Question

If $\alpha$ and $ \beta$ be the values of x in $m^2 (x^2 - x) + 2mx + 3 = 0$ and $m_1$ and $m_2$ be two values of m for which $\alpha$ and $ \beta$ are connected by the relation $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{3}.$ Then the value of $\frac{m^{2}_{1}}{m_{2}}+\frac{m^{2}_{2}}{m_{1}}$ is

• A. $6$
• B. $68$
• C. $\frac{3}{68}$
• D. $-\frac{68}{3}$

Answer 1

Answer: (D)

The given equation is $m^2 x^2 + (2m - m^2 ) x + 3 = 0$
$\therefore \alpha + \beta = \frac{2m-m^{2}}{m^{2}} = \frac{m-2}{m}$ and $\alpha\beta = \frac{3}{m^{2}}$
Now $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{3} \Rightarrow \frac{\alpha ^{2} + \beta^{2} }{\alpha \beta } = \frac{4}{3} \Rightarrow \frac{\left(\alpha + \beta \right)^{2}- 2\alpha \beta }{\alpha\beta } = \frac{4}{3}$
Substituting the values, we get
$\frac{\left(\frac{m-2}{m}\right)^{2}-2. \frac{3}{m^{2}}}{\frac{3}{m^{2}}} = \frac{4}{3}$
$\Rightarrow \frac{m^{2}-4m+4-6}{3} = \frac{4}{3} \Rightarrow m^{2}-4m-6 = 0$
$m_{1}$ and $m_{2}$ are roots of this equation, therefore
$m_{1}+ m_{2} = 4$ and $m_{1}m_{2} = -6$
The given expression is, $\frac{m^{2}_{1}}{m_{2}} + \frac{m^{2}_{2}}{m_{1}} = \frac{m^{3}_{1} + m^{3}_{2}}{m_{1}m_{2}}$
$ = \frac{\left(m_{1}+m_{2}\right)^{3}-3m_{1}m_{2}\left(m_{1}+m_{2}\right)}{m_{1}m_{2}}$
$= \frac{\left(4\right)^{3}-3.\left(-6\right).\left(4\right)}{-6} = -\frac{68}{3}$