If the ratio of the roots of the equation $\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0$ is $a: b$, then $\frac{\mathrm{ab}}{(\mathrm{a}+\mathrm{b})^2}=$

Question

If the ratio of the roots of the equation $\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0$ is $a: b$, then $\frac{\mathrm{ab}}{(\mathrm{a}+\mathrm{b})^2}=$, $\frac{\mathrm{p}^2}{\mathrm{qr}}$, $\frac{\mathrm{pr}}{\mathrm{q}^2}$, $\frac{q^2}{\mathrm{pr}}$, $\frac{\mathrm{pq}}{\mathrm{r}^2}$

MARKS App · Accepted Answer

Hints: Let roots are $a \alpha$ and $b \alpha$ $\Rightarrow(\mathrm{a}+\mathrm{b}) \alpha=\frac{-\mathrm{q}}{\mathrm{p}}$ $\begin{aligned} & a b \alpha^2=\frac{r}{p} \ & \frac{a b \alpha^2}{(a+b)^2 \alpha^2}=\frac{r}{p} \cdot \frac{p^2}{q^2} \ & \frac{a b}{(a+b)^2}=\frac{1 p}{q^2}\end{aligned}$

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