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If x < 0, then root({-x} •|x|) is

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Re: modes and square root [#permalink]

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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]

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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]

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08 Mar 2011, 01:52
Merging similar topics.
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Re: Square root and Modulus [#permalink]

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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]

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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]

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29 Nov 2012, 13:51
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 $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

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]

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29 Nov 2012, 13:55
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]

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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]

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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]

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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]

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15 Jun 2013, 08:23
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?

The answer is A, which is $$-x$$, since $$x$$ is negative then $$-x=-(negative)=positive$$.
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Re: Square root and Modulus [#permalink]

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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?

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]

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25 Sep 2014, 08:07
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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29 Sep 2015, 09:45
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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10 Nov 2016, 05:44
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Re: If x < 0, then root({-x} •|x|) is   [#permalink] 10 Nov 2016, 05:44

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