If `alpha, beta` are the roots of `x^(2)+x+1=0`, then `alpha^(-2)+beta^(-2)` is

Options

A. `sqrt(-1)`
B. `-i sqrt3`
C. -1
D. `-i sqrt3`

Answer

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

### Step-by-Step Solution:

1. **Identify the roots of the quadratic equation**:
The given equation is \( x^2 + x + 1 = 0 \). We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \). Plugging in these values:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]
Therefore, the roots are:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

2. **Calculate \( \alpha^{-2} + \beta^{-2} \)**:
We can use the identity \( x^{-2} = \frac{1}{x^2} \) to rewrite \( \alpha^{-2} + \beta^{-2} \):
\[ \alpha^{-2} + \beta^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

3. **Find \( \alpha^2 + \beta^2 \)**:
We can use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
From Vieta's formulas, we know:
\[ \alpha + \beta = -1, \quad \alpha \beta = 1 \]
Therefore:
\[ \alpha^2 + \beta^2 = (-1)^2 - 2 \cdot 1 = 1 - 2 = -1 \]

4. **Find \( \alpha^2 \beta^2 \)**:
We can calculate \( \alpha^2 \beta^2 \) as:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = 1^2 = 1 \]

5. **Combine results**:
Now substitute back into the expression for \( \alpha^{-2} + \beta^{-2} \):
\[ \alpha^{-2} + \beta^{-2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{-1}{1} = -1 \]

### Final Answer:
Thus, the value of \( \alpha^{-2} + \beta^{-2} \) is \( -1 \).