AnswersFall 24 GMS Advanced Algebra Concepts and Connections AOperations with Complex Numbers

Completing The Square Answers

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1
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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
2
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Use completing the square to solve for in the equation .

Question illustration
A
x = –4 or 6
B
x = –2 or 14
C
x = 1 plus-or-minus StartRoot 89 EndRoot
Option C
D
x = 1 plus-or-minus StartRoot 87 EndRoot
Option D
3

Solve for in the equation .

Question illustration
A
x = –6 or x = 2
B
x = negative 2 plus-or-minus 2 StartRoot 2 EndRoot
Option B
C
x = –2 or x = 6
D
x = 2 plus-or-minus 2 StartRoot 2 EndRoot
Option D
4

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
5

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
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 = 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
9

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
10

Solve for in the equation .

Question illustration
A
x = –16 or x = –4
B
x = –10
C
x = –8
D
x = 4 or x = 16
11

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
12

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

A
One-fourth
Option A
B
One-half
Option B
C
1
D
4
13

What are the solutions of ?

Question illustration
A
or
Option A
B
or
Option B
C
or
Option C
D
or
Option D
14

Brian is solving the equation . What value must be added to both sides of the equation to make the left side a perfect-square trinomial?

Question illustration
A
StartFraction 9 Over 64 EndFraction
Option A
B
StartFraction 9 Over 16 EndFraction
Option B
C
Three-fourths
Option C
D
Nine-fourths
Option D
15

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
16

Which value must be added to the expression x2 + 16x to make it a perfect-square trinomial?

A
8
B
32
C
64
D
256

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