Question

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

• A. $ 2 - \sqrt{3}$
• B. $ 4 - 3 \sqrt{2}$
• C. $ - 2 + \sqrt{2}$
• D. $4 - 2 \sqrt{3}$

Answer 1

Answer: (B)

$3m^{2}x^{2} + m\left(m - 4\right) x + 2 = 0$
$ \lambda +\frac{1}{\lambda} = 1, \frac{\alpha}{\beta} +\frac{\beta}{\alpha} =1 , \alpha^{2} +\beta^{2} =\alpha\beta $
$ \left(\alpha+\beta\right)^{2} = 3\alpha\beta$
$ \left(- \frac{m\left(m-4\right)}{3m^{2}}\right)^{2} = \frac{3\left(2\right)}{3m^{2}} , \frac{\left(m-4\right)^{2}}{9m^{2}} = \frac{6}{3m} $
$ \left(m-4\right)^{2} = 18, m = 4\pm\sqrt{18} , 4\pm3\sqrt{2} $