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by rohangupta83 » Mon Jul 18, 2011 3:17 am
IMO: E

i'm a bit stumped on this one but this is my thought process.

Question: Is sqrt((x-5)^2) = 5-x?

sqrt((x-5)^2) = x-5 or 5-x

Choice (i) -x|x| > 0 implies that x is negative.
Thus both x-5 or 5-x are possible as we dont have any information to reject any one of them. Thus, I cannot come to a conclusion here and choice(i) is not sufficient.

Choice (ii) 5-x>0
All it says is x<5. Not enough information for me to reject either x-5 or 5-x.

And together its the same as x is negative

So, for this question I am unable to conclude whether sqrt((x-5)^2) = 5-x or x-5.

IMO: E

Experts please comment.. thanks!
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by Frankenstein » Mon Jul 18, 2011 3:35 am
dhannya wrote:Is root of ( x-5)square = 5-x

(1) -xlxl > 0
(2) 5-x > 0
Hi,
square root of a number is always non-negative
So, sqrt((x-5)^2) = 5-x ? => Is|x-5| = 5-x => Is 5-x >= 0 => Is x<= 5
From(1):
x<0
So, definitely, x<=5
Sufficient
From(2):
5-x > 0
Sufficient

Hence, D
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by rohangupta83 » Mon Jul 18, 2011 3:49 am
Hi,

Isn't sqrt(4) = +2 or -2?
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by Frankenstein » Mon Jul 18, 2011 3:52 am
rohangupta83 wrote:Hi,

Isn't sqrt(4) = +2 or -2?
Hi,
No sqrt(4) = 2 only
If we are given x^2 = 4, what values of x satisfy this equation?, then x = +2 or -2.
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by amit2k9 » Mon Jul 18, 2011 6:48 am
[(x-5)^2]^1/2 = |x-5|

hence the question is |x-5| = 5-x possible only when x<5 as can be seen.

a x will always have negative values hence x<5. sufficient.

b x<5 hence true too.

D is clean.
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by rohangupta83 » Mon Jul 18, 2011 7:22 am
Thanks Frankenstein,

Just read up and clarified my confusion. You're right. sqrt of a number is always positive.
I just ended up thinking too much about the problem.

Cheers !
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