Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.
• Recurrence Relations: If $\alpha$ and $\beta$ are roots of a polynomial, expressions of the form $a_n = C_1 \alpha^n + C_2 \beta^n$ often satisfy a linear recurrence relation.
• Quadratic Roots Substitution: If $\alpha$ is a root of $ax^2 + bx + c = 0$, then $a\alpha^2 + b\alpha + c = 0$.

Step-by-Step Solution

Step 1: Identify Coefficients and Apply Vieta's Formulas

Given the quadratic equation $x^2 - 6x - 2 = 0$, we identify the coefficients as $a = 1$, $b = -6$, and $c = -2$. Applying Vieta's formulas, we have:

• Sum of the roots: $\alpha + \beta = -\frac{b}{a} = -\frac{-6}{1} = 6$
• Product of the roots: $\alpha \beta = \frac{c}{a} = \frac{-2}{1} = -2$

These values will be used later to simplify the expression.

Step 2: Derive a Recurrence Relation for $a_n$

Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$, they satisfy the equation: $\alpha^2 - 6\alpha - 2 = 0 \quad \cdots (1)$ $\beta^2 - 6\beta - 2 = 0 \quad \cdots (2)$

Multiply equation (1) by $\alpha^{n-2}$ and equation (2) by $\beta^{n-2}$: $\alpha^{n-2}(\alpha^2 - 6\alpha - 2) = 0 \implies \alpha^n - 6\alpha^{n-1} - 2\alpha^{n-2} = 0 \quad \cdots (3)$ $\beta^{n-2}(\beta^2 - 6\beta - 2) = 0 \implies \beta^n - 6\beta^{n-1} - 2\beta^{n-2} = 0 \quad \cdots (4)$

Subtract equation (4) from equation (3) to utilize the definition $a_n = \alpha^n - \beta^n$: $(\alpha^n - \beta^n) - 6(\alpha^{n-1} - \beta^{n-1}) - 2(\alpha^{n-2} - \beta^{n-2}) = 0$

Substitute $a_n$ based on its definition: $a_n - 6a_{n-1} - 2a_{n-2} = 0$

Rearrange to obtain the recurrence relation: $a_n = 6a_{n-1} + 2a_{n-2}$ This relation holds for all $n \ge 3$.

Step 3: Simplify the Numerator of the Expression

We want to find the value of $\frac{a_{10} - 2a_8}{2a_9}$. Let's analyze the numerator: $a_{10} - 2a_8$.

Using the recurrence relation $a_n = 6a_{n-1} + 2a_{n-2}$ with $n = 10$: $a_{10} = 6a_9 + 2a_8$

Now, rearrange this to isolate the term $a_{10} - 2a_8$: $a_{10} - 2a_8 = 6a_9$

Step 4: Evaluate the Final Expression

Substitute the simplified numerator ($a_{10} - 2a_8 = 6a_9$) back into the original expression: $\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9}$

Since $\alpha$ and $\beta$ are distinct and non-zero, $a_n = \alpha^n - \beta^n$ will not be zero for any positive integer $n$. Therefore, we can safely cancel out $a_9$ from the numerator and the denominator: $= \frac{6}{2} = 3$

Common Mistakes & Tips:

• Recurrence Relation is Key: Recognizing and using the recurrence relation simplifies the problem considerably. Avoid trying to directly calculate the powers of roots.
• Vieta's Formulas are Essential: Always start by applying Vieta's formulas to find the sum and product of the roots.
• Careful Algebra: Double-check your algebraic manipulations to avoid sign errors and incorrect substitutions.

Summary

The problem involves simplifying an expression with powers of roots of a quadratic equation. The most efficient approach involves deriving and utilizing a linear recurrence relation. This relation allows us to express $a_{10}$ in terms of $a_9$ and $a_8$, leading to a simplified numerator. The final expression evaluates to 3.

Final Answer

The final answer is $\boxed{3}$, which corresponds to option (A).