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Re: Square root and Modulus [#permalink]
29 Nov 2012, 13:51

Expert's post

Bunuel wrote:

udaymathapati wrote:

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.

Re: Square root and Modulus [#permalink]
15 Jun 2013, 08:19

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}.