Let `alpha` and `beta` be the toots of `x^2-6x-2=0`, with `alphagtbeta` . If `a_n=alpha^n-beta^n` for `nge1`, then the value of the following rational function is: `f=(a_(10)-2a_8)/(2a_9)`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will first find the roots of the equation and then use the defined sequence to compute the required value of the rational function.

### Step 1: Find the roots of the equation
The given equation is:
\[ x^2 - 6x - 2 = 0 \]

We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -6, c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot (-2) = 36 + 8 = 44 \]

Now substituting back into the quadratic formula:
\[ x = \frac{6 \pm \sqrt{44}}{2} = \frac{6 \pm 2\sqrt{11}}{2} = 3 \pm \sqrt{11} \]

Thus, the roots are:
\[ \alpha = 3 + \sqrt{11} \quad \text{and} \quad \beta = 3 - \sqrt{11} \]
with \( \alpha > \beta \).

### Step 2: Define the sequence \( a_n \)
The sequence is defined as:
\[ a_n = \alpha^n - \beta^n \]

### Step 3: Compute \( a_{10}, a_9, \) and \( a_8 \)
Using the definition of the sequence:
- \( a_{10} = \alpha^{10} - \beta^{10} \)
- \( a_9 = \alpha^9 - \beta^9 \)
- \( a_8 = \alpha^8 - \beta^8 \)

### Step 4: Substitute into the function \( f \)
The function we need to evaluate is:
\[ f = \frac{a_{10} - 2a_8}{2a_9} \]

Substituting the expressions for \( a_{10}, a_9, \) and \( a_8 \):
\[ f = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)} \]

### Step 5: Simplify the numerator
Rearranging the numerator:
\[ f = \frac{\alpha^{10} - 2\alpha^8 - \beta^{10} + 2\beta^8}{2(\alpha^9 - \beta^9)} \]

Factoring out common terms:
\[ = \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)} \]

### Step 6: Use the original equation
From the original equation \( x^2 - 6x - 2 = 0 \), we can express \( \alpha^2 \) and \( \beta^2 \):
\[ \alpha^2 = 6\alpha + 2 \quad \text{and} \quad \beta^2 = 6\beta + 2 \]

Thus:
\[ \alpha^2 - 2 = 6\alpha \quad \text{and} \quad \beta^2 - 2 = 6\beta \]

Substituting these into the expression:
\[ f = \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)} \]
\[ = \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)} \]

### Step 7: Final simplification
The terms \( \alpha^9 - \beta^9 \) cancel out:
\[ f = \frac{6}{2} = 3 \]

### Final Answer
Thus, the value of the rational function \( f \) is:
\[ \boxed{3} \]