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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 :
Option 1)
$4-2\sqrt{3}$

Option 2)
$2-\sqrt{3}$
Option 3)
$4-3\sqrt{2}$
Option 4)
$-2+\sqrt{2}$

Answers (1)

best_answer

Roots of Quadratic Equation -

$\alpha = \frac{-b+\sqrt{b^{2}-4ac}}{2a}$

$\beta = \frac{-b-\sqrt{b^{2}-4ac}}{2a}$

- wherein

$ax^{2}+bx+c= 0$

is the equation

$a,b,c\in R,\: \: a\neq 0$

Sum of Roots in Quadratic Equation -

$\alpha +\beta = \frac{-b}{a}$

- wherein

$\alpha \: and\beta$ are root of quadratic equation

$ax^{2}+bx+c=0$

$a,b,c\in C$

Product of Roots in Quadratic Equation -

$\alpha \beta = \frac{c}{a}$

- wherein

$\alpha \: and\ \beta$ are roots of quadratic equation:

$ax^{2}+bx+c=0$

$a,b,c\in C$

Let roots of equation be $\alpha$ and $\beta$

Given

$\lambda +\frac{1}{\lambda }=1$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=1$

$\alpha ^{2}+\beta ^{2}=\alpha \beta$

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

$=>[\frac{-m(m-4)}{3m^{2}}]^{2}=3\cdot \frac{2}{3m^{2}}$

$m^{2}-8m-2=0$

$m=4\pm 3\sqrt 2$

So, least value of m is $4- 3\sqrt 2$ .

Option 1)

$4-2\sqrt{3}$

Option 2)

$2-\sqrt{3}$

Option 3)
$4-3\sqrt{2}$
Option 4)

$-2+\sqrt{2}$

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