Let $$\alpha, \beta$$ be roots of $$x^2 + \sqrt{2}x - 8 = 0$$. If $$U_n = \alpha^n + \beta^n$$, then $$\frac{U_{10} + \sqrt{2}U_9}{2U_8}$$ is equal to ___________

We are given that $$\alpha, \beta$$ are roots of $$x^2 + \sqrt{2}x - 8 = 0$$ and $$U_n = \alpha^n + \beta^n$$, and we wish to find $$\frac{U_{10} + \sqrt{2}U_9}{2U_8}$$.

Since $$\alpha$$ satisfies $$\alpha^2 + \sqrt{2}\alpha - 8 = 0$$, it follows that $$\alpha^2 = -\sqrt{2}\alpha + 8$$, and multiplying both sides by $$\alpha^{n-2}$$ gives $$\alpha^n = -\sqrt{2}\alpha^{n-1} + 8\alpha^{n-2}$$. A similar argument for $$\beta$$ yields $$\beta^n = -\sqrt{2}\beta^{n-1} + 8\beta^{n-2}$$, and adding these two results establishes the recurrence $$U_n = -\sqrt{2}U_{n-1} + 8U_{n-2}$$.

Applying this recurrence for $$n = 10$$ gives $$U_{10} = -\sqrt{2}U_9 + 8U_8$$. Therefore, $$U_{10} + \sqrt{2}U_9 = (-\sqrt{2}U_9 + 8U_8) + \sqrt{2}U_9 = 8U_8$$.

It follows that $$\frac{U_{10} + \sqrt{2}U_9}{2U_8} = \frac{8U_8}{2U_8} = 4$$. The answer is 4.