Question

If for some $p , q , r \in R$, not all have same sign, one of the roots of the equation $\left(p^2+q^2\right) x^2-2 q(p+r) x+q^2+r^2=0$ is also $a$ root of the equation $x^2+2 x-8=0$, then $\frac{ q ^2+ r ^2}{ p ^2}$ is equal to-

Answer 1

$ ( px - q )^2+( qx - r )^2=0 $
$ \Rightarrow x =\frac{ q }{ p }=\frac{ r }{ q }=-4$
$ \Rightarrow \frac{ q ^2+ r ^2}{ p ^2}=272$