Question

If α, β are the roots of the equation $a{x}^2+bx+c=0,\mathrm{then}\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=$
(a) c / ab
(b) a / bc
(c) b / ac
(d) none of these.

Answer 1

(c) b / ac
Given equation: $a{x}^2+bx+c=0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.

Then, sum of the roots = $\alpha +\beta =-\frac{b}{a}$
Product of the roots = $\alpha \beta =\frac{c}{a}$


$$\begin{gathered} \therefore \frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=\frac{a\beta +b+a\alpha +b}{(a\alpha +b)(a\beta +b)} \\ =\frac{a(\alpha +\beta )+2b}{a^2\alpha \beta +ab\alpha +ab\beta +b^2} \\ =\frac{a(\alpha +\beta )+2b}{a^2\alpha \beta +ab\left(\alpha +\beta \right)+b^2} \\ =\frac{a\left(-{\displaystyle \frac{b}{a}}\right)+2b}{a^2\left({\displaystyle \frac{c}{a}}\right)+ab\left(-{\displaystyle \frac{b}{a}}\right)+b^2} \\ =\frac{b}{ac} \end{gathered}$$