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Is root{(x-3)^2}=3-x?

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Re: Is root{(x-3)^2}=3-x? [#permalink] New post 16 Jan 2013, 03:04
vaivish1723 wrote:
Is \sqrt{(x-3)^2}=3-x?

(1) x\neq{3}
(2) -x|x| >0


\sqrt{(x-3)^2} = 3-x
|x-3| = 3-x
3-x >= 0
3>=x x is less than or equal to 3
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Re: [square_root](X-3)^2[/square_root][/m] = 3 -X ? [#permalink] New post 06 Mar 2013, 20:19
sujit2k7 wrote:
This is a DS question ..

Is \sqrt{(X-3)^2} = 3 -X ?

1) X # 3
2) -X|X| > 0


Only thing which the question is asking is 3-X positive
As Sqrt (X-3)^2
= X-3 if X-3 is positive
= 3-X if 3-X is positive

STAT1
is INSUFFICIENT as X# 3 doesn't tell anything about whether 3-X is positive or not.

STAT2
-X|X| > 0
since |X| is positive so
-X > 0
=> X <0
and if X< 0 then 3-X will be positive.
So, Answer will be B
Hope it helps!
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Re: [square_root](X-3)^2[/square_root][/m] = 3 -X ? [#permalink] New post 06 Mar 2013, 20:43
Expert's post
sujit2k7 wrote:
This is a DS question ..

Is \sqrt{(X-3)^2} = 3 -X ?

1) X # 3
2) -X|X| > 0


We know that \sqrt{X^2} = |X|. Thus, the question stem is asking whether |X-3| = 3-X. This is possible only if (X-3) is negative or X<3.

From F.S 1, we have x is not equal to 3. Clearly Insufficient.

From F.S 2, we have -X|X|>0. Thus, as |X| is always positive, X has to be negative. Thus, if X is negative, it will always be less than 3. Sufficient.

B.
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Re: Is root{(x-3)^2}=3-x? [#permalink] New post 18 Apr 2013, 15:04
vaivish1723 wrote:
Is \sqrt{(x-3)^2}=3-x?

(1) x\neq{3}
(2) -x|x| >0


Here is how i solved it:

1) x [/s]=[/s] 3. Not sufficient
2) -x|x|>0 => x<0 only then the given inequality holds

so [/square_root](s-3)^2[/square_root] => -(x-3) (as we know [/square_root]x^2[/square_root] = - (x) if x<0) and 3-x = -(x-3) so sufficient.

IMO B.
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Re: Is root{(x-3)^2}=3-x? [#permalink] New post 16 May 2014, 00:01
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Re: Is root{(x-3)^2}=3-x?   [#permalink] 16 May 2014, 00:01
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