Let `alpha` and `beta` are roots of `x^(2) - 17x - 6 = 0` with `alpha gt beta, if a_(n) = alpha^(n+2)+beta^(n+2)` for `n ge 5` then the value of `(a_(10)-6a_(8)-a_(9))/(a_(9))`

Options

A. `6`
B. `16`
C. `12`
D. `8`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - 17x - 6 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -17, c = -6 \).

### Step 2: Calculate the discriminant
Calculate the discriminant:
\[ D = b^2 - 4ac = (-17)^2 - 4 \cdot 1 \cdot (-6) = 289 + 24 = 313 \]

### Step 3: Find the roots
Now substituting the values into the quadratic formula:
\[ x = \frac{17 \pm \sqrt{313}}{2} \]
Let:
\[ \alpha = \frac{17 + \sqrt{313}}{2}, \quad \beta = \frac{17 - \sqrt{313}}{2} \]
where \( \alpha > \beta \).

### Step 4: Define the sequence \( a_n \)
The sequence is defined as:
\[ a_n = \alpha^{n+2} + \beta^{n+2} \]
We need to find \( \frac{a_{10} - 6a_{8} - a_{9}}{a_{9}} \).

### Step 5: Express \( a_{10}, a_{9}, a_{8} \) using the recurrence relation
From the quadratic equation, we can derive a recurrence relation:
\[ a_n = 17a_{n-1} + 6a_{n-2} \]
Using this, we can calculate:
- \( a_{10} = 17a_{9} + 6a_{8} \)
- \( a_{9} = 17a_{8} + 6a_{7} \)

### Step 6: Substitute into the expression
Now substituting \( a_{10} \) into the expression:
\[ a_{10} - 6a_{8} - a_{9} = (17a_{9} + 6a_{8}) - 6a_{8} - a_{9} \]
This simplifies to:
\[ 16a_{9} \]

### Step 7: Final calculation
Now we can substitute back into the expression:
\[ \frac{a_{10} - 6a_{8} - a_{9}}{a_{9}} = \frac{16a_{9}}{a_{9}} = 16 \]

Thus, the final answer is:
\[ \boxed{16} \]