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Question:

Let $\lambda \neq 0$ be in $\mathbf{R}$. If $\alpha$ and $\beta$ are the roots of the equation $\mathrm{x}^{2}-\mathrm{x}+2 \lambda=0$, 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__________

Solution:

$3 \alpha^{2}-10 \alpha+27 \lambda=0$      …..(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 $\alpha=3 \lambda \Rightarrow \lambda=\frac{1}{3}$

$\alpha+\beta=1 \Rightarrow \beta=2 / 3$

$\alpha+\gamma=\frac{10}{3} \Rightarrow \gamma=3$

$\frac{\beta \gamma}{\lambda}=\frac{\frac{2}{3} \times 3}{\frac{1}{9}}=18$

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