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. $27$
• B. $9$
• C. $18$
• D. $36$

Answer 1

The correct option is C $18$

$x^2-x+2\lambda =0\cdots \left(1\right)$
$3x^2-10x+27\lambda =0\cdots \left(2\right)$
Multiplying $\left(1\right)$ by $3,$ we get
$3x^2-3x+6\lambda =0\cdots \left(3\right)$
From equation $\left(2\right)$ and $\left(3\right),$ we get
$-7x+21\lambda =0$
Since $\alpha$ is the common root.
$\therefore \alpha =3\lambda$
Now, $\alpha \beta =2\lambda$
$\Rightarrow \beta =\frac{2\lambda }{3\lambda }$
$\therefore \beta =\frac{2}{3}$
and $\alpha \gamma =9\lambda$
$\Rightarrow \gamma =\frac{9\lambda }{3\lambda }$
$\therefore \gamma =3$

Put the value of $\alpha$ in equation $\left(1\right),$ we get
$9{\lambda }^2-\lambda =0$
$\Rightarrow \lambda =\frac{1}{9}$
$\therefore \frac{\beta \gamma }{\lambda }=\frac{\frac{2}{3}\cdot 3}{\frac{1}{9}}=18$