Question

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

Answer 1

Consider the given equation

$p{x}^2+qx+r=0$ ratio $=l:m$

Let $l\alpha$ and $m\alpha$ be the roots given equation

Then, we know that,

$sumsofroots=-\frac{cofficentsofx}{cofficentsof{x}^2}$

$l\alpha +m\alpha =-\frac{q}{p}$

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

$\Rightarrow \alpha =-\frac{q}{p(l+m)}......\left(1\right)$

$productofroots=\frac{\text{constant}\text{term}}{cofficentof{x}^2}$

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

${\alpha }^2=\frac{r}{plm}......\left(2\right)$

By equation (1) and (2) to, we get,

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

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

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

$\Rightarrow q^{2}lm=rp{(l+m)}^2$

$\Rightarrow q^{2}lm=rp{(l+m)}^2$

$\Rightarrow q^{2}lm=rp(l^2+m^2+2lm)$

$\Rightarrow q^{2}lm=rp(l^2+m^2)+2rplm$

$\Rightarrow q^{2}lm-2rplm=rp(l^2+m^2)$

$\Rightarrow lm(q^2-2pr)=(l^2+m^2)$

$\Rightarrow (l^2+m^2)-lm(q^2-2pr)=0$

$\Rightarrow (l^2+m^2)+lm(2pr-q^2)=0$

Hence, this is the answer.