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

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(B) Given equations are $x^{2}-x+2\lambda=0$ and $3x^{2}-10x+27\lambda=0$.Since $\alpha$ is a common root,it satisfies both equations:$\alpha^{2}-\alp

Vedclass · Accepted Answer

$18$

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(B) Given equations are $x^{2}-x+2\lambda=0$ and $3x^{2}-10x+27\lambda=0$.Since $\alpha$ is a common root,it satisfies both equations:$\alpha^{2}-\alpha+2\lambda=0 \quad ...(1)$$3\alpha^{2}-10\alpha+27\lambda=0 \quad ...(2)$Multiply equation $(1)$ by $3$: $3\alpha^{2}-3\alpha+6\lambda=0 \quad ...(3)$Subtract $(3)$ from $(2)$: $(3\alpha^{2}-10\alpha+27\lambda) - (3\alpha^{2}-3\alpha+6\lambda) = 0$$-7\alpha+21\lambda=0 \Rightarrow \alpha=3\lambda$.Substitute $\alpha=3\lambda$ into equation $(1)$:$(3\lambda)^{2}-(3\lambda)+2\lambda=0 \Rightarrow 9\lambda^{2}-\lambda=0$.Since $\lambda 
eq 0$,we have $\lambda=\frac{1}{9}$.Then $\alpha=3\lambda=3(\frac{1}{9})=\frac{1}{3}$.From the first equation,$\alpha+\beta=1 \Rightarrow \beta=1-\frac{1}{3}=\frac{2}{3}$.From the second equation,$\alpha\gamma=\frac{27\lambda}{3}=9\lambda=9(\frac{1}{9})=1 \Rightarrow \gamma=\frac{1}{\alpha}=\frac{1}{1/3}=3$.Finally,$\frac{\beta\gamma}{\lambda} = \frac{(2/3)(3)}{1/9} = \frac{2}{1/9} = 18$.

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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