Question

If the roots are in the ratio $m:n$ then, show that ${(m+n)}^{2}ac=mn{b}^2$

Answer 1

Let the roots of the equation be $m\alpha ,n\alpha$

$\therefore (m+n)\alpha =\frac{-b}{a},mn{\alpha }^2=\frac{c}{a}$

$\Rightarrow \alpha =\frac{-b}{a(m+n)},mn{\alpha }^2=\frac{c}{a}$

$\Rightarrow {\alpha }^2=\frac{b^2}{a^2{(m+n)}^2},mn{\alpha }^2=\frac{c}{a}$

$\Rightarrow mn{\alpha }^2=\frac{c}{a}$

$\Rightarrow mn\frac{b^2}{a^2{(m+n)}^2}=\frac{c}{a}$ where ${\alpha }^2=\frac{b^2}{a^2{(m+n)}^2}$

$\Rightarrow mn.\frac{b^2}{a^2{(m+n)}^2}=\frac{c}{a}$

$\Rightarrow mn.\frac{b^2}{a{(m+n)}^2}=c$

$\Rightarrow mn{b}^2=ac{(m+n)}^2$

$\therefore ac{(m+n)}^2=mn{b}^2$

Hence proved.