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# If x < 0, then (-x *|x|)^0.5 is A. -x B. -1 C. 1 D.

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Intern
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If x < 0, then (-x *|x|)^0.5 is A. -x B. -1 C. 1 D. [#permalink]

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07 May 2007, 22:42
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If x < 0, then (-x *|x|)^0.5 is

A. -x
B. -1
C. 1
D. x
E. x^0.5
SVP
Joined: 01 May 2006
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08 May 2007, 00:31
(A) for me

(-x *|x|)^0.5
= sqrt(x^2)
= |x|
= -x because x < 0 and abs must always be positive or nul.
Manager
Joined: 04 Oct 2006
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08 May 2007, 05:56
Lets take x=-3.

(-x *|x|)^0.5 = (3 *|-3|)^0.5 = 3

VP
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11 May 2007, 10:05
I Second (A)

So happy that the site is working ! I was getting worried !

Director
Joined: 26 Feb 2006
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11 May 2007, 13:30
PPGJ wrote:
If x < 0, then (-x *|x|)^0.5 is

A. -x
B. -1
C. 1
D. x
E. x^0.5

lets solve it this way:

x = -3

= (-x *|x|)^0.5
= (-(-3) *|-3|)^0.5
= (3*3 )^0.5
= 3 which is equivalant to - (-3) or -x.

so A.
Director
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13 May 2007, 04:11
Fig wrote:
(A) for me

(-x *|x|)^0.5
= sqrt(x^2)
= |x|
= -x because x < 0 and abs must always be positive or nul.

Hey Fig,
I am a bit confused.

Is it not (-x * |x| ) = sqrt(-x^2) instead of sqrt(x^2) .

Thanks..
SVP
Joined: 01 May 2006
Posts: 1796
Followers: 9

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13 May 2007, 04:46
Amit05 wrote:
Fig wrote:
(A) for me

(-x *|x|)^0.5
= sqrt(x^2)
= |x|
= -x because x < 0 and abs must always be positive or nul.

Hey Fig,
I am a bit confused.

Is it not (-x * |x| ) = sqrt(-x^2) instead of sqrt(x^2) .

Thanks..

Ok, I can try

As x < 0, we have -x > 0 and -x = |x|.

Then,
-x * |x|
= -x * (-x)
= x^2.

Notice also that, no matter x > 0 or x <0, x^2 > 0 (if x !=0). Thus, -x^2 < 0 and we cannot have a sqrt( Negative ) in GMAT.

Hope that helps
13 May 2007, 04:46
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