If be the ratio of the roots of 3m^2x^2+m(m-4)x+2=0, and +1=1, then the least value of m i…

Question

If $\lambda$ be the ratio of the roots of $3m^2x^2+m(m-4)x+2=0$, and $\lambda+\dfrac1\lambda=1$, then the least value of $m$ is:

Accepted Answer

$\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=1$, with $\alpha+\beta=\dfrac{4-m}{3m}$, $\alpha\beta=\dfrac2{3m^2}$:

$$\left(\frac{4-m}{3m}\right)^2\cdot\frac{3m^2}2-2=1 \implies (m-4)^2=18 \implies m=4\pm3\sqrt2.$$

Least value: $m=4-3\sqrt2.$

Solutions · 1