Problem Solving

BTGmoderatorDC wrote:If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

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

OA E

Source: Official Guide

GIVEN: √(3-2x) = √(2x) +1

Square both sides to get: [√(3-2x)]² = [√(2x) +1]²

Expand and simplify to get: 3 - 2x = 2x + 2√(2x) + 1

Subtract 1 from both sides to get: 1 - 2x = 2x + 2√(2x)

Subtract 2x from both sides to get:2 - 4x = 2√(2x)

Divide both sides by 2 to get: 1 - 2x = √(2x)

Square both sides to get: (1 - 2x)² = [√(2x)]²

Expand and simplify to get: 1 - 4x + 4x² = 2x

Add 4x to both sides: 1 + 4x² = 6x

Subtract 1 from both sides to get: 4x² = 6x - 1

Answer: E

Cheers,
Brent

BTGmoderatorDC wrote:If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

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

OA E

Source: Official Guide

Squaring both sides of the equation, we have:

3 - 2x = 2x + 1 + 2√(2x)

2 - 4x = 2√(2x)

1 - 2x = √(2x)

Squaring both sides of the equation again, we have:

1 + 4x^2 - 4x = 2x

4x^2 = 6x - 1

Answer: E