Journey with alpha and beta

Since $\alpha$ and $\beta$ are roots of $x^2 - 6x-2=0$ or $x^2 = 6x +2$ . Then, $\alpha^2 = 6\alpha +2$ and $\beta^2 = 6\beta +2$ . Therefore,

$\begin{aligned} \frac{a_{10}-2a_8}{2a_9} & = \frac{\alpha^{10}-\beta^{10}-2(\alpha^8-\beta^8)}{2a_9} \\ & = \frac{\alpha^8(\alpha^2-2)-\beta^8(\beta^2-2)}{2a_9} \\ & = \frac{\alpha^8(6\alpha+2-2)-\beta^8(6\beta+2-2)}{2a_9} \\ & = \frac{6\alpha^9-6\beta^9}{2a_9} = \frac{6a_9}{2a_9} = \boxed{3} \end{aligned}$

We have $a_{n+2}=6a_{n+1}+2a_n$ . For $n=8$ this gives $\frac{a_{10}-2a_8}{2a_9}=\boxed{3}$ .

Let me spell it out step by step.

We have $\alpha^2=6\alpha+2$ since $\alpha$ is a root. Multiplying with $\alpha^n$ we see that $\alpha^{n+2}=6\alpha^{n+1}+2\alpha^n$ (Equation I) Likewise, $\beta^{n+2}=6\beta^{n+1}+2\beta^n$ (Equation II). Subtracting (I)-(II) gives $a_{n+2}=6a_{n+1}+2a_n$ .

From the equation, product of roots is $-2$ $\frac{a_{10} - 2(a_8)}{2(a_9)} = \frac{(\alpha ^{10} - \beta ^{10}) - (-\alpha\beta)(\alpha ^8- \beta ^8)}{2 \times (\alpha ^9 - \beta ^9)}$ Opening the bracket and taking common terms, $\frac{\alpha ^9(\alpha+\beta) - \beta^9(\alpha + \beta) }{2 \times (\alpha ^9 - \beta ^9)}$ $\frac{\alpha + \beta}{2}$ From the equation, sum of roots is $6$ . Therefore the answer is $3$

Let $a_n={\alpha}^n+{\beta}^n$ . From the equation we get that $\alpha+\beta=a_1=6,{\alpha}{\beta}=-2$ using vieta's formula.

$6 a_{n-1}=(\alpha+\beta) ({\alpha}^{n-1}+{\beta}^{n-1})={\alpha}^n+{\beta}^n+ {\alpha}{\beta}({\alpha}^{n-2}+{\beta}^{n-2}))=a_n-2a_{n-2}$

So $6 a_{n-1}=a_n-2 a_{n-2} \Rightarrow 6 a_9=a_{10}-2a_8$

$\frac{a_{10}-2 a_{8}}{2 a_9}=\frac{6 a_9}{2 a_9}=3$