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

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 1: Find the roots \( \alpha \) and \( \beta \)
We start with the quadratic equation:
\[ x^2 - 2x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ x = \frac{2 \pm \sqrt{8}}{2} \]
\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ x = 1 \pm \sqrt{2} \]
Thus, the roots are:
\[ \alpha = 1 + \sqrt{2}, \quad \beta = 1 - \sqrt{2} \]

### Step 2: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
We know:
\[ \alpha + \beta = 2 \quad \text{(sum of roots)} \]
\[ \alpha \beta = -1 \quad \text{(product of roots)} \]
Now substituting these values:
\[ \alpha^2 + \beta^2 = (2)^2 - 2(-1) = 4 + 2 = 6 \]

### Step 3: Calculate \( \alpha^4 + \beta^4 \)
Using the identity:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha^2 \beta^2) \]
We already found \( \alpha^2 + \beta^2 = 6 \) and \( \alpha^2 \beta^2 = (\alpha \beta)^2 = (-1)^2 = 1 \):
\[ \alpha^4 + \beta^4 = 6^2 - 2(1) = 36 - 2 = 34 \]

### Step 4: Calculate \( \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 \)
We can rewrite the expression:
\[ \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 = \frac{\alpha^4 + \beta^4}{\alpha^2 \beta^2} \]
Substituting the known values:
\[ \alpha^2 \beta^2 = 1 \quad \text{and} \quad \alpha^4 + \beta^4 = 34 \]
Thus:
\[ \alpha^2 \beta^{-2} + \alpha^{-2} \beta^2 = \frac{34}{1} = 34 \]

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