Question

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

• A. clab
• B. albc
• C. blac
• D. none of these

Answer 1

The correct option is C

blac

$\text{Given equation:}a{x}^2+bx+c=0Also,\alpha ,\beta aretherootsoftheequationThen,sumoftheroots=\alpha +\beta =-\frac{b}{a}Productoftheroots=\alpha \beta =\frac{c}{a}\therefore \frac{1}{a\alpha +b}+\frac{1}{a\alpha +b}=\frac{a\beta +b+a\alpha +b}{\left(a\alpha \right)(a\beta +b)}=\frac{a(\alpha +\beta )+2b}{a^2\alpha \beta +ab\alpha +ab\beta +b^2}=\frac{\alpha (-\frac{b}{a})+2b}{a^2\left(\frac{c}{a}\right)+ab(-\frac{b}{a})+b^2}=\frac{b}{ac}$