If 0 1, x + y − 2 xy + x + y + 2 xy =

If 0 < x < 1 and y > 1, x + y − 2 xy + x + y… | GMAT Sprint

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PS · Problem Solving

705-805

Roots

If $0 < x < 1$ and $y > 1$ , $\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =$

Answer Choices

Correct answer marked below

√x

2√x

√y +√x

√y

2√y

Correct

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Explanation

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Official Explanation

Official

It's probably not immediately apparent, but

x - 2\sqrt{xy} + y

is a form of the SPECIAL PRODUCT

x^2 - 2xy + y^2

.

Notice that

(\sqrt{x} - \sqrt{y})^2 = (\sqrt{x} - \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - 2\sqrt{xy} + y

.

Likewise,

(\sqrt{x} + \sqrt{y})^2 = (\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y}) = x + 2\sqrt{xy} + y

.

So,

\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} = \sqrt{(\sqrt{x} - \sqrt{y})^2} + \sqrt{(\sqrt{x} + \sqrt{y})^2}

.

IMPORTANT: before we conclude that

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

, we must recognize that, since

0 < x < y

\sqrt{x} - \sqrt{y}

is NEGATIVE.

Since the square root of a value cannot be NEGATIVE, it must be the case that

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

.…

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What this question tests

What it tests

Your handling of square and higher roots, including simplifying radicals.

Common trap

Forgetting the ± when taking a square root to solve an equation, or assuming √(a²) = a.

If 0<x<10 < x < 10<x<1 and y>1y > 1y>1, x+y−2xy+x+y+2xy=\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =x+y−2xy​​+x+y+2xy​​=