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

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(D) For the equation $x^{2}-x+2 \lambda=0$,the sum of roots is $\alpha+\beta=1$ and the product of roots is $\alpha \beta=2 \lambda$.For the equation

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$18$

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

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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