Mathematics - Quadratic Equation Question with Solution | TestHub

Satisfying $\alpha$ we have,

$3{\alpha }^2-10\alpha +27\lambda =0 \dots \left(\mathrm{i}\right)$ and
$3{\alpha }^2-3\alpha +6\lambda =0 \dots \left(\mathrm{ii}\right)$
Subtract

$-7\alpha +21\lambda =0$
$3\lambda =\alpha$
by $\left(\mathrm{ii}\right)$, $9{\lambda }^2-3\lambda +2\lambda =0$
$\Rightarrow \lambda =0,\frac{1}{9} \text{but} \lambda \ne 0$
$\therefore$ Given equation are $x^2-x+\frac{2}{9}=0$ and $3x^2-10x+3=0$
$\therefore \alpha =\frac{1}{3},\beta =\frac{2}{3},\alpha =\frac{1}{3},\gamma =3$
$\therefore \frac{\beta \gamma }{\lambda }=\frac{\frac{2}{3}·3}{\frac{1}{9}}=18$