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

Answer 1

The correct option is A $4-3\sqrt{2}$

$3m^2{x}^2+m(m-4)x+2=0$

Let $\alpha ,\beta$ be the roots of quadratic equation
$\therefore \alpha +\beta =-\frac{m(m-4)}{3m^2},\alpha .\beta =\frac{2}{3m^2}$

As per the question, $\lambda =\frac{\alpha }{\beta }$
$\therefore \lambda +\frac{1}{\lambda }=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=1$

$\Rightarrow \frac{(\alpha +\beta {)}^2}{\alpha \beta }-2=1$

$\Rightarrow (\alpha +\beta {)}^2=3\alpha \beta$

$\Rightarrow \frac{m^2(m-4{)}^2}{9m^4}=\frac{2}{m^2}$

$\Rightarrow (m-4{)}^2=18$

$\Rightarrow m^2-8m-2=0$

$\therefore m=4\pm 3\sqrt{2}$

Hence, least value of $m=4-3\sqrt{2}$