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# is sqrt ((x-3)^2) = 3-x? 1) x not equal to 3 2) -x|x| > 0

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is sqrt ((x-3)^2) = 3-x? 1) x not equal to 3 2) -x|x| > 0 [#permalink]  24 Apr 2008, 17:42
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is sqrt ((x-3)^2) = 3-x?

1) x not equal to 3
2) -x|x| > 0

The oa is B, but why is A alone not enough?
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  24 Apr 2008, 17:50
Nevermind, I've got it.

Here is the answer, part of the clue is from http://gmatclub.com/forum/7-t61120:

sqrt ((x-3)^2) is |x-3|

if x = 1 or 2 or 0, then |x-3| = 3-x

if x = 5, etc, then |x-3| is not equal to 3-x
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  24 Apr 2008, 21:45
From the que: 3-x is always >0 --> x has to be less than 3.

Option 1: X can also be >3 when ans fails so insufficient
Option 2: -x|x|>0 implies x is always < 0 which means x is less than 3 hence sufficient.

Ans : B
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  25 Apr 2008, 00:23
gmatnub wrote:
is sqrt ((x-3)^2) = 3-x?

1) x not equal to 3
2) -x|x| > 0

The oa is B, but why is A alone not enough?

given |x-3| can be equal to 3-x for x < 3,

1) X can be greater than 3
2) X is less than 0, i.e x < 3, for all x.

so B
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  13 Jun 2010, 02:57
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  13 Jun 2010, 03:41
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

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

Remember: \sqrt{x^2}=|x|. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}.

So \sqrt{x^2}\geq{0}. But what does \sqrt{x^2} equal to?

Let's consider following examples:
If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive;
If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive.

So we got that:
\sqrt{x^2}=x, if x\geq{0};
\sqrt{x^2}=-x, if x<0.

What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x|

Back to the original question:

So \sqrt{(x-3)^2}=|x-3| and the question becomes is: |x-3|=3-x?

When x>3, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true.

(1) x\neq{3}. Clearly insufficient.

(2) -x|x| >0, basically this inequality implies that x<0, hence x<3. Sufficient.

Hope it helps.
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x?   [#permalink] 13 Jun 2010, 03:41
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