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lordw
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If x is < 0, then -x (root of -x multiplied by absolute 23 Oct 2008, 06:42
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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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maratikus
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If x is < 0, then -x (root of -x multiplied by absolute 23 Oct 2008, 07:23
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lordw
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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i assume it's sqrt[(-x)*abs(x)]=sqrt(x^2)=abs(x)=-x -> A
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linau1982
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If x is < 0, then -x (root of -x multiplied by absolute 23 Oct 2008, 07:52
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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...
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lordw
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If x is < 0, then -x (root of -x multiplied by absolute 25 Oct 2008, 06:15
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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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bigtreezl
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If x is < 0, then -x (root of -x multiplied by absolute 25 Oct 2008, 20:58
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maratikus
lordw
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
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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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wanwinnie
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If x is < 0, then -x (root of -x multiplied by absolute 28 Oct 2008, 03:12
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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



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