If a and `beta` are the zeroes of the polynomial `x^(2) + 2x + 1`, then `(1)/(alpha)+(1)/(beta)` is equal to :

Options

A. `-2`
B. 2
C. 0
D. 1

Answer

To solve the problem, we need to find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\), where \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(x^2 + 2x + 1\).

### Step-by-Step Solution:

1. **Identify the Polynomial**: The given polynomial is \(x^2 + 2x + 1\).

2. **Find the Zeroes**:
- The zeroes of a quadratic polynomial \(ax^2 + bx + c\) can be found using the relationships:
- Sum of the zeroes (\(\alpha + \beta\)) = \(-\frac{b}{a}\)
- Product of the zeroes (\(\alpha \cdot \beta\)) = \(\frac{c}{a}\)
- Here, \(a = 1\), \(b = 2\), and \(c = 1\).

3. **Calculate the Sum of the Zeroes**:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{2}{1} = -2 \]

4. **Calculate the Product of the Zeroes**:
\[ \alpha \cdot \beta = \frac{c}{a} = \frac{1}{1} = 1 \]

5. **Calculate \(\frac{1}{\alpha} + \frac{1}{\beta}\)**:
- We can rewrite \(\frac{1}{\alpha} + \frac{1}{\beta}\) as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \cdot \beta} \]
- Substitute the values we found:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta} = \frac{-2}{1} = -2 \]

6. **Final Answer**:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = -2 \]