Equation with the discriminant

Question

I'm stuck on my maths homework, and would appreciate help.

The question is:

Show that if the equation $(m + n)x^2 - 2mnx - (m-n) = 0$ has equal roots, then $$m^2 = \frac{n^2}{1-n^2}$$

I've worked out that I need to use the discriminant to prove this. This is my progress so far:

$b^2 - 4ac = 0$

$(-2mn)^2 - (4(m+n)(-(m-n)) = 0$

$4m^2n^2 - ((4m + 4n)(-m+n)) = 0$

$4m^2n^2 - (-4m^2 + 4mn - 4mn + 4n^2) = 0$

$4m^2n^2 - (4n^2 - 4m^2) = 0$

$m^2n^2 - (n^2 - m^2) = 0$

$m^2n^2 - n^2 + m^2 = 0$

$m^2n^2 + m^2 = n^2$

$m^2(n^2 + 1) = n^2$

$m^2 = \frac{n^2}{n^2 + 1}$

I can't find a way to rearrange it to get the required proof. Can someone solve this for me?

Answer 1

For equal roots, $$\begin{align}b^2&=4ac\\ (2mn)^2&=4(m+n)[-(m-n)]\\ m^2n^2&=-(m^2-n^2)\\ m^2(n^2+1)&=n^2\\ m^2&=\frac{n^2}{1+n^2}\end{align}$$

which seems to be the same as your answer.

Answer 2

$$\Delta:=b^2-4ac=(2mn)^2\color{red}-4(m+n)\cdot(\color{red}-(m-n))=4m^2n^2+4(m^2-n^2)=0\iff$$

$$\iff m^2n^2+m^2-n^2=0\iff m^2(n^2+1)=n^2\iff m^2=\frac{n^2}{n^2+1}$$

Thus, the solution the OP provides as given is wrong, and hypergeometric (and the solution the OP himself provides) is right.