If $\lambda$ be the ratio of the roots of the quadratic equation in $x , 3 \mathrm { m } ^ { 2 } \mathrm { x } ^ { 2 } + \mathrm { m } ( \mathrm { m } - 4 ) \mathrm { x } + 2 = 0 ,$ then the least value of $\mathrm { m }$ for which $\lambda + \dfrac { 1 } { \lambda } = 1 ,$ is :

Given lambda + 1/lambda = 1, we have lambda^2 - lambda + 1 = 0, which implies lambda is a complex cube root of -1. For a quadratic ax^2+bx+c=0, the ratio of roots lambda = alpha/beta satisfies (alpha+beta)^2 / (alpha*beta) = (lambda+1)^2 / lambda = (lambda^2+2lambda+1)/lambda = (lambda+1+2lambda+1)/lambda = (3lambda)/lambda = 3. Using the equation coefficients, (b/a)^2 / (c/a) = b^2/ac = 3. Here, m^2(m-4)^2 / (3m^2 * 2) = 3. Solving (m-4)^2 / 6 = 3 gives (m-4)^2 = 18, so m-4 = +/- 3*sqrt(2). m = 4 +/- 3*sqrt(2). The least value is 4 - 3*sqrt(2).