If $$\alpha$$ and $$\beta$$ are the roots of the equation $$ax^{2}+bx+c=0$$ then the value of $$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}$$ is:

$$\alpha$$ is given as the root of the equation $$ax^2+bx+c=0$$. Thus -

$$a\alpha^2+b\alpha+c=0$$

$$a\alpha+b=-\dfrac{c}{\alpha}\rightarrow1$$

$$\beta$$ is given as the root of the equation $$ax^2+bx+c=0$$. Thus -

$$a\beta^2+b\beta+c=0$$

$$a\beta+b=-\dfrac{c}{\beta}\rightarrow2$$

We need to find the value of $$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}$$.

$$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}=-\dfrac{\alpha}{c}-\dfrac{\beta}{c}$$

$$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}=-\dfrac{\alpha+\beta}{c}$$

$$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}=-\dfrac{-b}{ac}$$

$$\dfrac{1}{a\alpha+b}+\dfrac{1}{a\beta+b}=\dfrac{b}{ac}$$