If alpha, beta are the roots of the equation | vedclass

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(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.Multiplying by $\alpha^8$,w

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$3$

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(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.Multiplying by $\alpha^8$,we get $\alpha^{10} = 6\alpha^9 + 2\alpha^8$,which implies $\alpha^{10} - 2\alpha^8 = 6\alpha^9$.Similarly,for $\beta$,we have $\beta^{10} - 2\beta^8 = 6\beta^9$.Subtracting the two equations:$(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = 6(\alpha^9 - \beta^9)$.By definition,$a_n = \alpha^n - \beta^n$,so $a_{10} - 2a_8 = 6a_9$.Therefore,$\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$.

If $\alpha, \beta$ are the roots of the equation $x^2 - 6x - 2 = 0$,$\alpha > \beta$ and $a_n = \alpha^n - \beta^n$,$n > 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to

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