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If x < 0, then root({-x} •|x|) is

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Re: modes and square root [#permalink] New post 11 Feb 2011, 00:42
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"
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Re: Square root and Mod Problem [#permalink] New post 07 Mar 2011, 22:32
sqrt(x^2) = |x|
if x <0 |x| = -x
x^2 = -x|x|

Hence A
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Re: Square root and Mod Problem [#permalink] New post 08 Mar 2011, 01:52
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Re: Square root and Modulus [#permalink] New post 14 Jun 2011, 01:07
A as root will always give positive value.
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Re: If x < 0, then root({-x} •|x|) is [#permalink] New post 29 Nov 2012, 13:10
x is < 0.

This makes sense, if it was positive, then it would be the square root of (-)(+)(|+|)= -. We can't have the square of a negative, not a real #

so x is -. Let's pick x= -5. (-1)(-5)(|-5|)= √25 = 5. answer is 5 which is -x.
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Re: Square root and Modulus [#permalink] New post 29 Nov 2012, 13:51
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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 \(- X\).

or is wrong this simple way ))

thanks
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Re: modes and square root [#permalink] New post 29 Nov 2012, 13:55
Expert's post
fluke wrote:
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 \(+ or -\).........
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Re: If x < 0, then root({-x} •|x|) is [#permalink] New post 03 Dec 2012, 19:59
Since \(x < 0\),
\(\sqrt{-x|x|}\)
\(\sqrt{x^2}=|x|\)

Answer: -x since x is of negative value.
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Re: If x < 0, then root({-x} •|x|) is [#permalink] New post 08 May 2013, 20:43
I said X before I read the very important point that x<-1

I need to stop that
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Re: Square root and Modulus [#permalink] New post 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 \(\sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}\).

Hope it's clear.
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Re: Square root and Modulus [#permalink] New post 15 Jun 2013, 08:23
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Re: Square root and Modulus [#permalink] New post 15 Jun 2013, 08:25
Ha! I got it just before I read your response. That is a very tricky problem - it's -(-x). Thanks!

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


Please read the solution carefully: if-x-0-then-root-x-x-is-100303.html#p773754

The answer is A, which is \(-x\), since \(x\) is negative then \(-x=-(negative)=positive\).
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Re: If x < 0, then root({-x} •|x|) is [#permalink] New post 25 Sep 2014, 08:07
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Re: If x < 0, then root({-x} •|x|) is [#permalink] New post 29 Sep 2015, 09:45
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Re: If x < 0, then root({-x} •|x|) is   [#permalink] 29 Sep 2015, 09:45

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