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(D) Given the quadratic equation $x^2-10x-8=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^2 - 10\alpha - 8 = 0 \implies

Vedclass · Accepted Answer

$2$

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(D) Given the quadratic equation $x^2-10x-8=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^2 - 10\alpha - 8 = 0 \implies \alpha^2 - 8 = 10\alpha$ $\beta^2 - 10\beta - 8 = 0 \implies \beta^2 - 8 = 10\beta$ We need to evaluate $\frac{a_{10}-8a_8}{5a_9}$. Substitute $a_n = \alpha^n - \beta^n$: $\frac{(\alpha^{10}-\beta^{10}) - 8(\alpha^8-\beta^8)}{5(\alpha^9-\beta^9)}$ $= \frac{(\alpha^{10}-8\alpha^8) - (\beta^{10}-8\beta^8)}{5(\alpha^9-\beta^9)}$ $= \frac{\alpha^8(\alpha^2-8) - \beta^8(\beta^2-8)}{5(\alpha^9-\beta^9)}$ Using the relations $\alpha^2-8=10\alpha$ and $\beta^2-8=10\beta$: $= \frac{\alpha^8(10\alpha) - \beta^8(10\beta)}{5(\alpha^9-\beta^9)}$ $= \frac{10\alpha^9 - 10\beta^9}{5(\alpha^9-\beta^9)}$ $= \frac{10(\alpha^9-\beta^9)}{5(\alpha^9-\beta^9)} = 2$.

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

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