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I remember seeing this problem before but I cannot find it

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I remember seeing this problem before but I cannot find it [#permalink] New post 08 Dec 2006, 23:46
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I remember seeing this problem before but I cannot find it anymore in the posts.

Is \sqrt{(x-3)^2} = 3-x

(1) x is not equal to 3

(2) - x | x | > 0

Please explain your answers!
[Reveal] Spoiler: OA

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 [#permalink] New post 09 Dec 2006, 00:40
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Is sqrt of (x-3)^2 = 3-x

(1) x is not equal to 3

(2) - x | x | > 0

REPHRASE STEM

SQRT(X-3)^2 = / X-3/ = 3-X
/X-3/ = 3-X ONLY IF X< = 3

FROM ONE

INSUFF COULD BE < OR > 3

FROM TWO

-X/X/ >0 WE ARE SURE /X/ IS +VE THUS -X IS +VE TOO

ie: x is -ve this staisfys x<=3 thus suff

answer is B
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 [#permalink] New post 09 Dec 2006, 00:43
Same logic as Yezz :).... (B) :)
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 [#permalink] New post 09 Dec 2006, 04:07
Yezz, Fig, please help me out :oops: Showing you below how I solved it but it seems I have a different approach, which I think is wrong...

stem is asking if |x-3| = 3 - x

if x>0 then x-3 = x-3 ---> you're left with nothing
if x<0 then x-3 = 3-x ---> 2x=6 ---> x=3

is this the correct way???

so frustrating these absolute value problems!!!
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 [#permalink] New post 09 Dec 2006, 05:26
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A kind of rule to remember is to search the roots of the expression in each absolute value. It's similar to what I said in my response to u on another topic :)

As we have |x-3|, so we search the roots of x-3. Hence,
x-3 = 0 <=> x = 3

Thus, the domains to consider are x > 3 and x < 3.

Each domain implies a simplified equation that removes the absolute.

We do not consider x > 0 or x < 0 because nothing in the equation asks for it :)
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Re: DS: Absolute Value [#permalink] New post 09 Dec 2006, 06:04
Hermione wrote:
I remember seeing this problem before but I cannot find it anymore in the posts.

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

(1) x is not equal to 3

(2) - x | x | > 0

Please explain your answers!


The q asked here is if the sq root is -ve

So if |x| = -x

From (1) we do not know anything

From (2)

-x|x| > 0 means

since |x| is positive for the eq -x|x| to be > 0 x should be -ve

So B
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 [#permalink] New post 09 Dec 2006, 06:41
Thanks Fig! I got it!

Thank you for your help!!

Will keep that in mind for test day :-D
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Absolute value DS [#permalink] New post 04 Jun 2009, 20:45
hey guys was wondering if anyone could offer a way to solve this problem:


Is \sqrt{(X-3)^2} = 3-X?

(1) X is not equal to 3

(2) -X*|X| is greater than 0

answer: B.
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Re: Data Sufficiency problem [#permalink] New post 05 Jun 2009, 01:34
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Is square root (X-3)^2= 3-X ?

(1) X is not equal to 3
Since for +ve valuues of x ,
(X-3)^2= 3-X
and for negative values of x.,
(X-3)^2!= 3-X
INSUFFICIENT

(2) -X|X| is greater than 0

-X|X| > 0
this clearly means x <0

for x<0,
(X-3)^2!= 3-X

SUFFICIENT
IMO B
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Re: Data Sufficiency problem [#permalink] New post 05 Jun 2009, 01:35
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And yes , I assumed "(X-3)*2" as "square of (x-3)"
I have seen this problem before , I guess its GmatPrep one...
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Re: Absolute value DS [#permalink] New post 22 Jun 2010, 16:41
WE have [x-3]=3-x [] is absolute value
if x>= 3 we have x-3 = 3-x. Solve x-3 = 3-x by itself we have x = 3. Put togethere with x>= 3 we have x= 3

if x=<3 we have 3-x = 3-x. it if right with all x =<3
==> the solution without any additional condition is x =< 3

Let use the condition 1 only we can have x> 3 doesn't belong to the area x=<3==> insufficient
Let use the condition 2 only we have x< 0 belong to area x=< 3 ==> sufficient

SO I think the answer is B
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Re: DS: Absolute Value [#permalink] New post 28 Jun 2010, 00:15
i. Equality in question is TRUE if x is negative and FALSE if x is positive. Dual solution, hence NOT SUFFICIENT
ii. |x| always positive. For -x to be greater than 0, x has to be negative. Single solution, hence SUFFICIENT.
Re: DS: Absolute Value   [#permalink] 28 Jun 2010, 00:15
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