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If x is < 0, then -x (root of -x multiplied by absolute

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Manager
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Joined: 22 Jun 2008
Posts: 101
Schools: Darden School of Business (Class of 2012)
If x is < 0, then -x (root of -x multiplied by absolute  [#permalink]

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New post 23 Oct 2008, 06:42
If x is < 0, then √-x [x] (root of -x multiplied by absolute value of x)

a) -x
b) -1
c) 1
d) x
e) √x

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Director
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Joined: 01 Jan 2008
Posts: 601
Re: root + asbolute value  [#permalink]

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New post 23 Oct 2008, 07:23
lordw wrote:
If x is < 0, then √-x [x] (root of -x multiplied by absolute value of x)

a) -x
b) -1
c) 1
d) x
e) √x


i assume it's sqrt[(-x)*abs(x)]=sqrt(x^2)=abs(x)=-x -> A
Manager
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Joined: 12 Oct 2008
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Re: root + asbolute value  [#permalink]

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New post 23 Oct 2008, 07:52
that's a tough one....
if x<0, that means √-x * -x=-x√-x
I see no way I could simplify the above and come up with one of the given answers...
Manager
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Joined: 22 Jun 2008
Posts: 101
Schools: Darden School of Business (Class of 2012)
Re: root + asbolute value  [#permalink]

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New post 25 Oct 2008, 06:15
Hi guys
I appreciate very much your answers, but without proper explanation I can not have a logic idea of the problem.
Tks. Lw.
Director
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Joined: 23 May 2008
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Re: root + asbolute value  [#permalink]

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New post 25 Oct 2008, 20:58
maratikus wrote:
lordw wrote:
If x is < 0, then √-x [x] (root of -x multiplied by absolute value of x)

a) -x
b) -1
c) 1
d) x
e) √x


i assume it's sqrt[(-x)*abs(x)]=sqrt(x^2)=abs(x)=-x -> A


ok i get why A could be the answer

the sqrt(-x*lxl) = sqrt(-x^2)= -x, theoretically

but isnt this impossible with real numbers? GMAT is based on real numbers


if you take sqrt(-4^2) you get sqrt16=4, but question says x<0
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Joined: 14 Jul 2008
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Re: root + asbolute value  [#permalink]

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New post 28 Oct 2008, 03:12
sq root of a number can't have a negative value. It should be D.
sq root of -X* absolute value of X = (x is less than 0) sq root of
sq root of = X

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: root + asbolute value &nbs [#permalink] 28 Oct 2008, 03:12
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