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Re: DS- squareroot [#permalink]
16 May 2008, 17:05

should be 'B'.

Stmt1: for x = 2, sqrt[(x-3)^2] = (3-x)^2 for x=5, sqrt[(x-3)^2] does not equal to (3-x)^2 insuff Stmt2: deduces x< 0 for x<0, sqrt[(x-3)^2] does not equal to (3-x)^2, so suff

Re: DS- squareroot [#permalink]
17 May 2008, 08:26

vshaunak@gmail.com wrote:

should be 'B'.

Stmt1: for x = 2, sqrt[(x-3)^2] = (3-x)^2 for x=5, sqrt[(x-3)^2] does not equal to (3-x)^2 insuff Stmt2: deduces x< 0 for x<0, sqrt[(x-3)^2] does not equal to (3-x)^2, so suff

Can you please explain how the 2nd stmnt deduces to x<0?

Re: DS- squareroot [#permalink]
18 May 2008, 06:49

goalsnr wrote:

ventivish wrote:

Hi I agree with stat B

Stat1. Both x=2 and x=4 give a YES for the question but 5 gives a NO.So insifficient

Stat2. For all values of x<0 ; sqrt[(x-3)^2] is not equal to (3-x)^2 . So sufficient.

I think the trick is to plug in numbers

My question is why the solving for x will not work?

You need to be careful with solving for variables. If you solve for sqrt[(x-3)^2]=(3-x)^2 your calculations assume that (x-3) is positive, it could be that x-3 is negative in which case your equation would look like this: -(x-3)=(3-x)^2 This would solve for x=2 and x=3

So you still have 2 options, x=2 and x=4 which is insufficient. If you tried plugging in numbers you would just save time. Hope this helps!