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by tohellandback » Thu Jun 18, 2009 12:40 am
IMO B
√(x-3)² = x-3 when x-3>=0
=3-x when x-3<0
A) not sufficient
B)–x|x|=0, so x must be negative

B is suffiecient
The powers of two are bloody impolite!!
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by rah_pandey » Thu Jun 18, 2009 2:04 am
x not equal to 3 => answer is not zero and so a conscious decision has to be made to either put answer as x-3 or 3-x

&#8730;(x-3)² = +/-(x-3)

in second condition it is given
-x|x|>0
since |x|>0 always
therefore -x>0
=> x<0
now
let x=-k where k>0
&#8730;(-k-3)² =k+3 (here we are not taking -ve value as by convention &#8730;X=> positive root.
=3-x (replace k by -x)

B is sufficient
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