Question

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

• A. $27$
• B. $9$
• C. $18$
• D. $36$

Answer 1

The correct option is C

$18$

Explanation for the correct option:

Finding $\frac{\beta \gamma }{\lambda }$is equal to :

Given: $\alpha ,\beta$are the roots of the equation, $x^2-x+2\lambda =0$ and $\alpha ,\gamma$ are the roots of the equation,$3x^2-10x+27\lambda =0$

$\begin{array}{rcl}{x}^2-x+2\lambda & =& 0....\left(i\right)\\ 3x^2-10x+27\lambda & =& 0.....\left(ii\right)\\ & & 3\left(i\right)\\ 3x^2-3x+6\lambda & =& 0....\left(iii\right)\\ & & \text{(ii)-(iii)}\\ -7x+21\lambda & =& 0\\ \alpha & =& 3\lambda \end{array}$

Putting this value in equation (i)

$\begin{array}{rcl}9{\lambda }^2-3\lambda +2\lambda & =& 0\\ 9{\lambda }^2-\lambda & =& 0\\ & \Rightarrow & \lambda =\frac{1}{9}\\ & \Rightarrow & \alpha =\frac{1}{3}\\ \alpha \beta & =& \frac{2}{9}\\ & \Rightarrow & \beta =\frac{2}{3}\\ \alpha \gamma & =& 1\\ & \Rightarrow & \gamma =3\\ \frac{\beta \gamma }{\lambda }& =& \frac{2}{3}\times 3÷\frac{1}{9}\end{array}$

$\Rightarrow \frac{\beta \gamma }{\lambda }=18$

Therefore, the correct answer is option (C)