# 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.

• I think you have missed a factor of $x$ out, should it be $(\sigma_A^2 - 2\rho\sigma_A\sigma_B +\sigma_B^2)x^2 +2(\rho\sigma_A\sigma_B - \sigma_B^2) {\bf x} +\sigma_B^2 = 0$? As well, could you give some background to the equations, are the $\sigma 's$ standard deviations of some experiment and $\rho$ some correlation? – jim Mar 25 '16 at 11:02
• If $\rho$ is a correlation then $| \rho | \le 1$. But for $| \rho | < 1$ you have no (real) solution for $x$, so you must have $| \rho | = 1$. – jim Mar 25 '16 at 11:16
• Sorry, yes i missed a factor of $x$ out, the $\sigma's$ are standard deviations and $\rho$ is a correlation so $0\le\rho\le1$. – Kai_M Mar 25 '16 at 11:43
• The idea behind the question is that we have a portfolio of two risky assets A and B with variance of returns equal to $\sigma_A^2$ and $\sigma_B^2$ and an amount $x$ invested in asset A and $1-x$ in asset B. The portfolio return is thus $w=(1+r_A)x+(1+r_B)(1-x)$ and the variance of the portfolio return is $V[w]= x^2\sigma_A^2 +2x(1-x)\rho\sigma_A\sigma_B +(1-x)^2\sigma_B^2$. The portfolio is riskless if $V[w] =0$, which is the equation we are trying to solve. – Kai_M Mar 25 '16 at 11:56

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 $\rho = \pm 1$.

• Thanks for your answer. What you and Jim suggested was indeed correct. – Kai_M Mar 25 '16 at 11:58

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:

$$\frac{2c}{-b\pm\sqrt{b^2-4ac}}.$$