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Director
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Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Is sqrt((x-3)^2)=3-x
1) x not equal to 3
2) -xlxl>0
Show the evidence please 
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Director
Joined: 12 Jun 2006
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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
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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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SimaQ wrote: Is sqrt((x-3)^2)=3-x 1) x not equal to 3 2) -xlxl>0 Show the evidence please 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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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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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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I'm puzzled by this one - as mentioned, the stem gives an equivalent equation
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Director
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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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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
_________________
Regards,
Ahamdevam
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GMAT Club Legend
Joined: 07 Jul 2004
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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
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SimaQ wrote: Is sqrt((x-3)^2)=3-x 1) x not equal to 3 2) -xlxl>0 Show the evidence please 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
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OA is B.
I needed the clearence on -xlxl>0 part....
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Intern
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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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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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