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Algebra Difficulty 2.7 Junior Find the answer

The equality (x+m)2−(x+n)2=(m−n)2(x+m)^2-(x+n)^2=(m-n)^2, where mm and nn are unequal non-zero constants, is satisfied by x=am+bnx=am+bn, where:
$\textbf{

Pick one

Solution

Expand binomials, combine like terms, and subtract terms from both sides.
x2+2xm+m2−x2−2xn−n2=m2−2mn+n2x^2 + 2xm + m^2 - x^2 - 2xn - n^2 = m^2 - 2mn + n^2
2xm+m2−2xn−n2=m2−2mn+n22xm + m^2 - 2xn - n^2 = m^2 - 2mn + n^2
2xm−2xn−n2=−2mn+n22xm - 2xn - n^2 = -2mn + n^2
Get all the x-terms on one side and factor to solve for x.
2xm−2xn=−2mn+2n22xm - 2xn = -2mn + 2n^2
2x(m−n)=−2n(m−n)2x(m-n) = -2n(m-n)
Since m≠nm \not= n, both sides can be divided by m−nm-n.
x=−nx = -n
That means a=0a = 0 and b=−1b = -1, so the answer is (A)\boxed{\textbf{(A)}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.