Let α and β be the roots of equation x ^ 2 - 6x - 2 = 0 . If a _ n...

Given, $\alpha, \beta$ are roots of ${x}^{2} - 6x - 2 = 0$.

$⇒ {\alpha}^{2} - 6\alpha - 2 = 0$ and ${\beta}^{2} - 6\beta - 2 = 0$

$⇒ {\alpha}^{2} - 2 = 6\alpha$ and ${\beta}^{2} - 2 = 6\beta . . . \left(1\right)$

$\dfrac{{a}_{10} - 2{a}_{8}}{2{a}_{9}} = \dfrac{\left({\alpha}^{10} - {\beta}^{10}\right) - 2\left({\alpha}^{8} - {\beta}^{8}\right)}{2\left({\alpha}^{9} - {\beta}^{9}\right)}$

$= \dfrac{\left({\alpha}^{10} - 2{\alpha}^{8}\right) - \left({\beta}^{10} - 2{\beta}^{8}\right)}{2\left({\alpha}^{9} - {\beta}^{9}\right)}$

= \dfrac{{\alpha}^{8} \left({\alpha}^{2} - 2\right) - {\beta}^{8} \left({\beta}^{2} - 2\right)}{2 \left({\alpha}^{9} - {\beta}^{9}\right)}

$= \dfrac{{\alpha}^{8} \left(6 \alpha\right) - {\beta}^{8} \left(6 \beta\right)}{2 \left({\alpha}^{9} - {\beta}^{9}\right)}$ [$∵$ From (1)]

$= \dfrac{6{\alpha}^{9} - 6{\beta}^{9}}{2\left({\alpha}^{9} - {\beta}^{9}\right)} = 3$