If λ be the ratio of the roots of the quadratic equation in x, 3 m...

Let $\alpha and$\beta$are the roots of given equation$3{m}^{2}{x}^{2} + m\left(m - 4\right)x + 2 = 0$.</p><p>$∴ \alpha + \beta = - \dfrac{m\left(m - 4\right)}{3{m}^{2}} \text{$and$} \alpha\beta = \dfrac{2}{3{m}^{2}} . . . . . . . . . \left(i\right)$</p><p>Given$\dfrac{\alpha}{\beta} = \lambda \text{$and$} \lambda + \dfrac{1}{\lambda} = 1$</p><p>$⇒ \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = 1$</p><p>$⇒ {\left(\alpha + \beta\right)}^{2} = 3\alpha\beta$</p><p>Using equation \left(i\right)$, we get

${\left(\dfrac{- m\left(m - 4\right)}{3{m}^{2}}\right)}^{2} = \dfrac{3 \times 2}{3{m}^{2}}$

$⇒ m = 4 \pm 3\sqrt{2}$

Hence, the least value of $m$ is $4 - 3\sqrt{2}$.