Let $\lambda \neq 0$ be in $\mathbf{R}$. If $\alpha$ and $\beta$ are roots of the equation, $x^{2}-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x^{2}-10 x+27 \lambda=0$, then $\frac{\beta \gamma}{\lambda}$ is equal to :
Correct Option: , 2
Since $\alpha$ is common root of $x^{2}-x+2 \lambda=0$ and
$3 x^{2}-10 x+27 \lambda=0$
$\therefore 3 \alpha^{2}-10 \alpha+27 \lambda=0$...(i)
$3 \alpha^{2}-3 \alpha+6 \lambda=0$..(ii)
$\therefore$ On subtract, we get $\alpha=3 \lambda$
Now, $\alpha \beta=2 \lambda \Rightarrow 3 \lambda \cdot \beta=2 \lambda \Rightarrow \beta=\frac{2}{3}$
$\Rightarrow \alpha+\beta=1 \Rightarrow 3 \lambda+\frac{2}{3}=1 \Rightarrow \lambda=\frac{1}{9}$ and
$\alpha \gamma=9 \lambda \Rightarrow 3 \lambda \cdot \gamma=9 \lambda \Rightarrow \gamma=3$
$\therefore \frac{\beta \gamma}{\lambda}=18$
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