<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] We need $a\ne0$. The equation can be written as $(x-p)(x-q)=a^2(2x-p-q)$, or $x^2-(2a^2+p+q)x+a^2(p+q)+pq=0$.
The discriminant is 
$(2a^2+p+q)^2-4a^2(p+q)-4pq=(p-q)^2+4a^4>0$
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