If `alpha, beta` are the zeros of the polynomial `f(x) = ax^2 + bx + C`, then find `1/alpha^(2) +1/beta^(2)`.

Answer

To find \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \) given that \( \alpha \) and \( \beta \) are the zeros of the polynomial \( f(x) = ax^2 + bx + c \), we can follow these steps:

### Step 1: Use the relationship of zeros
From Vieta's formulas, we know:
- The sum of the zeros \( \alpha + \beta = -\frac{b}{a} \)
- The product of the zeros \( \alpha \beta = \frac{c}{a} \)

### Step 2: Rewrite the expression
We can rewrite \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \) using the identity:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

### Step 3: Find \( \alpha^2 + \beta^2 \)
To find \( \alpha^2 + \beta^2 \), we can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) \]
\[ = \frac{b^2}{a^2} - \frac{2c}{a} \]

### Step 4: Find \( \alpha^2 \beta^2 \)
Now, we need \( \alpha^2 \beta^2 \):
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{c}{a}\right)^2 = \frac{c^2}{a^2} \]

### Step 5: Substitute back into the expression
Now we can substitute \( \alpha^2 + \beta^2 \) and \( \alpha^2 \beta^2 \) back into the rewritten expression:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} \]
\[ = \frac{b^2 - 2ac}{c^2} \]

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

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