Question:
If $\lambda$ be the ratio of the roots of the quadratic equation in $x, 3 m^{2} x^{2}+m(m-4) x+2=0$, then the
least value of $m$ for which $\lambda+\frac{1}{\lambda}=1$, is :
Correct Option: , 2
Solution:
$3 m^{2} x^{2}+m(m-4) x+2=0$
$\lambda+\frac{1}{\lambda}=1, \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=1, \alpha^{2}+\beta^{2}=\alpha \beta$
$(\alpha+\beta)^{2}=3 \alpha \beta$
$\left(-\frac{m(m-4)}{3 m^{2}}\right)^{2}=\frac{3(2)}{3 m^{2}}, \frac{(m-4)^{2}}{9 m^{2}}=\frac{6}{3 m}$
$(m-4)^{2}=18, m=4 \pm \sqrt{18}, 4 \pm 3 \sqrt{2}$
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