Question

If $\alpha ,\beta$ are roots of the equation,$a{x}^2+bx+c=0$, then $\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=$

Answer 1

Given: $\alpha ,\beta$ are the roots of the equation

$a{x}^2+bx+c=0$

Now,

$\text{sum of roots}=-\frac{\text{coefficient of x}}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha +\beta =-\frac{b}{a}\dots \left(1\right)$

$\text{Product of roots}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha \beta =\frac{c}{a}\dots \left(2\right)$

Now,

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

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

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

$=\frac{-b+2b}{(ac-b^2+b^2)}$ (from (1) and (2) )

$=\frac{b}{ac}$

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

Hence, option (C) is correct.