Let , be roots of x^2+√2x-8=0.

Question

Let $\alpha ,\beta $ be roots of ${x}^{2}+\sqrt{2}x-8=0$. If ${U}_{n}={\alpha }^{n}+{\beta }^{n}$, then $\frac{{U}_{10}+\sqrt{2}{U}_{9}}{2{U}_{8}}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 4

Accepted Answer

$\Rightarrow {\alpha }^{2}+\sqrt{2}\alpha =8$

$\begin{gathered} \alpha +\beta =\sqrt{2},\, \alpha \beta =-8, \Rightarrow \alpha +\sqrt{2}=\frac{8}{\alpha } \ \Rightarrow \beta +\sqrt{2}=\frac{8}{\beta } \end{gathered}$

$\begin{gathered} \frac{{U}_{10}+\sqrt{2}{U}_{9}}{2{U}_{8}}=\frac{{\alpha }^{10}+{\beta }^{10}+\sqrt{2}{\alpha }^{9}+\sqrt{2}{\beta }^{9}}{2{\alpha }^{8}+2{\beta }^{8}} \ =\frac{{\alpha }^{9}(\alpha +\sqrt{2})+{\beta }^{9}(\beta +\sqrt{2})}{2({\alpha }^{8}+{\beta }^{8})} \ =\frac{{\alpha }^{9}\cdot (\frac{8}{\alpha })+{\beta }^{9}(\frac{8}{\beta })}{2({\alpha }^{8}+{\beta }^{8})}=\frac{8}{2}=4 \end{gathered}$

Solutions · 1