AnswersVA-Mathematical AnalysisPiecewise Defined Functions

Completing The Square Answers

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1
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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
2
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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
3

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
4

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
5

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
6

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
7

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
8

What are the solutions of ?

Question illustration
A
or
Option A
B
or
Option B
C
or
Option C
D
or
Option D
9

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
10

Solve for in the equation .

Question illustration
A
x = negative 1 plus-or-minus StartRoot 15 EndRoot
Option A
B
x = negative 1 plus-or-minus StartRoot 17 EndRoot
Option B
C
x = negative 2 plus-or-minus 2 StartRoot 5 EndRoot
Option C
D
x = negative 1 plus-or-minus StartRoot 13 EndRoot
Option D

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