Question on the principles of square roots in algebraic eqs

[OneTwoThreeFour]

From OG 12th Ed pg. 22 #16

sqrt(3-2x) = sqrt(2x) +1
then 4x^2 =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Answer is E

My real equation is whether or not there is some set of rules that limit this question from being solved other than the OG way. For example, I solved it by:

sqrt(3-2x)^2 = (sqrt(2x)+1)^2
3-2x= 2x + 2sqrt(2x) +1
2-4x= 2 sqrt(2x)
1-2x= sqrt(2x)

This is where I decided to deviate from OG's way:

1- sqrt(2x) =2x
(1-sqrt(2x))^2= (2x)^2
1- 2sqrt(2x) + 2x= 4x^2

However, what I found out is that 1- 2sqrt(2x) + 2x does not equal to 6x-1! For example by using quadratic formula, one of the solutions for x is about 1.309.

4(1.309^2) = 6(1.309)-1

But, 1 - 2sqrt(2(1.309)) + 2(1.309) does not equal to 4(1.309)^2.

Why is that? I want to say I made a mathematical error, but I checked my work numerous times and still got the same result. What am I missing here?
Thanks!

[GMATGuruNY]

From OG 12th Ed pg. 22 #16

sqrt(3-2x) = sqrt(2x) +1
then 4x^2 =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Answer is E

My real equation is whether or not there is some set of rules that limit this question from being solved other than the OG way. For example, I solved it by:

sqrt(3-2x)^2 = (sqrt(2x)+1)^2
3-2x= 2x + 2sqrt(2x) +1
2-4x= 2 sqrt(2x)
1-2x= sqrt(2x)

This is where I decided to deviate from OG's way:

1- sqrt(2x) =2x
(1-sqrt(2x))^2= (2x)^2
1- 2sqrt(2x) + 2x= 4x^2

However, what I found out is that 1- 2sqrt(2x) + 2x does not equal to 6x-1! For example by using quadratic formula, one of the solutions for x is about 1.309.

4(1.309^2) = 6(1.309)-1

But, 1 - 2sqrt(2(1.309)) + 2(1.309) does not equal to 4(1.309)^2.

Why is that? I want to say I made a mathematical error, but I checked my work numerous times and still got the same result. What am I missing here?
Thanks!

When we square both sides of an equation -- as you did above -- we introduce solutions that might not work in the original equation.

For example:
x-1 = 1.
The only possible solution is x=2.
But if we square both sides of the equation, we get:
(x-1)^2 = 1^2.
x^2 - 2x + 1 = 1.
x^2 - 2x = 0.
x(x-2) = 0.
x = 0 or x=2.
But x = 0 is not a solution of the original equation.

Your error was in performing algebra that was not going to lead to one of the answer choices. Since none of the answer choices contains a √, we need to perform algebra that will clear all the √ symbols:

√(3-2x) = √2x + 1

[√(3-2x)]² = (√2x + 1)²

3 - 2x = (√2x + 1)(√2x + 1)

3 - 2x = 2x + √2x + √2x + 1

2 - 2x = 2x + 2√2x

1 - x = x + √2x

1 - 2x = √2x

(1 - 2x)² = (√2x)²

(1 - 2x)(1 - 2x) = 2x

1 - 2x - 2x + 4x² = 2x

4x² = 6x - 1.

The correct answer is E.

[OneTwoThreeFour]

Thank you again Mitch, that was extremely helpful!