Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Linear Recurrence Relations: If $\alpha$ is a root of a polynomial equation, then powers of $\alpha$ satisfy a linear recurrence relation derived from the polynomial equation.

Step 1: Applying Vieta's Formulas

We are given the quadratic equation $x^2 - 6x - 2 = 0$. Let its roots be $\alpha$ and $\beta$. We use Vieta's formulas to find the sum and product of the roots.

• Why: Vieta's formulas provide the fundamental relationship between the roots and the coefficients of the quadratic equation. These relationships are used to build the recurrence relation.

Sum of roots: $\alpha + \beta = -(-6)/1 = 6$. Product of roots: $\alpha \beta = -2/1 = -2$.

So, we have: $\alpha + \beta = 6$ $\alpha \beta = -2$

Step 2: Deriving the Linear Recurrence Relation

Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$, they satisfy the equation. We use this to derive a recurrence relation for $a_n = \alpha^n - \beta^n$.

• Why: Since $\alpha$ and $\beta$ satisfy the quadratic equation, we can derive a recurrence relation for $a_n$ that allows us to express higher-order terms in terms of lower-order terms.

Since $\alpha$ is a root: $\alpha^2 - 6\alpha - 2 = 0$ Multiply by $\alpha^{n-2}$ (for $n \ge 2$): $\alpha^n - 6\alpha^{n-1} - 2\alpha^{n-2} = 0$ $\alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2}$ Similarly, for the root $\beta$: $\beta^2 - 6\beta - 2 = 0$ $\beta^n = 6\beta^{n-1} + 2\beta^{n-2}$ Now, we use the definition $a_n = \alpha^n - \beta^n$: $a_n = \alpha^n - \beta^n = (6\alpha^{n-1} + 2\alpha^{n-2}) - (6\beta^{n-1} + 2\beta^{n-2})$ $a_n = 6(\alpha^{n-1} - \beta^{n-1}) + 2(\alpha^{n-2} - \beta^{n-2})$ $a_n = 6a_{n-1} + 2a_{n-2} \quad \text{for } n \ge 2$

Step 3: Simplifying the Expression

We need to find the value of $\frac{a_{10} - 2a_8}{3a_9}$. We simplify the numerator using the recurrence relation.

• Why: The recurrence relation allows us to rewrite $a_{10}$ in terms of $a_9$ and $a_8$, which simplifies the expression.

From the recurrence relation, with $n = 10$: $a_{10} = 6a_9 + 2a_8$ $a_{10} - 2a_8 = 6a_9$

Step 4: Final Calculation

Substitute the simplified numerator into the expression:

• Why: Substituting the simplified numerator allows for direct cancellation and easy calculation of the final result.

$\frac{a_{10} - 2a_8}{3a_9} = \frac{6a_9}{3a_9}$ Since $\alpha$ and $\beta$ are distinct and non-zero, $a_9 = \alpha^9 - \beta^9 \neq 0$. Therefore, we can cancel $a_9$. $\frac{6a_9}{3a_9} = \frac{6}{3} = 2$

Common Mistakes & Tips

• Coefficient Errors: Pay close attention to the signs and coefficients when applying Vieta's formulas and deriving the recurrence relation.
• Recurrence Relation Application: Ensure the recurrence relation is applied correctly, especially when substituting values of n.

Summary

By using Vieta's formulas and deriving a linear recurrence relation, we simplified the given expression. The recurrence relation allowed us to rewrite $a_{10}$ in terms of $a_9$ and $a_8$, which led to a simple cancellation and a final answer of 2.

The final answer is $\boxed{2}$, which corresponds to option (B).