<s>[INST] Prove that for $a,p,q \in \Bbb R$ the solutions of: $\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$ are real numbers. Prove that for $a,p,q \in \Bbb R$ the solutions of: $$\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$$
are real numbers.
I tried manipulating the expression, getting rid of the denominators, but i can't factor $x$.
Any hints?
 [/INST] Observe that your equation is equivalent to 
$$ a^2(x-q)+a^2(x-p)=(x-p)(x-q)$$ $$\iff a^2x-a^2q+a^2x-a^2p-x^2+(p+q)x-pq=0$$ $$\iff-x^2+(2a^2+p+q)x-(a^2q+a^2p+pq)=0$$
Which is a quadratic equation with discriminant $$\color{red}{D=(2a^2+p+q)^2-4(a^2p+a^2q+pq)=4a^4+p^2+q^2-2pq>0}$$
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