Question

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

• A. $\frac{2}{b}$
• B. $\frac{2}{a}$
• C. $-\frac{2}{a}$
• D. $\frac{2}{c}$

Answer 1

The correct option is C

$-\frac{2}{a}$

Given: α, β are the roots of the equation $a{x}^2+bx+c=0$. Then we have
α + β = - $\frac{b}{a}$, αβ = $\frac{c}{a}$

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

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

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

$=\frac{b^2-ac-b^2}{a^{2}c-a{b}^2+a{b}^2}=\frac{-2ac}{a^{2}c}=-\frac{2}{a}$