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Re: How to solve this Question...I have no clue... [#permalink]

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27 Feb 2010, 21:47

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]

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18 Mar 2011, 20:49

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...

Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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24 Jul 2011, 23:47

Lets disect the question stem first.

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

Carrying out the squaring and then the square root operation, we get:

|(x-3)| = 3 - x

To solve the modulus sign, we will have two cases - Case 1: (x-3)>=0 => x = 3

Case 2: (x-3) < 0 => Every value of x less than 3 satisfies the equation

Therefore sqrt((x-3)^2) = 3 - x for all values of x<=3, but this equation is not satisfied for values of x>3

Now consider the statements of the DS.

Statement (1): x is not equal to 3 If x is not equal to 3, it could be greater than 3 or less than 3. If x is less than 3, the equation is satisfied. If x is greater than 3, the equation is not satisfied. Therefore we cannot say if the equation will be satisfied if x is not equal to 3. Statement (1) alone is therefore insufficient to answer the question.

Next lets consider statement (2): Statement (2): -x|x|>0 Again to remove the modulus sign make two cases. Case 1: x>=0 => -x(x) >0 => -(x^2) > 0, which is impossible because x^2 is always >=0, so -(x^2) cannot be >0 for any values of x. Therefore to satisfy this inequality, x cannot be >0 Case 2: x<0 => -x (-x) > 0 => x^2 > 0 , which is true for all values of x except 0. Therefore the inequality is satisfied for all values of x<0

Therefore from statement (2) we know that x<0. And from the stem of the question (after we solved it), we know that the equation given in the stem is satisfied for all values of x<=3. Therefore the equation will hold for all values of x specified by statement (2). Therefore statement (2) alone is enough to solve the question.

Re: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x [#permalink]

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28 Jul 2011, 22:48

The inequality sign is needed to work with the modulus. When we remove the square root sign, we must take the modulus of the answer because the answer could be positive or negative.

To know more about working with modulus operators, please search for 'Absolute value (Modulus'
_________________

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