AnswersMathematics III ACompleting The Square

Completing The Square Answers

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Amira is solving the equation x2 – 6x = 1. Which value must be added to both sides of the equation to make the left side a perfect-square trinomial?

A
–9
B
8
C
9
D
36
2
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Solve for in the equation .

Question illustration
A
x = negative 12 plus-or-minus StartRoot 85 EndRoot
Option A
B
x = negative 6 plus-or-minus StartRoot 23 EndRoot i
Option B
C
x = 6 plus-or-minus StartRoot 23 EndRoot i
Option C
D
x = 12 plus-or-minus StartRoot 85 EndRoot
Option D
3

What are the solution(s) of ?

Question illustration
A
or
Option A
B
or
Option B
C
x = 2
Option C
D
x = 4
Option D
4

Solve for in the equation .

Question illustration
A
x = five-halves plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction
Option A
B
x = five-halves plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction
Option B
C
x = five-fourths plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction
Option C
D
x = five-fourths plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction
Option D
5

Solve for in the equation .

Question illustration
A
x = 5 plus-or-minus 2 StartRoot 5 EndRoot
Option A
B
x = 5 plus-or-minus StartRoot 35 EndRoot
Option B
C
x = 10 plus-or-minus 2 StartRoot 5 EndRoot
Option C
D
x = 10 plus-or-minus StartRoot 35 EndRoot
Option D
6

Which value must be added to the expression x2 + 12x to make it a perfect-square trinomial?

A
6
B
36
C
72
D
144
7

Solve for in the equation .

Question illustration
A
x = 2 plus-or-minus StartRoot 42 EndRoot
Option A
B
x = 2 plus-or-minus StartRoot 33 EndRoot
Option B
C
x = 2 plus-or-minus StartRoot 34 EndRoot
Option C
D
x = 4 plus-or-minus StartRoot 42 EndRoot
Option D
8

Solve for in the equation .

Question illustration
A
x = –12 or x = 2
B
x = –11 or x = 1
C
x = –2 or x = 12
D
x = –1 or x = 11
9

Solve for in the equation .

Question illustration
A
x = negative 6 plus-or-minus StartRoot 82 EndRoot
Option A
B
x = negative 6 plus-or-minus 2 StartRoot 17 EndRoot
Option B
C
x = 1 plus-or-minus StartRoot 33 EndRoot
Option C
D
x = 1 plus-or-minus StartRoot 47 EndRoot
Option D
10

Solve for in the equation .

Question illustration
A
x = negative 7 plus-or-minus 4 StartRoot 6 EndRoot i
Option A
B
x = –7 ± 8i
C
x = 7 plus-or-minus 4 StartRoot 6 EndRoot i
Option C
D
x = 7 ± 8i

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