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.

Question

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?

Answer 1

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

$(2a^2+p+q)^2-4a^2(p+q)-4pq=(p-q)^2+4a^4>0$

Answer 2

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 $$\color{red}{D=(2a^2+p+q)^2-4(a^2p+a^2q+pq)=4a^4+p^2+q^2-2pq>0}$$

Answer 3

Define the function $f(x)=\frac{1}{x-p}+\frac{1}{x-q}$. $f$ is clearly decreasing function since the derivative is negative.Wlog we suppose $p\le q$ and keep in mind $a^2>0$, we use ITV.

$f(]-\infty;p[)=(-\infty,0) $therefore no solution. $f(p,q)=(-\infty, \infty)$, here we get one solution $f(q,\infty)=(0,\infty)$, here we get one solution

Since your equation is quadratic, it has at most two real roots, which end the proof.