Simplifying a solution to a quadratic equation I am solving: 
$(\sigma_A^2 - 2\rho\sigma_A\sigma_B +\sigma_B^2)x^2 +2(\rho\sigma_A\sigma_B - \sigma_B^2)x +\sigma_B^2 = 0$
I need to show that a real $x$ exists if and only if $\rho = \pm 1$
Using the quadratic formula I could only get as far as
$ x = \frac{-2(\rho\sigma_A\sigma_B - \sigma_B^2) \pm \sqrt{4\sigma_A^2\sigma_B^2(\rho^2-1)}}{2( \sigma_A^2 +\sigma_B^2 - 2\rho\sigma_A\sigma_B)}$
I have a more simplified solution, which is 
$x=[1- \frac{\sigma_A}{\sigma_B}(\rho\pm\sqrt{\rho^2-1})]^{-1}$
but I cannot see how to get there.
 A: I'm assuming you've missed an $x$ factor out, as jim has suggested in a comment above.
If so, I don't think that a real $x$ exists if and only if $\rho = \pm 1$ . The discriminant is indeed $\Delta = 4\sigma_A^2\sigma_B^2(\rho^2-1)$ , so that a real solution exists if and only if $\rho^2-1 \geqslant 0$, which is true if and only if $\rho \geqslant 1$ or $\rho \leqslant -1$.
However your exercise might have extra conditions given, and I could imagine something like the condition $\left|\rho\right| \leqslant 1 $ being given here. (Edit: for instance if $\rho$ is a correlation as jim has pointed out!) If this condition is given then combining it with the conditions above ( $\rho \geqslant 1$ or $\rho \leqslant -1$ ) yields that a real solutions exists iff $\rho = \pm 1$.
Another possibility would be that the question is when a unique real solution exists, in which case we need $\Delta = 0$ and so $\rho^2 = 1$, iff $\rho = \pm 1$.
A: Hint:
The usual formula for the roots of a quadratic equation is
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But there is a variant:
$$\frac{2c}{-b\pm\sqrt{b^2-4ac}}.$$
