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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
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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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Verified Answer
The correct answer is:
$2$
$x^2-10 x-8=0$
$\begin{aligned} & x^2-8=10 x \\ & \frac{x^2-8}{5 x}=2\end{aligned}$
$\alpha, \beta$ 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)
$\frac{a_{10}-8 a_8}{5 a_9}=\frac{\alpha^{10}-\beta^{10}-8\left(\alpha^8-\beta^8\right)}{5\left(\alpha^9-\beta^9\right)}$
$\begin{aligned} & =\frac{\left(\alpha^{10}-8 \alpha^8\right)-\left(\beta^{10}-8 \beta^8\right)}{5\left(\alpha^9-\beta^9\right)} \\ & =\frac{\alpha^8\left(\alpha^2-8\right)-\beta^8\left(\beta^2-8\right)}{5\left(\alpha^9-\beta^9\right)}\end{aligned}$
Put, $\alpha^2-8=10 \alpha$ and $\beta^2-8=10 \beta$ From (i) and (ii)
$\begin{aligned} & =\frac{\alpha^8 \cdot 10 \alpha-\beta^8 \cdot 10 \beta}{5\left(\alpha^9-\beta^9\right)} \\ & =\frac{10\left(\alpha^9-\beta^9\right)}{5\left(\alpha^9-\beta^9\right)}=2\end{aligned}$
$\begin{aligned} & x^2-8=10 x \\ & \frac{x^2-8}{5 x}=2\end{aligned}$
$\alpha, \beta$ 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)
$\frac{a_{10}-8 a_8}{5 a_9}=\frac{\alpha^{10}-\beta^{10}-8\left(\alpha^8-\beta^8\right)}{5\left(\alpha^9-\beta^9\right)}$
$\begin{aligned} & =\frac{\left(\alpha^{10}-8 \alpha^8\right)-\left(\beta^{10}-8 \beta^8\right)}{5\left(\alpha^9-\beta^9\right)} \\ & =\frac{\alpha^8\left(\alpha^2-8\right)-\beta^8\left(\beta^2-8\right)}{5\left(\alpha^9-\beta^9\right)}\end{aligned}$
Put, $\alpha^2-8=10 \alpha$ and $\beta^2-8=10 \beta$ From (i) and (ii)
$\begin{aligned} & =\frac{\alpha^8 \cdot 10 \alpha-\beta^8 \cdot 10 \beta}{5\left(\alpha^9-\beta^9\right)} \\ & =\frac{10\left(\alpha^9-\beta^9\right)}{5\left(\alpha^9-\beta^9\right)}=2\end{aligned}$
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