Algebra problem?

Question

How do you solve...?
Sqrt = Square root ^=to the power of (exponant)

This is the problem:

Sqrt(2x+2) + Sqrt(x+2) = 1 can be rewritten as:

A. (Sqrt(2x+2))^2 + (Sqrt(x+2))^2 = 1^2
B. x^2-6x-7=-0
C. Sqrt((2x+2)(x+2))=1
D. 2x^2 + 6x +3=0

A seems right to me, just want to make sure.

Thanks.

Answer 1

A is not correct because you should raise the whole expression by the power of two, that means [Sqrt(2x+2) + Sqrt(x+2)]^2, so A is definitely wrong.

First you transpose any of the two sqrt to the other side
(I really don't square an expression with two radicals in it, so you transpose)
For example

Sqrt(2x+2) = 1 - Sqrt(x+2)

Then square both sides, we have

2x + 2 = 1 - 2Sqrt(x+2) + (x + 2)

Transpose all expressions that doesn't have any radicals in it

2x + 2 - 1 - x - 2 = - 2Sqrt(x+2)

Simplifying,

x - 1 = -2Sqrt(x+2)

Squaring both sides again, we have

x² - 2x +1 = 4(x+2)

x² - 2x +1 = 4x + 8

Transposing all on the other side, we have

x² - 2x +1 -4x - 8 = 0

Simplifying, we have

x² - 6x - 7 = 0

B is the correct answer...

Answer 2

The answer is not A, you must square the left side as a whole expression rather than each term individually.

The answer is B. Solve this problem through the process of getting rid of one root at a time by isolating the root and squaring both sides of the equation. See work below:

√(2x+2) + √(x+2) = 1

√(2x+2) = 1 - √(x+2)    (isolate one of the roots)

[√(2x+2)]² = [1 - √(x+2)]²    (square both sides)

2x + 2 = 1 - 2√(x+2) + x + 2

2x + 2 = 3 + x - 2√(x+2)

x - 1 = - 2√(x+2)    (isolate the root and combine like terms)

(x - 1)² = [- 2√(x+2)]²    (square both sides)

x² - 2x + 1 = 4(x+2)

x² - 2x + 1 = 4x + 8

x² - 2x + 1 - 4x - 8 = 0

x² - 6x - 7 = 0    (answer B)

Answer 3

The answer is not A.

sqrt(2x+2) + sqrt(x+2) = 1
sqrt(2x+2) = 1 -sqrt(x+2)
square both sides
2x + 2 = 1 -2sqrt(x+2) + x + 2
2sqrt(x+2) = 1 - x
square both sides again
4(x+2) = 1 -2x + x^2
4x + 8 = 1 - 2x + x^2
x^2 - 6x -7 = 0

Answer B

Answer 4

The answer is A.