 EndRoot Over 2 EndFractionD
StartFraction negative 13 minus StartRoot 153 EndRoot Over 2 EndFraction comma StartFraction negative 13 + StartRoot 153 EndRoot Over 2 EndFraction3
What number should be added to both sides of the equation to complete the square?x2 + 8x = 4
A
4B
8C
16D
323
Which are the solutions of x2 = –11x + 4?
A
StartFraction negative 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFraction negative 11 + StartRoot 137 EndRoot Over 2 EndFraction
B
StartFraction negative 11 minus StartRoot 125 EndRoot Over 2 EndFraction comma StartFraction negative 11 + StartRoot 125 EndRoot Over 2 EndFraction
C
StartFraction 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFraction 11 + StartRoot 137 EndRoot Over 2 EndFraction
D
StartFraction 11 minus StartRoot 125 EndRoot Over 2 EndFraction comma StartFraction 11 + StartRoot 125 EndRoot Over 2 EndFraction
3
Which are the solutions of x2 = –11x + 4?
A
StartFraction negative 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFraction negative 11 + StartRoot 137 EndRoot Over 2 EndFractionB
StartFraction negative 11 minus StartRoot 125 EndRoot Over 2 EndFraction comma StartFraction negative 11 + StartRoot 125 EndRoot Over 2 EndFractionC
StartFraction 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFraction 11 + StartRoot 137 EndRoot Over 2 EndFractionD
StartFraction 11 minus StartRoot 125 EndRoot Over 2 EndFraction comma StartFraction 11 + StartRoot 125 EndRoot Over 2 EndFraction4
What value of c makes x2 − 24x + c a perfect square trinomial?
A
−144B
−48C
48D
1444
Which are the solutions of the quadratic equation?x2 = 7x + 4
A
StartFraction negative 7 minus StartRoot 65 EndRoot Over 2 EndFraction comma StartFraction negative 7 + StartRoot 65 EndRoot Over 2 EndFraction
B
–7, 0C
StartFraction 7 minus StartRoot 65 EndRoot Over 2 EndFraction comma StartFraction 7 + StartRoot 65 EndRoot Over 2 EndFraction
D
7, 05
Which are the solutions of x2 = 19x + 1?
A
StartFraction 19 minus StartRoot 19 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 19 EndRoot Over 2 EndFraction
B
StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction
C
StartFraction negative 19 minus StartRoot 19 EndRoot Over 2 EndFraction comma StartFraction negative 19 + StartRoot 19 EndRoot Over 2 EndFraction
D
StartFraction negative 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction negative 19 + StartRoot 365 EndRoot Over 2 EndFraction
6
Bao Yu is completing the square to solve the polynomial equation.x2 – x – 3 = 0
A
adding –x to both sides of the equationB
adding –3 to both sides of the equationC
isolating the first term, x2D
isolating the constant, –36
Which are the solutions of x2 = –5x + 8?
A
StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 57 EndRoot Over 2 EndFraction
B
StartFraction negative 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 7 EndRoot Over 2 EndFraction
C
StartFraction 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 57 EndRoot Over 2 EndFraction
D
StartFraction 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 7 EndRoot Over 2 EndFraction
7
Which are the solutions of x2 = –5x + 8?
A
StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 57 EndRoot Over 2 EndFractionB
StartFraction negative 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 7 EndRoot Over 2 EndFractionC
StartFraction 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 57 EndRoot Over 2 EndFractionD
StartFraction 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 7 EndRoot Over 2 EndFraction7
What value of c makes x2 − 24x + c a perfect square trinomial?
A
−144B
−48C
48D
1448
Which are the solutions of x2 = –7x – 8?
A
StartFraction 7 Over 2 EndFraction minus StartFraction StartRoot 17 EndRoot Over 4 EndFraction comma StartFraction 7 Over 2 EndFraction + StartFraction StartRoot 17 EndRoot Over 4 EndFraction 
B
StartFraction negative 7 Over 2 EndFraction minus StartFraction StartRoot 17 EndRoot Over 4 EndFraction comma StartFraction negative 7 Over 2 EndFraction + StartFraction StartRoot 17 EndRoot Over 4 EndFraction 
C
StartFraction 7 minus StartRoot 17 EndRoot Over 2 EndFraction comma StartFraction 7 + StartRoot 17 EndRoot Over 2 EndFraction
D
StartFraction negative 7 minus StartRoot 17 EndRoot Over 2 EndFraction comma StartFraction negative 7 + StartRoot 17 EndRoot Over 2 EndFraction
8
Solve x2 + 14x = −24 by completing the square.What is the solution set of the equation?
A
{−12, −2}B
{−7, 7}C
{−6, −4}D
{−5, 5}9
What number should be added to both sides of the equation to complete the square? x2 – 6x = 5
Answer not available
9
Which are the solutions of the quadratic equation?x2 = 9x + 6
A
StartFraction negative 9 minus StartRoot 105 EndRoot Over 2 EndFraction comma StartFraction negative 9 + StartRoot 105 EndRoot Over 2 EndFractionB
StartFraction negative 9 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction negative 9 + StartRoot 57 EndRoot Over 2 EndFractionC
StartFraction 9 minus StartRoot 105 EndRoot Over 2 EndFraction comma StartFraction 9 + StartRoot 105 EndRoot Over 2 EndFractionD
StartFraction 9 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction 9 + StartRoot 57 EndRoot Over 2 EndFraction10
Which are the solutions of the quadratic equation?x2 = 7x + 4
A
StartFraction negative 7 minus StartRoot 65 EndRoot Over 2 EndFraction comma StartFraction negative 7 + StartRoot 65 EndRoot Over 2 EndFractionB
–7, 0C
StartFraction 7 minus StartRoot 65 EndRoot Over 2 EndFraction comma StartFraction 7 + StartRoot 65 EndRoot Over 2 EndFractionD
7, 010
Bao Yu is completing the square to solve the polynomial equation.x2 – x – 3 = 0
A
adding –x to both sides of the equationB
adding –3 to both sides of the equationC
isolating the first term, x2D
isolating the constant, –3Did you find these answers helpful?
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