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Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink]
 20 Dec 2009, 07:50
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Is \sqrt{(x-3)^2}=3-x? (1) x\neq{3}(2) -x|x| > 3
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Ans: E.
First statement is not sufficient because if x=4, it may or may not be true. LHS = +1 or -1 while RHS = -1.
Second statement leads us to -3^1/2 < x < 0. For all the values between 0 and -3^1/2, which leads to |LHS| = RHS.
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jan4dday wrote: Is \sqrt{(x-3)^2}=3-x?
(1) x not = 3
(2) -x|x| > 3
will give OA as soon as 1st few replies come in Remember: \sqrt{x^2}=|x|. \sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x. When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true. When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Basically question asks is x\leq{3}? (1) x is not equal to 3. Clearly insufficient. (2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient. Answer: B.
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Re: How to solve this Question...I have no clue... [#permalink]
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
if x is negative, then \sqrt{(x-3)^2} = 3-x
B
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Re: How to solve this Question...I have no clue... [#permalink]
27 Feb 2010, 21:58
Thank you very much.
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Re: How to solve this Question...I have no clue... [#permalink]
17 Mar 2011, 03:58
Just a slight correction I think...
St2 is saying x<-sqrt of 3 which is still a sufficient condition.
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I got x<-1.732 from statement 2 This satisfies the x<3 requisite Hence B Posted from my mobile device 
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Re: How to solve this Question...I have no clue... [#permalink]
17 Mar 2011, 23:07
sqrt((x-3)^2) = |x-3| = x-3 if x > 3 or 3-x if x < 3 (1) is not enough (2) -x|x| is positive then So -x is also +ve and hence x is -ve so x < 3, hence answer is B
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Re: How to solve this Question...I have no clue... [#permalink]
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...
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Is square_root of (x-3)^2 = 3 - x ? [#permalink]
27 Jun 2012, 12:08
Is \sqrt{(x-3)^2} = 3 - x ? (1) x is not equal to 3 (2) -x |x| > 0According to my reasoning: \sqrt{(x-3)^2}= |x-3|So, the question is: Is |x-3| = 3 -x? In other words: Is x - 3 < 0 ? Is x < 3? So, let's see the clues: (1) x is not equal to 3 INSUFFICIENT. (2) -x |x| > 0Based on this info we know that x is not zero. So, -x > 0x < 0SUFFICIENT. IMO, the OA is : Please, confirm whether I am Ok.
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Re: Is square_root of (x-3)^2 = 3 - x ? [#permalink]
27 Jun 2012, 12:15
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Bunuel wrote: jan4dday wrote: Is \sqrt{(x-3)^2}=3-x?
(1) x not = 3
(2) -x|x| > 3
will give OA as soon as 1st few replies come in Remember: \sqrt{x^2}=|x|. \sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x. When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true. When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Basically question asks is x\leq{3}? (1) x is not equal to 3. Clearly insufficient. (2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient. Answer: B. Dear Bunuel, is it possible in this question to solve by taking conditions x>0 and x<0 for |x-3| = 3-x instead of x>3 ?
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Re: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink]
01 Jul 2012, 13:07
B it is!!Good explanation though!
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DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink]
21 Nov 2012, 20:11
I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help
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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink]
21 Nov 2012, 22:02
arvindbhat1887 wrote: I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help The question is Is \sqrt{(x-3)^2}= 3-x? (1) x<>3 (2) -x|x|>0 \sqrt{y} can only be positive or 0. So,The question is basically asking whether 3-x >= 0 1) Insufficient. If x=1, true, If x=10, false. 2) From this we get that x is negative. So -x is postivie. So, 3 + (-x) will always be a positive number. Sufficient. Answer is hence B. Kudos Please... If my post helped.
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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink]
22 Nov 2012, 05:48
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Bunuel wrote: jan4dday wrote: Is \sqrt{(x-3)^2}=3-x?
(1) x not = 3
(2) -x|x| > 3
will give OA as soon as 1st few replies come in Remember: \sqrt{x^2}=|x|. \sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x. When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true. When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true. Basically question asks is x\leq{3}? (1) x is not equal to 3. Clearly insufficient. (2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient. Answer: B. (2) Actually -x|x| > 3, implies that x<-sqrt(3), but still, as you mentioned, consequently implies that x<3. Sufficient.
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