Question

Let $\lambda \ne 0$ 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

• A. 18.00
• B. 18.0
• C. 18

Answer 1

Answer: (A), (B), (C)

$x^2-x+2\lambda =0\{\begin{array}{c}\alpha \\ \beta \end{array}\Rightarrow \alpha .\beta =2\lambda$
$3x^2-10x+27\lambda =0\{\begin{array}{c}\alpha \\ \gamma \end{array}\Rightarrow \alpha .\gamma =\frac{27\lambda }{3}=9\lambda$
Both equations have a common root $\alpha .$
$\therefore \frac{{\alpha }^2}{-27\lambda +20\lambda }=\frac{\alpha }{6\lambda -27\lambda }=\frac{1}{-10+3}$
$\Rightarrow \frac{{\alpha }^2}{-7\lambda }=\frac{\alpha }{-19\lambda }=\frac{1}{-7}$
$\Rightarrow {\alpha }^2=\lambda$

Now, $\left(\alpha \beta \right)\cdot \left(\alpha \gamma \right)=\left(2\lambda \right)\left(9\lambda \right)$
$\frac{\beta \cdot \gamma }{\lambda }=2\times 9\cdot \frac{\lambda }{{\alpha }^2}=18$