$\alpha, \beta$ are the roots of $x^2-10 x-8=0$ with $\alpha\beta$. If $\alpha_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}$ for $\mathrm{n} \in \mathbb{N}$, then the value of $\frac{a_{10}-8 a_8}{5 a_9}$ is

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$\alpha, \beta$ are the roots of $x^2-10 x-8=0$ with $\alpha>\beta$. If $\alpha_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}$ for $\mathrm{n} \in \mathbb{N}$, then the value of $\frac{a_{10}-8 a_8}{5 a_9}$ is

Aicharya · Accepted Answer

$x^2-10 x-8=0$ $\begin{aligned} & x^2-8=10 x \ & \frac{x^2-8}{5 x}=2\end{aligned}$ $\alpha, \beta$ are roots $\frac{\alpha^2-8}{5 \alpha}=2 \Rightarrow \alpha^2-8=10 \alpha$...(i) $\frac{\beta^2-8}{5 \beta}=2 \Rightarrow \beta^2-8=10 \beta$ ...(ii) $\frac{a_{10}-8 a_8}{5 a_9}=\frac{\alpha^{10}-\beta^{10}-8\left(\alpha^8-\beta^8ight)}{5\left(\alpha^9-\beta^9ight)}$ $\begin{aligned} & =\frac{\left(\alpha^{10}-8 \alpha^8ight)-\left(\beta^{10}-8 \beta^8ight)}{5\left(\alpha^9-\beta^9ight)} \ & =\frac{\alpha^8\left(\alpha^2-8ight)-\beta^8\left(\beta^2-8ight)}{5\left(\alpha^9-\beta^9ight)}\end{aligned}$ Put, $\alpha^2-8=10 \alpha$ and $\beta^2-8=10 \beta$ From (i) and (ii) $\begin{aligned} & =\frac{\alpha^8 \cdot 10 \alpha-\beta^8 \cdot 10 \beta}{5\left(\alpha^9-\beta^9ight)} \ & =\frac{10\left(\alpha^9-\beta^9ight)}{5\left(\alpha^9-\beta^9ight)}=2\end{aligned}$

$\alpha, \beta$ are the roots of $x^2-10 x-8=0$ with $\alpha>\beta$. If…

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