If `alpha,beta` are the roots of the equation `a x^2+b x+c=0`, then `1/(aalpha+b)+1/(abeta+b)=`

Options

A. `c/(ab)`
B. `a/(bc)`
C. `b/(ac)`
D. none of these

Answer

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

### Step-by-Step Solution:

1. **Identify the roots**: We know that \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \). By Vieta's formulas, we have:
- Sum of roots: \( \alpha + \beta = -\frac{b}{a} \)
- Product of roots: \( \alpha \beta = \frac{c}{a} \)

2. **Rewrite the expression**: We need to simplify the expression \( \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} \).
- We can combine the two fractions:
\[ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{(a\beta + b) + (a\alpha + b)}{(a\alpha + b)(a\beta + b)} \]

3. **Simplify the numerator**:
- The numerator becomes:
\[ (a\beta + b) + (a\alpha + b) = a(\alpha + \beta) + 2b \]
- Substituting \( \alpha + \beta = -\frac{b}{a} \):
\[ = a\left(-\frac{b}{a}\right) + 2b = -b + 2b = b \]

4. **Simplify the denominator**:
- The denominator is:
\[ (a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha + \beta) + b^2 \]
- Substituting \( \alpha\beta = \frac{c}{a} \) and \( \alpha + \beta = -\frac{b}{a} \):
\[ = a^2\left(\frac{c}{a}\right) + ab\left(-\frac{b}{a}\right) + b^2 = ac - b^2 + b^2 = ac \]

5. **Combine results**:
- Now we have:
\[ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{b}{ac} \]

### Final Answer:
Thus, the value of \( \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} \) is \( \frac{b}{ac} \).