Theory of Equations 1 Question 6
7. 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
(2019 Main, 12 Jan I)
(a) $-2+\sqrt{2}$
(b) $4-2 \sqrt{3}$
(c) $4-3 \sqrt{2}$
(d) $2-\sqrt{3}$
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Solution:
- Let the given quadratic equation in $x$,
$3 m^{2} x^{2}+m(m-4) x+2=0, m \neq 0$ have roots $\alpha$ and $\beta$, then
$$ \alpha+\beta=-\frac{m(m-4)}{3 m^{2}} \text { and } \alpha \beta=\frac{2}{3 m^{2}} $$
Also, let $\quad \frac{\alpha}{\beta}=\lambda$
Then, $\quad \lambda+\frac{1}{\lambda}=1 \Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=1$
(given)
$\Rightarrow \quad \alpha^{2}+\beta^{2}=\alpha \beta$
$\Rightarrow \quad(\alpha+\beta)^{2}=3 \alpha \beta$
$\Rightarrow \quad \frac{m^{2}(m-4)^{2}}{9 m^{4}}=3 \frac{2}{3 m^{2}}$
$\Rightarrow \quad(m-4)^{2}=18$
$[\because m \neq 0]$
$\Rightarrow \quad m-4= \pm 3 \sqrt{2}$
$$ \Rightarrow \quad m=4 \pm 3 \sqrt{2} $$
The least value of $m=4-3 \sqrt{2}$
