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# Ds - 1000Q

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Director
Joined: 06 Feb 2006
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16 Nov 2006, 11:56
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Is sqrt((x-3)^2)=3-x

1) x not equal to 3
2) -xlxl>0

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Director
Joined: 12 Jun 2006
Posts: 532

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16 Nov 2006, 12:38
A

1) (x-3)^2 = 3-x
2) x^2 - 6x + 9 = 3-x
3) x^2 + 9 = 3 + 5x
4) x^2 + 6 = 5x
5) x^2 - 5x + 6 = 0
6) x=2 and 3

Statement 1 says x doesn't equal 3. That leaves 2.

Statement 2 tells us that x is positive. That could be 2 or 3.

Is this right?

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Director
Joined: 06 Feb 2006
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16 Nov 2006, 12:41
ggarr wrote:
A

1) (x-3)^2 = 3-x
2) x^2 - 6x + 9 = 3-x
3) x^2 + 9 = 3 + 5x
4) x^2 + 6 = 5x
5) x^2 - 5x + 6 = 0
6) x=2 and 3

Statement 1 says x doesn't equal 3. That leaves 2.

Statement 2 tells us that x is positive. That could be 2 or 3.

Is this right?

Nope.

You forgot to raise (3-x)^2.....

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VP
Joined: 15 Jul 2004
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Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)

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16 Nov 2006, 12:54
SimaQ wrote:
Is sqrt((x-3)^2)=3-x

1) x not equal to 3
2) -xlxl>0

Going with B - stmt 2 alone is sufficient but 1 is not.

since -x|x| >0 ==> x < 0.

This means x - 3 < 0 and therefore we know that x - 3 is a negative number that is being squared. when you take the sqr root of the square of a negative number we should consider the negative root as the result.

sqrt(-3^2) = -3

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

What do you think??

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Intern
Joined: 11 Sep 2006
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16 Nov 2006, 13:24
My answer is D. From the stem we get:

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

Plug in any values for x and they are equal. It seems that you can solve this from the info. in the stem, is that right?

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Director
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16 Nov 2006, 13:27
Don't the 2 equations equal one another? If so, wouldn't the answer be D?

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Senior Manager
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16 Nov 2006, 14:47
I'm puzzled by this one - as mentioned, the stem gives an equivalent equation

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Director
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16 Nov 2006, 18:32
Maybe I'm really far off on this one, but...
the square root of (x-3)^2...isn't that just (x-3)? So isn't the question asking, is x-3 = 3-x? And if you solve the equation, then the stem is just, is x = 3? If that's the case, then only A is correct...
What is OA?
_________________

...there ain't no such thing as a free lunch...

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Intern
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16 Nov 2006, 19:03
From-1 : we get (x-3) or 3-x
From-2 : x < 0 means x-3 must be negative which when sqared.
on sqare-rooting it should give us 3 - x.

so Stmt-2 alone is sufficient..

B
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Regards,
Ahamdevam

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GMAT Club Legend
Joined: 07 Jul 2004
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16 Nov 2006, 19:37
St1:

x = 2, sqrt((x-3)^2) = 1, 3-x = 1 --> LHS = RHS
x = 7, sqrt((x-3)^2) = 4, 3-7 = -4 --> LSH != RHS

Insufficient.

St2:

|x| is alway positive, so x must be negative for -x|x| to be positive.

x = -1, sqrt((x-3)^2) = 4, 3-x = 4 --> LHS = RHS
x = -2, sqrt((x-3)^2) = 5, 3-x = 5 --> LHS = RHS
x = -7, sqrt((x-3)^2) = 10, 3-x = 10 --> LHS = RHS
x = -1/2, sqrt((x-3)^2) = 3.5, x-x = 3.5 --> LHS = RHS

I'll take B.

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VP
Joined: 28 Mar 2006
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16 Nov 2006, 19:46
SimaQ wrote:
Is sqrt((x-3)^2)=3-x

1) x not equal to 3
2) -xlxl>0

Straight B....

-x|x| > 0 if x is -ve.

|x| is always positive

for -x|x| > 0 then x has to be -ve.

1) Isnt needed

\\So B

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Director
Joined: 06 Feb 2006
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17 Nov 2006, 00:24
OA is B.

I needed the clearence on -xlxl>0 part....

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Intern
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17 Nov 2006, 04:38
I think I'm mixed up with simplifying. Can somebody explain why
sqrt((x-3)^2)=3-x isn't the same as (x-3)^2=(3-x)^2 ?

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VP
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17 Nov 2006, 11:43
Googlore wrote:
I think I'm mixed up with simplifying. Can somebody explain why
sqrt((x-3)^2)=3-x isn't the same as (x-3)^2=(3-x)^2 ?

(x-3)^2=(3-x)^2 are equal. But the if squares of two numbers are equal doesn't mean that the numbers too are equal. They could be opposite in sign and still their squares would be the same.

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17 Nov 2006, 11:43
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