Question

If $\lambda$ be the ratio of the roots of the quadratic equation in $x$, $3m^2{x}^2+m(m-4)x+2=0,$ 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)

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