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If x < 0, then \sqrt{-x} •|x|) is A. -x B. -1 C. 1 D. x E. \sqrt{x}

1. If x<0, then \(\sqrt{-x*|x|}\) equals:

A. \(-x\) B. \(-1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\)

Remember: \(\sqrt{x^2}=|x|\).

The point here is that square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=|x|\)

Back to the original question: \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\)

Or just substitute the value let \(x=-5<0\) --> \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x\).

Answer: A.

Hope it's clear.

Sorry Bunuel but the approach :\(\sqrt{-x* |x|}\) if we put this one ^2 then we have simply \(-x * |x|\). The latter is positive, so we have a quantity straight negative \(- X\).

the question is: If x < 0, what is the value for \(sqrt{-x*\mid x \mid}\):

\(sqrt{(+x)*(+x)}\) \(sqrt{x^2}\) \(\pm x\)

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"

Sorry fluke but the quantity \(\sqrt{x^2}\) is not + or - X but only + X, if we have \(x^2= something\) then we have \(+ or -\).........
_________________

How can the solution be negative if we're taking the square root of a positive number?

Bunuel wrote:

mbafall2011 wrote:

udaymathapati wrote:

If x < 0, then \sqrt{-x} •|x|) is A. -x B. -1 C. 1 D. x E. \sqrt{x}

what is the source of this question. I havent seen any gmat question testing imaginary numbers

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. So you won't see any question involving imaginary numbers.

This question also does not involve imaginary numbers as expression under the square root is non-negative (actually it's positive): we have \(\sqrt{-x*|x|}\) --> as \(x<0\) then \(-x=positive\) and \(|x|=positive\), so \(\sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}\).

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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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