If the value of real number $$\alpha \gt 0$$ for which $$x^2 - 5\alpha x + 1 = 0$$ and $$x^2 - \alpha x - 5 = 0$$ have a common real roots is $$\frac{3}{\sqrt{2\beta}}$$ then $$\beta$$ is equal to ______.

The two quadratics $$x^2 - 5\alpha x + 1 = 0$$ and $$x^2 - \alpha x - 5 = 0$$ are assumed to share a common real root $$p$$, and for $$\alpha\gt 0$$ this root is given in the form $$\frac{3}{\sqrt{2\beta}}\,. $$ To determine $$\beta$$, note that $$p$$ must satisfy $$p^2 - 5\alpha p + 1 = 0 \quad(1)$$ and $$p^2 - \alpha p - 5 = 0 \quad(2)\,. $$

Subtracting equation (2) from equation (1) gives $$(p^2 - 5\alpha p + 1) - (p^2 - \alpha p - 5) = 0\,, $$ so $$-4\alpha p + 6 = 0\quad\Longrightarrow\quad p = \frac{6}{4\alpha} = \frac{3}{2\alpha}\,. $$

Substituting $$p = \tfrac{3}{2\alpha}$$ back into equation (2) yields $$\Bigl(\frac{3}{2\alpha}\Bigr)^2 - \alpha\cdot\frac{3}{2\alpha} - 5 = 0\,, $$ which simplifies to $$\frac{9}{4\alpha^2} - \frac{3}{2} - 5 = 0\quad\Longrightarrow\quad \frac{9}{4\alpha^2} = \frac{13}{2}\,. $$ Hence $$\alpha^2 = \frac{9}{4\cdot\frac{13}{2}} = \frac{9}{26}\,, $$ and since $$\alpha\gt 0$$, $$\alpha = \frac{3}{\sqrt{26}}\,. $$

Writing this in the form $$\frac{3}{\sqrt{2\beta}}$$ gives $$\alpha = \frac{3}{\sqrt{26}} = \frac{3}{\sqrt{2\cdot 13}} = \frac{3}{\sqrt{2\beta}}\,, $$ therefore $$\therefore \beta = 13\,. $$