# If alpha,beta are roots of x^2 -6x-2=0 and alpha>beta , if a_n =(alpha)^n - (beta)^n , n geq 1 then (a_10 - 2a_8)/(2a_9)?

Mar 7, 2018

$3$

#### Explanation:

$\alpha$ and $\beta$ are roots of ${x}^{2} - 6 x - 2$

So, $\implies {\alpha}^{2} - 6 \alpha - 2 = 0$

$\implies {\alpha}^{2} = 6 \alpha + 2$

Multiply ${\alpha}^{8}$ both sides.

=>alpha^2×color (blue)(alpha^8)=color (blue)(alpha^8) (6alpha+2)

$\implies {\alpha}^{10} = 6 {\alpha}^{9} + 2 {\alpha}^{8} \text{ equation} 1$

Similarly
$\implies {\beta}^{2} = 6 \beta + 2$

$\implies {\beta}^{10} = 6 {\beta}^{9} + 2 {\beta}^{8} \text{ " "equation} 2$

We have
=>a_n=alpha^n-beta^n" "("where " alpha>beta)

we have to find
$\implies \frac{{a}_{10} - 2 {a}_{8}}{a} _ 9$

Or
$\implies \frac{{\alpha}^{10} - {\beta}^{10} - 2 \left({\alpha}^{8} - {\beta}^{8}\right)}{2 {a}_{9}}$

From equation $1 \text{ and } 2$

$\implies \frac{6 {\alpha}^{9} + \cancel{2 {\alpha}^{8}} - 6 {\beta}^{9} - \cancel{2 {\beta}^{8}} - \cancel{2 {\alpha}^{8}} + \cancel{2 {\beta}^{8}}}{2 {a}_{9}}$

$\implies \frac{6 \left({\alpha}^{9} - {\beta}^{9}\right)}{2 {a}_{9}}$

$\implies \frac{6 {a}_{9}}{2 {a}_{9}}$

$\implies 3$