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# Is |x|<1?

Author Message
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)
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Kudos [?]: 232 [0], given: 104516

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19 Aug 2004, 09:30
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Hello all

Is |x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 63 [0], given: 0

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19 Aug 2004, 10:23
These are my least favorite problems.

If we solve out the first one, and there are several ways to do it, you see that x could be 1/3 or it could be 3. Both work. So it's insufficient.

How do we do it? That's the sucky part. There are two ways. In the first way, you square both sides and solve for x, but when we do we get 0=3x^2 - 10x + 3, which means we've got to go to the quadratic equation, and we should always try to avoid that. But plugging those numbers into it yields both 1/3 and 3.

The other way is to set up two equations based on the original, with the second one showing the negative side of absolute value. The two equations would be:

a) x+1=2(x-1)
b) x+1=-[2(x-1)]

These are based on the original. If the absolute value of x+1 = x(x-1), then we can say the two equations above. Solving both, we get x=1/3 or x=3.

The second one only tells us that x isn't 3, so that's not sufficient.

Together, it must be that x is 1/3, so that's enough info.
19 Aug 2004, 10:23
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# Is |x|<1?

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