If $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$, then value of

$\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}$ is :

1. Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$:

$a\alpha^2 + b\alpha + c = 0$ ...(1)

$a\beta^2 + b\beta + c = 0$ ...(2)

2. We need to find $\frac{1}{a\alpha+b} + \frac{1}{a\beta+b}$

3. From (1), we can write:

$c = -a\alpha^2 - b\alpha$

$c = -a\beta^2 - b\beta$

4. Let's find LCM:

$\frac{1}{a\alpha+b} + \frac{1}{a\beta+b}$

= $\frac{(a\beta+b) + (a\alpha+b)}{(a\alpha+b)(a\beta+b)}$

= $\frac{a(\alpha+\beta) + 2b}{(a\alpha+b)(a\beta+b)}$

5. For quadratic equation $ax^2 + bx + c = 0$:

$\alpha + \beta = -\frac{b}{a}$

6. Substituting:

= $\frac{a(-\frac{b}{a}) + 2b}{(a\alpha+b)(a\beta+b)}$

= $\frac{-b + 2b}{(a\alpha+b)(a\beta+b)}$

= $\frac{b}{(a\alpha+b)(a\beta+b)}$

= $\frac{b}{a^2\alpha\beta+ab(\alpha+\beta)+b^2}$

Put $\alpha\beta=\frac{c}{a}$ and $\alpha+\beta=\frac{-b}{a}$ in the above equation, and we'll get $\frac{b}{ac}$

Therefore, $\frac{1}{a\alpha+b} + \frac{1}{a\beta+b} = \frac{b}{ac}$