Question

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

• A. $\frac{p^2}{qr}$
• B. $\frac{pr}{q^2}$
• C. $\frac{q^2}{pr}$
• D. $\frac{pq}{r^2}$

Answer 1

Answer: (B)

Let the roots be $a\alpha$ and $b\alpha$. Then,
$a\alpha +b\alpha =-\frac{q}{p}\text{ and }a\alpha \times b\alpha =\frac{r}{p}$
$\therefore \frac{{\alpha }^2(a+b{)}^2}{ab{\alpha }^2}=\frac{q^2/p^2}{r/p}$
$\Rightarrow \frac{ab}{(a+b{)}^2}=\frac{rp}{q^2}$