Let λ ≠ 0 be in R. If α and β are the roots of the equation ${\mathrm{x}}^2-\mathrm{x}+2\mathrm{\lambda }=0$ and α and γ are the roots of equation $3{\mathrm{x}}^2-10\mathrm{x}+27\mathrm{\lambda }=0,\text{ then }\frac{\mathrm{\beta \gamma }}{\mathrm{\lambda }}$ is equal to

We have, $\mathrm{\alpha }$ is common root of the equations ${\mathrm{x}}^2-\mathrm{x}+2\mathrm{\lambda }=0$ and $3{\mathrm{x}}^2-10\mathrm{x}+27\mathrm{\lambda }=0$.

Now, common root of these equations is

$$\begin{gathered} \left(3{\mathrm{\alpha }}^2-10\mathrm{\alpha }+27\mathrm{\lambda }\right)-\left(3{\mathrm{\alpha }}^2-3\mathrm{\alpha }+6\mathrm{\lambda }\right)=0 \\ \begin{array}{cl}\Rightarrow & -7\mathrm{\alpha }+21\mathrm{\lambda }=0\\ \Rightarrow & \mathrm{\alpha }=3\mathrm{\lambda }\end{array} \end{gathered}$$

Again, $\mathrm{\alpha }$ is root of  ${\mathrm{x}}^2-\mathrm{x}+2\mathrm{\lambda }=0$

$\begin{array}{ll}\therefore & {\mathrm{\alpha }}^2-\mathrm{\alpha }+2\mathrm{\lambda }=0\\ \Rightarrow & (3\mathrm{\lambda }{)}^2-3\mathrm{\lambda }+2\mathrm{\lambda }=0\\ \Rightarrow & 9{\mathrm{\lambda }}^2-\mathrm{\lambda }=0\\ \Rightarrow & \mathrm{\lambda }(9\mathrm{\lambda }-1)=0\\ \Rightarrow & \mathrm{\lambda }=0,\frac{1}{9}\\ \Rightarrow & \mathrm{\lambda }=\frac{1}{9} [\because \mathrm{\lambda }\ne 0]\\ \therefore & \mathrm{\alpha }=3\mathrm{\lambda }=3\times \frac{1}{9}=\frac{1}{3}\end{array}$

Again, $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ are roots of the equation

$\begin{array}{cl} & {\mathrm{x}}^2-\mathrm{x}+2\mathrm{\lambda }=0\\ \therefore & \mathrm{\alpha }+\mathrm{\beta }=\frac{-(-1)}{1}=1\\ \Rightarrow & \mathrm{\beta }=1-\mathrm{\alpha }=1-\frac{1}{3}=\frac{2}{3}\end{array}$

And $\mathrm{\alpha } \mathrm{and} \mathrm{\gamma }$ are the roots of the equation $3{\mathrm{x}}^2-10\mathrm{x}+27\mathrm{\lambda }=0$

$\begin{array}{l}\therefore \mathrm{\alpha }+\mathrm{\gamma }=\frac{-(-10)}{3}=\frac{10}{3}\\ \Rightarrow \mathrm{\gamma }=\frac{10}{3}-\mathrm{\alpha }=\frac{10}{3}-\frac{1}{3}=\frac{9}{3}=3\\ \therefore \frac{\mathrm{\beta \gamma }}{\mathrm{\lambda }}=\frac{\left(\frac{2}{3}\right)\times \left(3\right)}{\left(\frac{1}{9}\right)}=18\end{array}$