Let p,q,r be real numbers (p q, r0), such that the roots of 1x+p+1x+q=1r are equal in magn…

Question

Let $p,q,r$ be real numbers ($p
e q$, $r
e0$), such that the roots of $\dfrac1{x+p}+\dfrac1{x+q}=\dfrac1r$ are equal in magnitude but opposite in sign. Then the sum of squares of these roots equals

Accepted Answer

The equation reduces to $x^2+(p+q-2r)x+(pq-pr-qr)=0$. With roots $\alpha,-\alpha$: sum $=0 \implies p+q=2r$.

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=0-2(pq-pr-qr)\cdots$$

Working through with $p+q=2r$ substituted, this simplifies to $p^2+q^2$.

Solutions · 1