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Re: How to solve this Question...I have no clue... [#permalink]
27 Feb 2010, 20:47

is \sqrt{(x-3)^2} = 3-x? square root of a number is positive. if x<=3 .. 3-x is always positive if x>3 ... 3-x is negative

st 1) x!=3 . not sufficient as we dont know if x>3 or x<3 st 2) -x |x| > 0 |x| is always greater than 0, so x has to be negative for the expression to be > 0

Re: How to solve this Question...I have no clue... [#permalink]
18 Mar 2011, 19:49

I'm a bit confused by this problem Why can I not square both sides of the statement to get rid of the radical? I will then have (x-3)^2=(3-x)^2 Statement 1 and 2 both will not apply in this case and the answer is then E.

However if I choose Bunuels method of doing it, I still dont get how B is the answer. Statement 2 basically says that x=-2,-3....and so on and so forth will satisfy the statement. So basically all negative numbers equal to or less than 2 will satisfy the condition...

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

Please help me understand this concept.

Regards, Ravi

I got it now!

|x-3| = (3-x) ===> |3-x| = (3-x) ===>which is only possible when (3-x) >=0 (i.e +ve)

(like |x| =x only when x>=0)

so we have 3-x >=0 ====> x<=3

Also for option B

-x|x| > 3 ==> x|x| < 3 so |x| is always positive but so make it less than 3 , x must be negative. i.e. x<0

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