If `alpha, beta` are the roots of the equations `x^(2) - 2x - 1 = 0`. Then what is the value of `alpha^(2) beta^(-2) + alpha^(-2) beta^(2)`

Options

A. -2
B. 0
C. 30
D. 34

Answer

To solve the problem, we need to find the value of \( \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 \) given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x - 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is \( x^2 - 2x - 1 = 0 \). Here, \( a = 1 \), \( b = -2 \), and \( c = -1 \).

2. **Find the sum and product of the roots**:
- The sum of the roots \( \alpha + \beta \) is given by \( -\frac{b}{a} = -\frac{-2}{1} = 2 \).
- The product of the roots \( \alpha \beta \) is given by \( \frac{c}{a} = \frac{-1}{1} = -1 \).

3. **Calculate \( \alpha^2 + \beta^2 \)**:
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ \alpha^2 + \beta^2 = (2)^2 - 2(-1) = 4 + 2 = 6. \]

4. **Calculate \( \alpha^4 + \beta^4 \)**:
Using the identity \( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha^2 \beta^2) \):
- First, we need \( \alpha^2 \beta^2 \). Since \( \alpha \beta = -1 \):
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = (-1)^2 = 1. \]
- Now, substituting into the equation:
\[ \alpha^4 + \beta^4 = (6)^2 - 2(1) = 36 - 2 = 34. \]

5. **Calculate \( \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 \)**:
This expression can be rewritten as:
\[ \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 = \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2}. \]
Substituting the values we found:
\[ \alpha^2 \beta^2 = 1 \quad \text{and} \quad \alpha^4 + \beta^4 = 34. \]
Therefore,
\[ \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 = \frac{34}{1} = 34. \]

### Final Answer:
The value of \( \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 \) is \( \boxed{34} \).