Mathematics - Quadratic Equation Question with Solution | TestHub

The roots $\alpha ,\beta$ satisfy the equation $x^2-6x-2=0$.

This implies ${\alpha }^2-6\alpha -2=0$ and ${\beta }^2-6\beta -2=0$.

From these, we can write ${\alpha }^2=6\alpha +2$ and ${\beta }^2=6\beta +2$.

Multiply the first equation by ${\alpha }^{n-2}$ and the second by ${\beta }^{n-2}$:

${\alpha }^n=6{\alpha }^{n-1}+2{\alpha }^{n-2}$

${\beta }^n=6{\beta }^{n-1}+2{\beta }^{n-2}$

Subtracting the second equation from the first:

${\alpha }^n-{\beta }^n=6({\alpha }^{n-1}-{\beta }^{n-1})+2({\alpha }^{n-2}-{\beta }^{n-2})$

Given $a_n={\alpha }^n-{\beta }^n$, we can write the recurrence relation:

$a_n=6a_{n-1}+2a_{n-2}$

Now, substitute $n=10$ into the recurrence relation:

$a_{10}=6a_9+2a_8$

Rearrange the equation to find the required expression:

$a_{10}-2a_8=6a_9$

Finally, substitute this into the expression we need to evaluate:

$$\frac{a_{10}-2a_8}{3a_9}=\frac{6a_9}{3a_9}$$

Assuming $a_9\ne 0$, we can simplify the expression:

$$\frac{6a_9}{3a_9}=2$$

The final answer is 2.

The final answer is B.