Question

If $\lambda $ be the ratio of roots of the quadratic equation in x, $3{{m}^{2}}{{x}^{2}}+m\left( m-4 \right)x+2=0$, then find the least value of m, for which $\lambda +\dfrac{1}{\lambda }=1$?
(a) $2-\sqrt{3}$,
(b) $4-3\sqrt{2}$,
(c) $-2+\sqrt{2}$,
(d) $4-2\sqrt{3}$.

Answer 1

Hint: We start solving the problem by assuming variables for the roots of the given quadratic equation. We then use the sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ to get the sum and product of the roots of given quadratic equation. We then substitute these values in $\lambda +\dfrac{1}{\lambda }=1$ and make calculations which led us to a quadratic equation in m. We solve this quadratic equation to find the least value of m.

Complete step by step answer:
According to the problem, we have given that $\lambda $ is the ratio of roots of the quadratic equation in x, $3{{m}^{2}}{{x}^{2}}+m\left( m-4 \right)x+2=0$. We need to find the least value of m, given that $\lambda +\dfrac{1}{\lambda }=1$.
Let us assume $\alpha $, $\beta $ be the roots of the quadratic equation $3{{m}^{2}}{{x}^{2}}+m\left( m-4 \right)x+2=0$ ---(1).
We know that the sum and product of the roots of the equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$. Using this fact for the equation (1), we get.
$\Rightarrow \alpha +\beta =\dfrac{-m\left( m-4 \right)}{3{{m}^{2}}}$.
$\Rightarrow \alpha +\beta =\dfrac{-{{m}^{2}}+4m}{3{{m}^{2}}}$ ---(2).
$\Rightarrow \alpha \beta =\dfrac{2}{3{{m}^{2}}}$ ---(3).
According to the problem we have $\lambda +\dfrac{1}{\lambda }=1$ and $\lambda $ is the ratio of roots of the quadratic equation $3{{m}^{2}}{{x}^{2}}+m\left( m-4 \right)x+2=0$. So, we have $\lambda =\dfrac{\alpha }{\beta }$.
$\Rightarrow \dfrac{\alpha }{\beta }+\dfrac{1}{\dfrac{\alpha }{\beta }}=1$.
$\Rightarrow \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=1$.
\[\Rightarrow \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=1\].
\[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=\alpha \beta \].
\[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =\alpha \beta +2\alpha \beta \].
\[\Rightarrow {{\left( \alpha +\beta \right)}^{2}}=3\alpha \beta \].
From equation (2) and (3) we get,
\[\Rightarrow {{\left( \dfrac{-{{m}^{2}}+4m}{3{{m}^{2}}} \right)}^{2}}=3\left( \dfrac{2}{3{{m}^{2}}} \right)\].
\[\Rightarrow \dfrac{{{\left( -{{m}^{2}}+4m \right)}^{2}}}{{{\left( 3{{m}^{2}} \right)}^{2}}}=\left( \dfrac{2}{{{m}^{2}}} \right)\].
\[\Rightarrow \dfrac{{{\left( -{{m}^{2}} \right)}^{2}}+{{\left( 4m \right)}^{2}}+2\left( -{{m}^{2}} \right)\left( 4m \right)}{9{{m}^{4}}}=\left( \dfrac{2}{{{m}^{2}}} \right)\].
\[\Rightarrow \dfrac{{{m}^{4}}+16{{m}^{2}}-8{{m}^{3}}}{9{{m}^{2}}}=2\].
\[\Rightarrow {{m}^{2}}-8m+16=2\times 9\].
\[\Rightarrow {{m}^{2}}-8m+16=18\].
\[\Rightarrow {{\left( m-4 \right)}^{2}}=18\].
\[\Rightarrow \left( m-4 \right)=\pm \sqrt{18}\].
\[\Rightarrow m-4=\pm 3\sqrt{2}\].
\[\Rightarrow m=4\pm 3\sqrt{2}\].
The least value of m is $4-3\sqrt{2}$, as 4 is a positive number and subtracting it will reduce its value.
So, we have found the least value for m as $4-3\sqrt{2}$.
∴ The least value of m is $4-3\sqrt{2}$.

So, the correct answer is “Option B”.

Note: Whenever we get this type of problems, we first need to start by assuming the variables for roots. We need to know that whenever a number is subtracted from a number, then the resultant number moves back in the number line. We can also solve the problem by finding the roots of the equations in m and then substitute in $\lambda +\dfrac{1}{\lambda }=1$ to get the required answer.