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

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Is root{(x-3)^2}=3-x? [#permalink] New post 25 May 2011, 01:49
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Question Stats:

48% (01:35) correct 52% (00:43) wrong based on 29 sessions
Is \sqrt{(x-3)^2}=3-x?

(1) x\neq{3}
(2) -x|x| >0

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-root-x-3-2-3-x-92204.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Jul 2013, 01:44, edited 1 time in total.
Renamed the topic and edited the question.
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Re: tricky ? [#permalink] New post 25 May 2011, 02:36
Expert's post
\sqrt{(x-3)^2}=3-x ---> |x-3| = 3-x ---> x\le3

1) x\ne3 - x can be greater or lesser than 3. insufficient.

2) -x*|x| >0 --(|x| is non-negative)--> -x>0 ---> x<0 - x is always lesser than 3. sufficient.
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Re: tricky ? [#permalink] New post 30 May 2011, 18:10
Expert's post
Warlock007: As per your request, I am explaining the solution to this question. The solution is exactly as given by walker above. I will just elaborate on 'why'

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

1)x\ne3

2) -x*|x| >0

When we say square root, we mean just the positive square root. Another way to think about it is:

\sqrt{x^2} = |x| i.e. square root is only the positive root.

Therefore, \sqrt{(x-3)^2} = |x - 3|
Now the question is, is |x - 3| = 3 - x?

Now the definition of modulus can be used here.
|x| = x when x >= 0
and |x| = -x when x < 0

i.e. |5| = 5 since 5 > 0
and |-5| = -(-5) = 5 since -5 < 0 (Works for x = 0 too)

Therefore, |x - 3| = -(x - 3) = (3 - x) if (x - 3) <= 0

Let's go back to our question: Is |x - 3| = 3 - x?
|x - 3| = 3 - x only when (x - 3) <= 0 i.e. when x <= 3

Stmnt 1: x\ne3
No idea whether x is less than 3 so not sufficient.

Stmnt 2: -x*|x| >0
Now, mods are always greater than or equal to 0 (i.e. they can never be negative)
So |x| has to be positive.
Then -x must be positive too to make -x*|x| >0
This means x must be negative (only then will -x be positive)
If x is negative, it is certainly less than 3. Hence, this stmnt alone is sufficient.

Answer (B)

You could also do this question by plugging in numbers and checking (say try x = -1, 0, 1, 3, 4) but plugging in for such questions makes me worry that I have forgotten to check for some specific condition and hence I avoid doing it.
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Re: tricky ? [#permalink] New post 06 Jun 2011, 02:15
b gives x<0.

equation is always true.
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Re: tricky ?   [#permalink] 06 Jun 2011, 02:15
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