Let lambda neq 0 be in mathbb{R}. If alpha an | vedclass

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(D) For the equation $x^{2}-x+2\lambda=0$,we have $\alpha+\beta=1$ and $\alpha\beta=2\lambda$. For the equation $3x^{2}-10x+27\lambda=0$,we have $\alp

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$18$

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(D) For the equation $x^{2}-x+2\lambda=0$,we have $\alpha+\beta=1$ and $\alpha\beta=2\lambda$. For the equation $3x^{2}-10x+27\lambda=0$,we have $\alpha+\gamma=\frac{10}{3}$ and $\alpha\gamma=\frac{27\lambda}{3}=9\lambda$. Subtracting the sum of roots: $(\alpha+\gamma)-(\alpha+\beta)=\frac{10}{3}-1 \Rightarrow \gamma-\beta=\frac{7}{3}$. Dividing the product of roots: $\frac{\alpha\gamma}{\alpha\beta}=\frac{9\lambda}{2\lambda}$ $\Rightarrow \frac{\gamma}{\beta}=\frac{9}{2}$ $\Rightarrow \gamma=\frac{9}{2}\beta$. Substituting $\gamma$ into $\gamma-\beta=\frac{7}{3}$: $\frac{9}{2}\beta-\beta=\frac{7}{3}$ $\Rightarrow \frac{7}{2}\beta=\frac{7}{3}$ $\Rightarrow \beta=\frac{2}{3}$. Then $\gamma=\frac{9}{2} 	imes \frac{2}{3}=3$. Since $\alpha+\beta=1$,$\alpha=1-\frac{2}{3}=\frac{1}{3}$. Using $\alpha\beta=2\lambda$: $\frac{1}{3} 	imes \frac{2}{3}=2\lambda$ $\Rightarrow \frac{2}{9}=2\lambda$ $\Rightarrow \lambda=\frac{1}{9}$. Finally,$\frac{\beta\gamma}{\lambda}=\frac{(2/3) 	imes 3}{1/9}=\frac{2}{1/9}=18$.

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

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