If `alpha,beta " are the roots of the equation "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

Explanation

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. **Use Vieta's Formulas**:
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-2}{1} = 2 \).
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-1}{1} = -1 \).

3. **Rewrite the expression**:
We need to find \( \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 \). This can be rewritten as:
\[ \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2} \]

4. **Calculate \( \alpha^2 + \beta^2 \)**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = 2^2 - 2(-1) = 4 + 2 = 6 \]

5. **Calculate \( \alpha^4 + \beta^4 \)**:
We can use the identity:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \]
Substituting the known values:
\[ \alpha^4 + \beta^4 = 6^2 - 2(-1)^2 = 36 - 2 = 34 \]

6. **Calculate \( \alpha^2 \beta^2 \)**:
From the product of the roots:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = (-1)^2 = 1 \]

7. **Final Calculation**:
Now substituting back into our rewritten expression:
\[ \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 = \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2} = \frac{34}{1} = 34 \]

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