Let `alpha` and `beta` be the roots of the equation `x^2-6x-2=0` If `a_n=alpha^n-beta^n` for `ngt=0` then find the value of `(a_10-2a_8)/(2a_9)`

Options

A. 6
B. `-6`
C. `3`
D. `-3`

Answer

To solve the problem, we need to find the value of \((a_{10} - 2a_{8}) / (2a_{9})\) given that \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - 6x - 2 = 0\) and that \(a_n = \alpha^n - \beta^n\) for \(n \geq 0\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)
The roots of the quadratic equation \(x^2 - 6x - 2 = 0\) can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -6\), and \(c = -2\).

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

Now substituting into the quadratic formula:
\[ \alpha, \beta = \frac{6 \pm \sqrt{44}}{2} = \frac{6 \pm 2\sqrt{11}}{2} = 3 \pm \sqrt{11} \]
Thus, \(\alpha = 3 + \sqrt{11}\) and \(\beta = 3 - \sqrt{11}\).

### Step 2: Establish the recurrence relation
From the roots, we can derive the recurrence relation for \(a_n\):
\[ \alpha^2 = 6\alpha + 2 \quad \text{and} \quad \beta^2 = 6\beta + 2 \]
This gives us:
\[ \alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2} \quad \text{and} \quad \beta^n = 6\beta^{n-1} + 2\beta^{n-2} \]
Subtracting these equations, we get:
\[ a_n = 6a_{n-1} + 2a_{n-2} \]

### Step 3: Calculate \(a_{10}\), \(a_{9}\), and \(a_{8}\)
Using the recurrence relation, we can calculate \(a_n\) for \(n = 0, 1, 2, \ldots, 10\).

1. **Base cases**:
- \(a_0 = \alpha^0 - \beta^0 = 1 - 1 = 0\)
- \(a_1 = \alpha^1 - \beta^1 = \alpha - \beta = (3 + \sqrt{11}) - (3 - \sqrt{11}) = 2\sqrt{11}\)

2. **Using the recurrence relation**:
- \(a_2 = 6a_1 + 2a_0 = 6(2\sqrt{11}) + 2(0) = 12\sqrt{11}\)
- \(a_3 = 6a_2 + 2a_1 = 6(12\sqrt{11}) + 2(2\sqrt{11}) = 72\sqrt{11} + 4\sqrt{11} = 76\sqrt{11}\)
- \(a_4 = 6a_3 + 2a_2 = 6(76\sqrt{11}) + 2(12\sqrt{11}) = 456\sqrt{11} + 24\sqrt{11} = 480\sqrt{11}\)
- Continuing this process, we find:
- \(a_5 = 6a_4 + 2a_3 = 2880\sqrt{11}\)
- \(a_6 = 6a_5 + 2a_4 = 17280\sqrt{11}\)
- \(a_7 = 6a_6 + 2a_5 = 103680\sqrt{11}\)
- \(a_8 = 622080\sqrt{11}\)
- \(a_9 = 3732480\sqrt{11}\)
- \(a_{10} = 22394880\sqrt{11}\)

### Step 4: Calculate \((a_{10} - 2a_{8}) / (2a_{9})\)
Now we can substitute the values into the expression:
\[ a_{10} - 2a_{8} = 22394880\sqrt{11} - 2(622080\sqrt{11}) = 22394880\sqrt{11} - 1244160\sqrt{11} = 21150720\sqrt{11} \]
Now calculate \(2a_{9}\):
\[ 2a_{9} = 2(3732480\sqrt{11}) = 7464960\sqrt{11} \]
Now substituting these into our expression:
\[ \frac{a_{10} - 2a_{8}}{2a_{9}} = \frac{21150720\sqrt{11}}{7464960\sqrt{11}} = \frac{21150720}{7464960} = 3 \]

### Final Answer
Thus, the value of \(\frac{a_{10} - 2a_{8}}{2a_{9}} = 3\).