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by bblast » Sat Aug 06, 2011 10:55 am
If m and n are negative numbers, which of the following numbers CANNOT be negative?

m^2 - n^2

n^2 - m^2

mn - m^2

n^3 - n

m^2 - 2mn + n^2


this can be solved by plugging numbers - looking for a more subtle approach-E
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by edge » Sat Aug 06, 2011 10:58 am
m² - 2mn + n² = (m - n)² > 0. E is the answer.
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by Tani » Sat Aug 06, 2011 12:05 pm
m^2 - 2mn + n^2 = (m-n)^2 Therefore it must be positive.
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by MBA.Aspirant » Sat Aug 06, 2011 2:51 pm
m^2 - 2mn + n^2 = (m-n) (m-n), since both are -ve, whether result is +ve or -ve, product is +ve
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by Tani » Sat Aug 06, 2011 3:22 pm
all squares of real numbers are positive.
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by gmatboost » Sun Aug 07, 2011 10:08 pm
Since you may not have realized that E was a perfect square, let's look at the other 4 choices too:

A. m^2 - n^2
We don't know whether m or n has the bigger absolute value, so this could go either way.

B. n^2 - m^2
This is basically the same answer as A, since we know the same info about m as we know about n. There is no way one of these could be right while the other is wrong.

C. mn - m^2
Factor: m(n-m)
m is negative, and (n-m) could be positive or negative, so the product could be negative

D. n^3 - n
Factor: n(n^2 - 1)
n is negative, and (n^2 - 1) could easily be positive, so the product could be negative
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by bblast » Mon Aug 08, 2011 12:06 am
Thank u Greg, I approached the problem the way u have solved.


Missed that E is a perfect square. Gotta tighten the screws of my cranium. :shock:
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