Question:
If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then $\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=$
(a) c / ab
(b) a / bc
(c) b / ac
(d) none of these.
Solution:
(c) b / ac
Given equation: $a x^{2}+b x+c=0$
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 $=\alpha \beta=\frac{c}{a}$
$\therefore \frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{a \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$
$=\frac{a(\alpha+\beta)+2 b}{a^{2} \alpha \beta+a b \alpha+a b \beta+b^{2}}$
$=\frac{a(\alpha+\beta)+2 b}{a^{2} \alpha \beta+a b(\alpha+\beta)+b^{2}}$
$=\frac{a\left(-\frac{b}{a}\right)+2 b}{a^{2}\left(\frac{c}{a}\right)+a b\left(-\frac{b}{a}\right)+b^{2}}$
$=\frac{b}{a c}$