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is x>0

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is x>0 [#permalink] New post 10 Jan 2012, 19:07
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63% (01:06) correct 38% (00:37) wrong based on 8 sessions
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Can you help me understand why (1) is not sufficient? It seems that I can express it as (x-3)(x-3) and therefore x=3,3 which is >0.

Here's the question:

Is x positive?

(1) (x – 3)2 > 0

(2) x3 – 1 > 0
[Reveal] Spoiler: OA
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Re: is x>0 [#permalink] New post 10 Jan 2012, 20:12
It's a problem on inequality. You cannot expand the expression and equate it to zero!

1: Square of any positive or negative real number is always positive. Thus, (x-3) and x can be both positive or negative.
1) x-3>0 => x>3 => x is positive
2) x-3<0 => x<3 => x can have negative values too!
(x-3)^2 > 0 cannot imply that x is positive and thus is not sufficient.

2: x^3-1> => x^3>1. Cube of a positive number is always positive, and of a negative number is always negative. This is sufficient, as it gives us x to be positive.

Thus the correct answer is B, statement 2 is sufficient.
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Re: is x>0 [#permalink] New post 11 Jan 2012, 19:13
From one x can be +ve or Negative

From 2 x can only be +ve
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Re: is x>0 [#permalink] New post 11 Jan 2012, 19:32
a) [(x-3)]^2 is positive for any value of x other than 3; this means x can be negative or positive- Not sufficient

b) rewrite it as x^3 > 1; since this is an odd exponent, there is no danger; x has to be positive (already sufficient) and it has to be greater than 1 to satisfy condition;
Re: is x>0   [#permalink] 11 Jan 2012, 19:32
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