Question

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

Answer 1

$p{x}^2+qx+r=0$

$\alpha +\beta =-\frac{-q}{p},\alpha \beta =\frac{r}{p}$

Given

$\frac{\alpha }{\beta }=\frac{l}{m}$

$\Rightarrow r=\frac{l\beta }{m}$

$(\frac{l}{m}+1)\beta =\frac{-q}{p}$

$\Rightarrow \frac{(l+m)}{m}\times \frac{r}{\alpha p}=\frac{-q}{p}$

$\Rightarrow \frac{(l+m)}{m}r=-2q$.....(i)

$\alpha \beta =\frac{r}{p}$

$\frac{l\beta }{m}.\beta =\frac{r}{p}$

${\beta }^2=\frac{rm}{pl}\therefore \beta =\sqrt{\frac{rm}{pl}}$

$\alpha =\frac{l}{m}\times \frac{\sqrt{r}}{\sqrt{p}}\times \frac{\sqrt{m}}{\sqrt{l}}$

$\alpha =\frac{\sqrt{l}}{\sqrt{m}}\times \frac{\sqrt{r}}{\sqrt{p}}$

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

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