Mathematics - Quadratic Equation Question with Solution | TestHub

Solution:

The roots $\alpha$ and $\beta$ satisfy the quadratic equation $x^2-12x-2=0$.

This means:

${\alpha }^2-12\alpha -2=0$ (1)

${\beta }^2-12\beta -2=0$ (2)

From equation (1), we can write ${\alpha }^2=12\alpha +2$.

From equation (2), we can write ${\beta }^2=12\beta +2$.

To find a recurrence relation for $a_n={\alpha }^n-{\beta }^n$, we multiply equation (1) by ${\alpha }^{n-2}$ and equation (2) by ${\beta }^{n-2}$ (for $n\ge 2$):

${\alpha }^n=12{\alpha }^{n-1}+2{\alpha }^{n-2}$ (3)

${\beta }^n=12{\beta }^{n-1}+2{\beta }^{n-2}$ (4)

Subtracting equation (4) from equation (3):

${\alpha }^n-{\beta }^n=(12{\alpha }^{n-1}+2{\alpha }^{n-2})-(12{\beta }^{n-1}+2{\beta }^{n-2})$

${\alpha }^n-{\beta }^n=12({\alpha }^{n-1}-{\beta }^{n-1})+2({\alpha }^{n-2}-{\beta }^{n-2})$

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

$a_n=12a_{n-1}+2a_{n-2}$

Now, we need to evaluate the expression $\frac{a_{10}-2a_8}{3a_9}$.

Let's use the recurrence relation for $n=10$:

$a_{10}=12a_9+2a_8$

Rearrange this equation to match the numerator of the expression:

$a_{10}-2a_8=12a_9$

Now substitute this into the given expression:

$$\frac{a_{10}-2a_8}{3a_9}=\frac{12a_9}{3a_9}$$

Assuming $a_9\ne 0$ (which it is, as $\alpha >\beta$ and the roots are real and distinct), we can simplify the expression:

$$\frac{12a_9}{3a_9}=\frac{12}{3}=4$$

The value of the expression is 4.

The final answer is 4.