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(C) Given the quadratic equation $x^{2}-6x-2=0$. Since $\alpha$ and $\beta$ are roots, they satisfy the equation: $x^{n}-6x^{n-1}-2x^{n-2}=0$ $\Righta

Vedclass · Accepted Answer

$3$

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(C) Given the quadratic equation $x^{2}-6x-2=0$. Since $\alpha$ and $\beta$ are roots, they satisfy the equation: $x^{n}-6x^{n-1}-2x^{n-2}=0$ $\Rightarrow x^{n}-2x^{n-2}=6x^{n-1}$ For $x = \alpha$ and $x = \beta$, we have: $\alpha^{n}-2\alpha^{n-2}=6\alpha^{n-1}$ $\beta^{n}-2\beta^{n-2}=6\beta^{n-1}$ Subtracting the two equations: $(\alpha^{n}-\beta^{n})-2(\alpha^{n-2}-\beta^{n-2})=6(\alpha^{n-1}-\beta^{n-1})$ Using the definition $a_{n}=\alpha^{n}-\beta^{n}$, we get: $a_{n}-2a_{n-2}=6a_{n-1}$ For $n=10$: $a_{10}-2a_{8}=6a_{9}$ Therefore, the value of $\frac{a_{10}-2a_{8}}{2a_{9}} = \frac{6a_{9}}{2a_{9}} = 3$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

Let $\alpha, \beta$ be the roots of the equation $x^{2}-6x-2=0$ with $\alpha>\beta$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a_{10}-2a_{8}}{2a_{9}}$ is

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