Let alpha and beta be the roots of x^2-6x-2=0 | vedclass

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(C) Since $\alpha$ and $\beta$ are the roots of the equation $x^2-6x-2=0$,they satisfy the equation:$\alpha^2-6\alpha-2=0 \implies \alpha^2 = 6\alpha

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Since $\alpha$ and $\beta$ are the roots of the equation $x^2-6x-2=0$,they satisfy the equation:$\alpha^2-6\alpha-2=0 \implies \alpha^2 = 6\alpha + 2$$\beta^2-6\beta-2=0 \implies \beta^2 = 6\beta + 2$Given $a_n = \alpha^n - \beta^n$,we have $a_{10} = \alpha^{10} - \beta^{10}$ and $a_8 = \alpha^8 - \beta^8$.Multiplying the root equations by $\alpha^8$ and $\beta^8$ respectively:$\alpha^{10} = 6\alpha^9 + 2\alpha^8$$\beta^{10} = 6\beta^9 + 2\beta^8$Subtracting these equations:$a_{10} = \alpha^{10} - \beta^{10} = 6(\alpha^9 - \beta^9) + 2(\alpha^8 - \beta^8)$$a_{10} = 6a_9 + 2a_8$Rearranging the terms:$a_{10} - 2a_8 = 6a_9$Dividing by $2a_9$:$\frac{a_{10}-2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$,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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