Let alpha and beta be the roots of x^{2}-6x-2 | vedclass

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(A) Since $\alpha$ and $\beta$ are the roots of $x^{2}-6x-2=0$,we have $\alpha^{2}-6\alpha-2=0$ and $\beta^{2}-6\beta-2=0$.Multiplying the first equat

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$2$

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(A) Since $\alpha$ and $\beta$ are the roots of $x^{2}-6x-2=0$,we have $\alpha^{2}-6\alpha-2=0$ and $\beta^{2}-6\beta-2=0$.Multiplying the first equation by $\alpha^{8}$,we get $\alpha^{10}-6\alpha^{9}-2\alpha^{8}=0$.Similarly,multiplying the second equation by $\beta^{8}$,we get $\beta^{10}-6\beta^{9}-2\beta^{8}=0$.Subtracting the two equations,we obtain $(\alpha^{10}-\beta^{10})-6(\alpha^{9}-\beta^{9})-2(\alpha^{8}-\beta^{8})=0$.Using the definition $a_{n}=\alpha^{n}-\beta^{n}$,this simplifies to $a_{10}-6a_{9}-2a_{8}=0$.Rearranging the terms,we get $a_{10}-2a_{8}=6a_{9}$.Therefore,$\frac{a_{10}-2a_{8}}{3a_{9}} = \frac{6a_{9}}{3a_{9}} = 2$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}-6x-2=0$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$,then the value of $\frac{a_{10}-2a_{8}}{3a_{9}}$ is ..... .

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