Question

Let $\lambda \neq 0$ 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 $3 x^{2}-10 x+27 \lambda=0$, then $\frac{\beta \gamma}{\lambda}$ is equal to ___

Answer 1

$3 \alpha^{2}-10 \alpha+27 \lambda=0\,\,\, ...(1)$
$\alpha^{2}-\alpha+2 \lambda=0 \,\,\, ...(2)$
$(1)-3(2)$ 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 $\lambda \neq 0$
Now $\alpha=3 \lambda$
$ \Rightarrow \lambda=\frac{1}{3}$
$\alpha+\beta=1 $
$\Rightarrow \beta=2 / 3$
$\alpha+\gamma=\frac{10}{3}$
$ \Rightarrow \gamma=3$
$\frac{\beta \gamma}{\lambda}=\frac{\frac{2}{3} \times 3}{\frac{1}{9}}=18$