Question

Let $\lambda \ne 0$ be in $R$. If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+2\lambda =0$, and $\alpha$ and $\gamma$ are the roots of equation $3x^2-10x+27\lambda =0$, then $\frac{\beta \gamma }{\lambda }$ is equal to ___

Answer 1

Answer: 18

$3{\alpha }^2-10\alpha +27\lambda =0...(1)$
${\alpha }^2-\alpha +2\lambda =0...(2)$
$(1)-3(2)$ gives
$-7\alpha +21\lambda =0$
$\Rightarrow \alpha =3\lambda$
Put $\alpha =3\lambda$ in equation (1) we get
$9{\lambda }^2-3\lambda +2\lambda -0$
$9{\lambda }^2=\lambda$
$\Rightarrow \lambda =\frac{1}{9}$ as $\lambda \ne 0$
Now $\alpha =3\lambda$
$\Rightarrow \lambda =\frac{1}{3}$
$\alpha +\beta =1$
$\Rightarrow \beta =2/3$
$\alpha +\gamma =\frac{10}{3}$
$\Rightarrow \gamma =3$
$\frac{\beta \gamma }{\lambda }=\frac{\frac{2}{3}\times 3}{\frac{1}{9}}=18$