Question

If the roots of the equation $p{x}^2+qx+r=0$ are in the ratio $l:m$ prove that
$(l^2+m^2)pr+lm(2pr-q^2)$ =0

Answer 1

Given equation $p{x}^2+qx+r=0$ Let roots be $l\alpha$ and $m\alpha$

$l\alpha +m\alpha =\frac{-q}{p}$ and $\left(l\alpha \right)\left(m\alpha \right)=\frac{r}{p}$

$(l+m)\alpha =-\frac{p}{q}$ $\left(lm\right){\alpha }^2=\frac{r}{p}$ ........ $\left(2\right)$

$\alpha =\frac{q}{p(l+m{)}^2}=\frac{r}{9}$

or, $\frac{lm{q}^2}{p^2(l^2+m^2+2lm)}=\frac{r}{p}$

or, $(l^2+m^2+2lm)pr=lm{q}^2$

or, $(l^2+m^2)pr+lm(2pr-q^2)=0$ ___________ proved.