JEE-Mains 2015 (3/30)

From the equation, product of roots is $-2$ $\frac{a_{10} - 2(a_8)}{2(a_9)} = \frac{(\alpha ^{10} - \beta ^{10}) - (-\alpha\beta)(\alpha ^8- \beta ^8)}{2 \times (\alpha ^9 - \beta ^9)}$ Opening the bracket and taking common terms, $\frac{\alpha ^9(\alpha+\beta) - \beta^9(\alpha + \beta) }{2 \times (\alpha ^9 - \beta ^9)}$ $\frac{\alpha + \beta}{2}$ From the equation, sum of roots is $6$ . Therefore the answer is $3$

$\alpha^2 - 6\alpha - 2 =0$

Multiplying by $\alpha^{8}$

$\alpha^{10} - 6\alpha^{9} -2\alpha^{8} = 0$

Similarly

$\beta^{10} - 6\beta^{9} -2 \beta^{2} = 0$

Adding,

$\alpha^{10} + \beta^{10} - 2\alpha^{8} - \beta^{8} = 6( \alpha^{9} + \beta^{9})$

$\dfrac{a_{10}-2a_8}{2a_9} = 3$