Description for Correct answer:
Since, \( \Large \alpha and\ \beta \) are the roots of the equation \( \Large ax^{2}+bx+c=0 \)

\( \Large \alpha + \beta = -\frac{b}{a}, \alpha \beta = \frac{c}{a} \)

and \( \Large \alpha ^{2}+ \beta ^{2}= \left( \alpha + \beta \right)^{2}-2 \alpha \beta = \left(\frac{b^{2}-2ac}{a^{2}}\right) \)

\( \Large \frac{ \alpha }{ \alpha \beta +b}+\frac{ \beta }{a \alpha +b}= \frac{a \left(a \alpha +b\right)+ \beta \left(a \beta +b\right) }{ \left(a \beta +b\right) \left(a \alpha +b\right) } \)

= \( \Large \frac{a \left( \alpha ^{2}+ \beta ^{2}\right)+b \left( \alpha + \beta \right) }{ \alpha \beta a^{2}+ab \left( \alpha + \beta \right)+b^{2} } \)

= \( \Large \frac{a \left(\frac{b^{2}-2ac}{a^{2}}\right)+b \left(-\frac{b}{a}\right) }{ \left(\frac{c}{a}\right)a^{2}+ab \left(-\frac{b}{a}\right)+b^{2} } \)

= \( \Large \frac{b^{2}-2ac-b^{2}}{a^{2}c-ab^{2}+ab^{2}}=\frac{-2ac}{a2c}=\frac{-2}{a} \)