If `alpha, beta` are zeroes of the polynomial f(x) = `ax^(2)` + bx + c, the c, then `(1)/(alpha^(2))+ (1)/(beta^(2))` =

Options

A. `(b^(2) -2 ac)/(a^(2))`
B. `(b^(2) - 2ac)/(c^(2))`
C. `(b^(2)+2ac)/(a^(2))`
D. `(b^(2) + 2ac)/(c^(2))`

Answer

To solve the problem, we need to find the value of \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \) given that \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( f(x) = ax^2 + bx + c \).

### Step-by-Step Solution:

1. **Start with the expression**:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} \]

2. **Find a common denominator**:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

3. **Use the identity for \( \alpha^2 + \beta^2 \)**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
From Vieta's formulas, we have:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a} \]
Therefore:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) \]
Simplifying this gives:
\[ \alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2} \]

4. **Substituting back into the expression**:
Now substitute \( \alpha^2 + \beta^2 \) back into our expression:
\[ \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} = \frac{\frac{b^2 - 2ac}{a^2}}{\alpha^2 \beta^2} \]

5. **Calculate \( \alpha^2 \beta^2 \)**:
We know:
\[ \alpha \beta = \frac{c}{a} \implies \alpha^2 \beta^2 = \left(\frac{c}{a}\right)^2 = \frac{c^2}{a^2} \]

6. **Final expression**:
Substitute \( \alpha^2 \beta^2 \) into the expression:
\[ \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2} \]

Thus, the final result is:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{b^2 - 2ac}{c^2} \]