If `alpha and beta` are the roots of `ax^(2)+bx+c=0`, then the value of `(a alpha +b)^(-2)+ (a beta +b)^(-2)` is equal to

Options

A. `(b^(2)-2ac)/(a^(2)c^(2))`
B. `(c^(2)-2ab)/(a^(2)b^(2))`
C. `(a^(2)-2bc)/(b^(2)c^(2))`
D. none

Answer

To solve the problem, we need to find the value of \( (a\alpha + b)^{-2} + (a\beta + b)^{-2} \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step-by-Step Solution:

1. **Understanding the Roots**:
From Vieta's formulas, we know:
- The sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots: \( \alpha \beta = \frac{c}{a} \)

2. **Rewriting the Expression**:
We need to rewrite the expression:
\[ (a\alpha + b)^{-2} + (a\beta + b)^{-2} \]
This can be rewritten as:
\[ \frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2} \]

3. **Finding a Common Denominator**:
The common denominator for the two fractions is:
\[ (a\alpha + b)^2 (a\beta + b)^2 \]
Thus, we can write:
\[ \frac{(a\beta + b)^2 + (a\alpha + b)^2}{(a\alpha + b)^2 (a\beta + b)^2} \]

4. **Expanding the Numerator**:
Now, let's expand the numerator:
\[ (a\beta + b)^2 + (a\alpha + b)^2 = a^2\beta^2 + 2ab\beta + b^2 + a^2\alpha^2 + 2ab\alpha + b^2 \]
This simplifies to:
\[ a^2(\alpha^2 + \beta^2) + 2ab(\alpha + \beta) + 2b^2 \]

5. **Using the Identities**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]

6. **Substituting Back**:
Now substituting this back into our expression:
\[ a^2\left(\frac{b^2}{a^2} - \frac{2c}{a}\right) + 2ab\left(-\frac{b}{a}\right) + 2b^2 \]
This simplifies to:
\[ b^2 - 2ac - 2b^2 + 2b^2 = b^2 - 2ac \]

7. **Denominator Calculation**:
Now we need to calculate the denominator:
\[ (a\alpha + b)^2 (a\beta + b)^2 \]
This can be expressed as:
\[ (a^2\alpha^2 + 2ab\alpha + b^2)(a^2\beta^2 + 2ab\beta + b^2) \]
Using the values of \( \alpha \) and \( \beta \):
\[ = (a^2\alpha\beta + 2ab(\alpha + \beta) + b^2)^2 \]
Substituting \( \alpha + \beta \) and \( \alpha\beta \):
\[ = \left(a^2\frac{c}{a} + 2ab\left(-\frac{b}{a}\right) + b^2\right)^2 = \left(ac - 2b^2 + b^2\right)^2 = (ac - b^2)^2 \]

8. **Final Expression**:
Now we can combine everything:
\[ \frac{b^2 - 2ac}{(ac - b^2)^2} \]

### Final Answer:
Thus, the value of \( (a\alpha + b)^{-2} + (a\beta + b)^{-2} \) is:
\[ \frac{b^2 - 2ac}{(ac - b^2)^2} \]