If for some p,q,r R, not all having the same sign, one of the roots of the equation (p^2+q…

Question

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

Answer: 272

Accepted Answer

The roots of $x^2+2x-8=0$ are $x=2,-4$. Testing the common root: since $p,q,r$ not all same sign, we need $\alpha+\beta=\frac{2q(p+r)}{p^2+q^2}>0$ and product $>0$, pointing to the common root being $4$ (with sign considerations from the source).

Substituting $x=4$:
$$16(p^2+q^2)-8q(p+r)+q^2+r^2=0 \implies (16p^2-8pq+q^2)+(16q^2-8qr+r^2)=0$$
$$\implies (4p-q)^2+(4q-r)^2=0 \implies q=4p,\ r=16p.$$

$$\frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272.$$

Solutions · 1