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# Is srqt{(x-3)^2} = 3-x ? 1) x is not eq 0 2) -x |x| > 0

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Director
Joined: 20 Sep 2006
Posts: 630
Is srqt{(x-3)^2} = 3-x ? 1) x is not eq 0 2) -x |x| > 0  [#permalink]

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04 Sep 2008, 23:34
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Is srqt{(x-3)^2} = 3-x ?

1) x is not eq 0
2) -x |x| > 0

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

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

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Director
Joined: 14 Aug 2007
Posts: 692
Re: DS tough  [#permalink]

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04 Sep 2008, 23:42
What we have to find out is, if |x - 3| = 3 -x ,
This will be true if x-3<0 or x < 3

1) let x = 1, |-2| = 2 , true,
let x = 4, untrue 1 = -1

thus insuff

2) as |x| is always positive, x has to be nagative in order to satisfy
-x|x|>0

Which thus satisfies x<3

B should be the answer
VP
Joined: 17 Jun 2008
Posts: 1474
Re: DS tough  [#permalink]

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05 Sep 2008, 00:20
B should be the answer.

From Stmt 1, x can be +ve or -ve....i.e. it can be greater than 3 or less than 0 making x-3 positve or negative. Hence, not sufficient.

From stmt 2, -x|x| is positive. This is possible only when |x| = -x. That means, x is -ve. Thus, x-3 will always be negative and hence, sufficient.
Senior Manager
Joined: 16 Jul 2008
Posts: 279
Re: DS tough  [#permalink]

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05 Sep 2008, 00:24
Agree with B.

Although we are looking for x<=3
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VP
Joined: 17 Jun 2008
Posts: 1279
Re: DS tough  [#permalink]

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05 Sep 2008, 21:42
rao_1857 wrote:
Is srqt{(x-3)^2} = 3-x ?

1) x is not eq 0
2) -x |x| > 0

(1)sqrt root any square of a number is +ve of the number=> x INSUFFI

(2)SUFFI => x is negative hence x<3 hence 3-x is +ve

IMO B

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Re: DS tough &nbs [#permalink] 05 Sep 2008, 21:42
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