Let $\lambda \ne 0$ be in R. If $\alpha$ and $\beta$ are the roots of the
equation, x² - x + 2$\lambda$ = 0 and $\alpha$ and $\gamma$ are the roots of
the equation, $3{x^2} - 10x + 27\lambda = 0$, then ${{\beta \gamma } \over \lambda }$ is equal to:

$\alpha$ and $\beta$ are the roots of the <br>equation x² - x + 2$\lambda$ = 0 .....(1) <br><br>$\therefore$ $\alpha + \beta = 1,\,\alpha \beta = 2\lambda$ <br><br>$\alpha$ and $\gamma$ are the roots of <br>the equation, $3{x^2} - 10x + 27\lambda = 0$ ......(2) <br><br>$\therefore$ $\alpha + \gamma$ = ${{10} \over 3}$, $\alpha \gamma$ = ${{27\lambda } \over 3}$ = 9$\lambda$ <br><br>Multiplying equation (1) by 3 and subtracting form equation (2) we get <br><br>-7x + 21$\lambda$ = 0 <br><br>$\Rightarrow$ x = 3$\lambda$ <br><br>$\therefore$ $\alpha$ = 3$\lambda$ <br><br>As $\alpha$ is root of equation (1) so <br><br>$\alpha$² - $\alpha$ + 2$\lambda$ = 0 <br><br>$\Rightarrow$ 9$\lambda$² - 3$\lambda$ + 2$\lambda$ = 0 <br><br>$\Rightarrow$ $\lambda$ = ${1 \over 9}$ <br><br>$\Rightarrow$ $\alpha$ = 3$\times$ ${1 \over 9}$ = ${1 \over 3}$ <br><br>Also $\alpha \beta = 2\lambda$ = ${2 \over 9}$ <br><br>$\Rightarrow$ $\beta$ = ${2 \over 3}$ <br><br>Also $\alpha \gamma$ = 9$\lambda$ = 9$\times$ ${1 \over 9}$ = 1 <br><br>$\Rightarrow$ $\gamma$ = 3 <br><br>$\therefore$ ${{\beta \gamma } \over \lambda }$ = ${{{2 \over 3}.3} \over {{1 \over 9}}}$ = 18