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Senior Manager
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13 Jun 2005, 10:10
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Senior Manager
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13 Jun 2005, 10:12
feel like A is the answer

1) can be only 2
2) can be 2 or -2
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13 Jun 2005, 10:23
Looks like it's C.

A is not suff.: x can be =2 or -1

B illiminates x=-1

So it's C
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13 Jun 2005, 12:21
formal solution for (1)

|x-x^2|=2
can be divided into 2 equations
x-x^2=2 roots[2;-1]
x-x^2=-2 no real roots

What is a formal solution for (2)??

You should come up with the following results( Can Mathcad be wrong?):
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13 Jun 2005, 13:23
OK:

Solution for #1 is -1 or 2

So A isn't enough

Solution for # 2 is -2 or 2

So B isn't enough either

Combine A and B: the only way to satisfy both equations is to use 2.

What is wrong in my line of reasoning ?
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13 Jun 2005, 15:33
C. The easiest way is to graph it (without 2).

This problem is too hard for GMAT. Both functions have negative solutions that aren't too obvious, and probably not integer
Senior Manager
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13 Jun 2005, 22:45
katrin wrote:
OK:

Solution for #1 is -1 or 2

So A isn't enough

Solution for # 2 is -2 or 2

So B isn't enough either

Combine A and B: the only way to satisfy both equations is to use 2.

What is wrong in my line of reasoning ?

Nothing, I suppose!
Just wanted a formal solution in the sense that you build up the equations from which you get 2 and -2
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16 Jun 2005, 04:35
In (1), we know |x^2| = x^2, so it reduces to solving

x-x^2=2 or x-x^2=-2

First one has no real solutions, second one has 2 and -1

So (1) gives 2 and -1 as possible values of x. Not suff

(2) alone does not tell us anything either

But together, they give x=2

If I saw this in my GMAT, I'd likely have solved (1) and assumed that (2) would not give us a unique solution and chosen C.
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16 Jun 2005, 11:50
(1)|x-|x^2||=2
Since x^2>=0 it redueces to |x-x^2|=2
(x-x^2) is always less than zero if x is an integer (very easy to prove)
So it reduces to
x^2-x-2=0
x=2 or x=-1
Insufficient

(2)|x^2-|x||=2
If x>0
|x^2-x|=2
x=2 (solve as (1))

If x<0
x=-2 (by symmetry. You can solve it too like before)
Insufficient

Combined:
x=2
sufficient

I disagree this is too hard for GMAT. I say it is a perfect GMAT question.
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16 Jun 2005, 11:50
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