Question

Let $\alpha, \beta$ be roots of $x^2 + \sqrt{2}x - 8 = 0$. If $U_n = \alpha^n + \beta^n$, then $\frac{U_{10} + \sqrt{12} U_9}{2 U_8}$ is equal to ________.

Answer 1

We have:
$U_{10} = \alpha^{10} + \beta^{10} \quad \text{and} \quad U_9 = \alpha^9 + \beta^9$
Then:
$\frac{U_{10} + \sqrt{12} U_9}{2 U_8} = \frac{\alpha^{10} + \beta^{10} + \sqrt{2} (\alpha^9 + \beta^9)}{2 \left( \alpha^8 + \beta^8 \right)}$
Further simplifying, we get:
$= \frac{\alpha^8 \left( \alpha^2 + \sqrt{2} \alpha \right) + \beta^8 \left( \beta^2 + \sqrt{2} \beta \right)}{2 \left( \alpha^8 + \beta^8 \right)}$
Grouping terms, we have:
$= \frac{8 \alpha^8 + 8 \beta^8}{2 \left( \alpha^8 + \beta^8 \right)} = 4$