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If m<n<0, x=m^2+n^2, and y=2mn,

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If m<n<0, x=m^2+n^2, and y=2mn,

by Max@Math Revolution » Tue Jan 30, 2018 2:09 am
If m<n<0, x=m2+n2, and y=2mn, what is the value of $$\sqrt{x^2-y^2}$$, in terms of m and n?

A. m^2+n^2
B. 2m^2+n^2
C.m^2+2n^2
D. m^2-n^2
E. n^2-m^2
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Watch out for negative values!

by DrMaths » Wed Jan 31, 2018 8:57 am
Test by using simple integers such as n = -1, m = -2
x=m^2+n^2 = 2^2 + 1^2 = 5
y = 2mn = 2 * 2 * 1 = 4

Now substitute...
x^2-y^2 = 25 - 16 = 9
So sqrt(x^2-y^2) = 3 (This is what we are looking for)

Now test the options:
A. m^2+n^2 = 4 + 1 = 5
B. 2m^2+n^2 = 8 + 1 = 9
C.m^2+2n^2 = 4 + 2
D. m^2-n^2 = 4 - 1 = 3 YES
E. n^2-m^2 = 1 - 4 = -3

Answer = D
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by DrMaths » Wed Jan 31, 2018 9:16 am
Alternative method:
We should know that the difference of 2 squares x^2-y^2 = (x+y)(x-y)
Now substitute x=m^2+n^2, and y=2mn:
(x+y)(x-y) = (m^2+n^2+2mn)(m^2+n^2-2mn)
This gives a result that includes negative terms, so only answers D and E are valid.

Also, the factor (m^2+n^2+2mn)>0 since all negative-term pairs produce positive multiples.
Furthermore, the factor (m^2+n^2-2mn)> 0 because (m^2+n^2)>2mn due to m^2>n^2>0
Hence we are looking for a positive answer.
Only D produces a positive answer containing negative terms.
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