Let be in R. If and

Let $\lambda \neq 0$ be in 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:
Option: 1 27
Option: 2 18
Option: 3 9
Option: 4 36

Answers (1)

H Himanshu Meshram

$\\\alpha+\beta=1, \alpha \beta=2 \lambda \\ \alpha+\beta=\frac{10}{3}, \quad \alpha \gamma=\frac{27 \lambda}{3}=9 \lambda \\ \gamma-\beta=\frac{7}{3}$

$\\\frac{\gamma}{\beta}=\frac{9}{2} \Rightarrow \gamma=\frac{9}{2} \beta=\frac{9}{2} \times \frac{2}{3} \Rightarrow \gamma=3 \\ \frac{9}{2} \beta-\beta=\frac{7}{3} \\ \frac{9}{2} \beta=\frac{7}{3} \Rightarrow \beta=\frac{2}{3}$

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

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Let be in R. If and are the roots of the equation, and and are the roots of the equation, , then is equal to: Option: 1 27 Option: 2 18 Option: 3 9 Option: 4 36

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Let $\lambda \neq 0$ be in 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:
Option: 1 27
Option: 2 18
Option: 3 9
Option: 4 36