Bunuel wrote:

x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?

A. \(\sqrt{x}-\sqrt{y}\)

B. \(\sqrt{x}+\sqrt{y}\)

C. \(\frac{\sqrt{x}}{\sqrt{y}}\)

D. \(2\sqrt{xy}\)

E. \(2\sqrt{x-y}\)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:To solve this problem we can focus on solving for b as a first step:

\(2b(\sqrt{x}-\sqrt{y})=x-y\)

\(2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}\)

Note that it is OK to divide by \(\sqrt{x}-\sqrt{y}\), since x > y , which implies that \(\sqrt{x}-\sqrt{y}\neq{0}\).

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

\(2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}\)

Cancel \(\sqrt{x}-\sqrt{y}\) in the numerator and denominator to get \(2b = \sqrt{x}+\sqrt{y}\).

The correct answer is B.could you show me the difference of 2 squares? I cannot see it...