if `alpha , beta` are root of `ax^2+bx+c=0` then `(1/alpha^2+1/beta^2)^2` (a) `(b^(2)(b^(2)-4ac))/(c^(2)a^(2))` (b) `(b^(2)(b^(2)-4ac))/(ca^(3))` (c) `(b^(2)(b^(2)-4ac))/(a^(4))` (d) `(b^(2)-2ac)^2/(c^(4))`

Options

A. `(b^(2)(b^(2)-4ac))/(c^(2)a^(2))`
B. `(b^(2)(b^(2)-4ac))/(ca^(3))`
C. `(b^(2)(b^(2)-4ac))/(a^(4))`
D. '(b^(2)-2ac)^2/(c^(4))`

Answer

To solve the problem, we need to find the value of \( \left( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \right)^2 \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step-by-Step Solution:

1. **Identify the Roots**:
Given the quadratic equation \( ax^2 + bx + c = 0 \), the roots \( \alpha \) and \( \beta \) satisfy:
- Sum of roots: \( \alpha + \beta = -\frac{b}{a} \)
- Product of roots: \( \alpha \beta = \frac{c}{a} \)

2. **Express \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)**:
We can rewrite \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \) as:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

3. **Use the Identity for \( \alpha^2 + \beta^2 \)**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from step 1:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]

4. **Substituting Back**:
Now substituting \( \alpha^2 + \beta^2 \) into the expression:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\left(\frac{c}{a}\right)^2} = \frac{b^2 - 2ac}{c^2} \cdot a^2 \]

5. **Final Expression**:
Thus, we have:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{b^2 - 2ac}{c^2} \cdot a^2 \]

6. **Square the Result**:
Now we need to square this expression:
\[ \left( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \right)^2 = \left( \frac{b^2 - 2ac}{c^2} \cdot a^2 \right)^2 = \frac{(b^2 - 2ac)^2}{c^4} \]

### Conclusion:
The final result is:
\[ \left( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \right)^2 = \frac{(b^2 - 2ac)^2}{c^4} \]

Thus, the correct option is:
**(d) \( \frac{(b^2 - 2ac)^2}{c^4} \)**.