If λ be the ratio of the roots of the quadratic equation
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 :

  1. $2-\sqrt{3}$

  2. $4-3 \sqrt{2}$

  3. $-2+\sqrt{2}$

  4. $4-2 \sqrt{3}$


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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