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Gmat Prep - sqrt [(x-3)^2]

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Source: — Data Sufficiency |
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by PAB2706 » Fri Jun 05, 2009 6:41 am
statement 2 tells you that x is negative..

If you take any negative value and substitute in the given equation you wil get LHS=RHS...

Thus B is sufficient
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by agoyal2 » Fri Jun 05, 2009 6:50 am
Aaah.. now I get it.. if x is positive, then the RHS might be -ve or +ve, LHS is always +ve...hence we need to know if x is -ve to be able to determine if a stmt is sufficient.. Thanks mate!
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by abhinav85 » Fri Jun 05, 2009 8:04 am
Can anyone explain this ques in detail???
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by ssmiles08 » Fri Jun 05, 2009 8:11 am
Can anyone explain this ques in detail???


https://www.beatthegmat.com/gmat-prep-ques-t37241.html

Similar problem should give you a better idea
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by Domnu » Sat Jun 06, 2009 2:18 pm
First of all,

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

Now, we need to look at |x - 3| = 3 - x. For the first part, x != 3 doesn't give us anything... if x = 2 they are equal, and for x = 10 they are not equal.

The second part is sufficient.. particularly, it says that x < 0. If this is the case, then |x - 3| = 3 - x, since the rhs is always going to be positive.

So, the answer is B.
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by yogami » Sun Jun 07, 2009 7:02 am
"statement 2 tells you that x is negative..

If you take any negative value and substitute in the given equation you wil get LHS=RHS...

Thus B is sufficient"



Hey man, why is LHS always positive? A square root of a positive number can be a +ve or -ve isn't it?. I know that the square root of a negative number is complex.
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