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I remember seeing this problem before but I cannot find it

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Senior Manager
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I remember seeing this problem before but I cannot find it [#permalink] New post 04 May 2007, 01:52
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
I remember seeing this problem before but I cannot find it anymore in the posts.

Is sqrt of (x-3)^2 = 3-x

(1) x is not equal to 3

(2) - x | x | > 0

Please explain your answers!


Javed.

Cheers!
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 [#permalink] New post 04 May 2007, 01:58
(B) for me :)

sqrt of (x-3)^2 = 3-x ?
<=> |x-3| = 3-x ?
<=> x =< 3 ?

From 1
x != 3
<=> X > 3 or X < 3

INSUFF.

From 2
- x*| x | > 0
<=> x < 0 < 3

SUFF.
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 [#permalink] New post 04 May 2007, 04:29
Is sqrt of (x-3)^2 = 3-x ?

(1) if x=1 then (x-3)^2 = 3-x
if x=10 then (x-3)^2 <3> 0 means that x < 0
then as sqrt(x-3)^2 = 3-x is equivalent to |x-3| = 3-x
In conclusion, (2) is sufficient

Answer B.
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 [#permalink] New post 04 May 2007, 05:23
I second (B)

sqrt[(x-3)^2]=x-3

Statement 1

x≠3

dosen't help - x=2 or x=6

sqrt[(2-3)^2]=2-3
sqrt[1]≠-1

or

sqrt[(6-3)^2]=6-3
sqrt[9]=3

insufficient

Statement 2

-x*| x |>0

meaning x has to be negative !

and sqrt[(x-3)^2] will never be equal to x-3

sufficient
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 [#permalink] New post 04 May 2007, 06:43
yeah all you need is statement 2 to solve it.

-x*|x|>0
|x| will always be positive
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 [#permalink] New post 04 May 2007, 07:03
yup B it is..

we first need to know if X is greater than 3 or not..

1) is not sufficient

2) this is suff..x is always less than 3...
  [#permalink] 04 May 2007, 07:03
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