If λ be the ratio of the roots of the quadratic equation in x ,   3 m 2 x 2 + m m - 4 x + 2 = 0 ,…
If be the ratio of the roots of the quadratic equation in , then the least value of for which , is :
Solution
Let $\alpha$ and $\beta$ be the roots of the given equation $3m^2x^2+mx(m-4)+2=0$.
Then, $\alpha + \beta = -\frac{m(m-4)}{3m^2}$ and $\alpha\beta = \frac{2}{3m^2}$.
Given $\frac{\alpha}{\beta} = \lambda$ and $\lambda + $\frac{1}{\lambda}$ = 1$.
This implies $\frac{\alpha}{\beta} + $\frac{\beta}{\alpha}$ = 1$.
Also, $(\alpha + \beta)^2 = 3\alpha\beta$.
Using equation (i), we get $\left(\frac{-m(m-4)}{3m^2}\right)^2 = \frac{3\times 2}{3m^2}$.
Therefore, $m = 4 \pm 3\sqrt{2}$.
Hence, the least value of $m$ is $4-3\sqrt{2}$.