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# Is (x-3)^2 = 3 - x ? (1) x 3 (2) x | x | > 0 Some

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Joined: 01 May 2009
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Is (x-3)^2 = 3 - x ? (1) x 3 (2) x | x | > 0 Some  [#permalink]

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20 Jul 2009, 01:49
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Is \sqrt{(x-3)^2} = 3 - x ?
(1) x ≠ 3
(2) – x | x | > 0

Some help & explanations needed please.

Many thanks!

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20 Jul 2009, 08:22
cialit0506 wrote:
Is \sqrt{(x-3)^2} = 3 - x ?
(1) x ≠ 3
(2) – x | x | > 0

Some help & explanations needed please.

Many thanks!

the square root of the (x-3)^2 basically makes whatever's in it positive all the time. on the left it is different though..
(1). INSUFFICIENT. Since the left side will ALWAYS be positive if I plug in 4 on the left i get 1 and the right get -1. If i plug in 1 i get 2 on the left and 2 on the right as well so no good.
(2). only way to have that statement true is to have a NEGATIVE x (so the - negates it..). Now if you plug in any negative number on the left you will still get a positive result and on the right you will always get positive result as well and they always equal to eachother so SUFFICIENT.

B. kudos
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Joined: 07 Apr 2009
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20 Jul 2009, 08:30
lets attack the option B.

-x|x| > 0

$$-x *(x) >0$$ or $$-x*(-x) > 0$$
$$x^2 < 0$$or $$x^2 > 0$$

$$x^2$$ can never be less than 0, so x cannot be positive. so from the above statement we have to conclude that x is negative.

for any negative value of X the given statement will hold good. so B is SUFF.

A is not SUFF because when x > 3, the given statement doesnt hold good.

hence 'B'
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20 Jul 2009, 08:37
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?
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Joined: 07 Apr 2009
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20 Jul 2009, 10:24
1
Quote:
let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?

IMO, while considering square root , we consider only positive square root.

16 = $$(+4)^2$$ or $$(-4)^2$$
$$sqrt (16) = 4$$
Senior Manager
Joined: 04 Jun 2008
Posts: 276

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20 Jul 2009, 10:40
1
pleonasm wrote:
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing
something ?

I was thinking the same

Actually its a confusion betwn sqrt x and x^2

if X^2 = 16; then x can be +4 or -4
but if asked to find out sqrt, sqrt 16 will always be 4
Senior Manager
Joined: 01 Mar 2009
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20 Jul 2009, 13:20
So sqrt of a number is always a positive value ... hmmmm .. ok I guess I now get it..

sqrt (16 ) = sqrt (4 *4) = sqrt(4) * sqrt(4) = 2*2 = 4
sqrt(16) = sqrt(-4*-4) = sqrt (-4) * sqrt (-4) .. oh well there's no square root for a -ve number .. Thanks guys
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23 Jul 2009, 08:16
Dear ppl,

Thanks!

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Re: Inequalities - Get Answers by Filling Up the Blanks? &nbs [#permalink] 23 Jul 2009, 08:16
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