If for some $$p, q, r \in \mathbb{R}$$, all have positive 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 the equation $$x^2 + 2x - 8 = 0$$, then $$\dfrac{q^2 + r^2}{p^2}$$ is equal to ______.

Given that $$(p^2 + q^2)x^2 - 2q(p + r)x + q^2 + r^2 = 0$$ and $$x^2 + 2x - 8 = 0$$ share a common root, with $$p, q, r \in \mathbb{R}$$ all positive.

First, analyse the discriminant: $$\Delta = 4q^2(p + r)^2 - 4(p^2 + q^2)(q^2 + r^2)$$. Expanding: $$\dfrac{\Delta}{4} = 2pq^2r - p^2r^2 - q^4 = -(q^2 - pr)^2 \le 0$$. For real roots: $$\Delta = 0$$, giving $$q^2 = pr$$.

Next, the repeated root is $$x = \dfrac{q(p + r)}{p^2 + q^2} = \dfrac{q(p + r)}{p(p + r)} = \dfrac{q}{p}$$.

Next, solve $$x^2 + 2x - 8 = 0$$. This gives $$(x + 4)(x - 2) = 0 \implies x = 2 \text{ or } x = -4$$.

Since $$p, q, r > 0$$, the root $$x = q/p > 0$$. Therefore $$x = 2$$ (rejecting $$x = -4$$). From this, $$\dfrac{q}{p} = 2 \implies q = 2p, \quad r = \dfrac{q^2}{p} = 4p$$.

Next, compute $$\dfrac{q^2 + r^2}{p^2} = \dfrac{4p^2 + 16p^2}{p^2} = 20$$.

The answer is $$20$$.