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

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Manager
Manager
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Joined: 22 Jun 2008
Posts: 104

Kudos [?]: 48 [0], given: 1

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, 05:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 48 [0], given: 1

Director
Director
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Joined: 01 Jan 2008
Posts: 617

Kudos [?]: 207 [0], given: 1

Re: root + asbolute value [#permalink]

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New post 23 Oct 2008, 06:23
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

Kudos [?]: 207 [0], given: 1

Manager
Manager
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Joined: 12 Oct 2008
Posts: 104

Kudos [?]: 10 [0], given: 0

Re: root + asbolute value [#permalink]

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New post 23 Oct 2008, 06:52
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...

Kudos [?]: 10 [0], given: 0

Manager
Manager
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Joined: 22 Jun 2008
Posts: 104

Kudos [?]: 48 [0], given: 1

Schools: Darden School of Business (Class of 2012)
Re: root + asbolute value [#permalink]

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New post 25 Oct 2008, 05:15
Hi guys
I appreciate very much your answers, but without proper explanation I can not have a logic idea of the problem.
Tks. Lw.

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Director
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Joined: 23 May 2008
Posts: 800

Kudos [?]: 87 [0], given: 0

Re: root + asbolute value [#permalink]

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New post 25 Oct 2008, 19:58
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

Kudos [?]: 87 [0], given: 0

Intern
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Joined: 14 Jul 2008
Posts: 12

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Re: root + asbolute value [#permalink]

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New post 28 Oct 2008, 02:12
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 <-(-x)*[-x]>
sq root of <x*x>= X

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Re: root + asbolute value   [#permalink] 28 Oct 2008, 02:12
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