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Is sqrt[(x-3)^2] = (3-x)^2 ? (1) x ≠ 3 (2) – x | x | > 0

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Is sqrt[(x-3)^2] = (3-x)^2 ? (1) x ≠ 3 (2) – x | x | > 0 [#permalink] New post 16 May 2008, 16:43
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A
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C
D
E

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20. Is sqrt[(x-3)^2] = (3-x)^2 ?
(1) x ≠ 3
(2) – x | x | > 0

Please explain your answers.
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Re: DS- squareroot [#permalink] New post 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
VP
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Re: DS- squareroot [#permalink] New post 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?
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Re: DS- squareroot [#permalink] New post 17 May 2008, 08:48
goalsnr wrote:

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


2nd statement -x |x| > 0
|x| >0 for all x
thus -x > 0 => x <0

hope this helps
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Re: DS- squareroot [#permalink] New post 18 May 2008, 05:39
goalsnr wrote:
20. Is sqrt[(x-3)^2] = (3-x)^2 ?
(1) x ≠ 3
(2) – x | x | > 0

Please explain your answers.


Can't we solve this problem by solving for x ? Iam confused what the question really asks for.

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

By solvingthe above equation
(x-3) = (3-x)^2

-> x^2 -7x +12 = 0

solving further
(x-3)(x-4)=0

x=3 0r x=4

Stat 1: since x ≠ 3 we can conclude x =4 Suff
stat2 : x<0, suff

answer is D.

Is my approach correct?
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Re: DS- squareroot [#permalink] New post 18 May 2008, 06:04
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
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Re: DS- squareroot [#permalink] New post 18 May 2008, 06:17
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?
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Re: DS- squareroot [#permalink] New post 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!
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Re: DS- squareroot [#permalink] New post 18 May 2008, 16:36
Finally I get it :)

num to plug in sqrt[(x-3)^2] (3-x)^2

0 3 9
1 2 4
2 1 1
-1 4 16
-2 5 25

values 0,1,2 show why stat1 is insuff

values -1, -2 show why stat2 is suff

OA is B
Re: DS- squareroot   [#permalink] 18 May 2008, 16:36
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