If for some p, q, r ∈ R , all have positive sign, one of the roots...

Given, $\left({p}^{2} + {q}^{2}\right){x}^{2} - 2q\left(p + r\right)x + {q}^{2} + {r}^{2} = 0$

On simplifying we get, ${\left(px - q\right)}^{2} + {\left(qx - r\right)}^{2} = 0$

$⇒ px - q = 0 \text{\text{ and } }$ qx - r = 0

$⇒ x = \dfrac{q}{p} = \dfrac{r}{q}$

$⇒ x = \dfrac{q}{p} = \dfrac{r}{q} = 4$ [because roots of equation ${x}^{2} - 2x - 8 = 0$ are $4 or - 2$]

As $p, q, r$ are positive, so $x$ must be $4$.

Now, $q = 4p and r = 4q = 16p$

So, $\dfrac{{q}^{2} + {r}^{2}}{{p}^{2}} = \dfrac{{\left(4p\right)}^{2} + {\left(4 \times 4p\right)}^{2}}{{p}^{2}} = 16 + 256 = 272$.