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 $\frac{m_1^2}{m_2}+\frac{m_2^2}{m_1}$ 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, $\frac{m_1^2}{m_2}+\frac{m_2^2}{m_1}=\frac{m_1^3+m_2^3}{m_1{m}_2}$
$=\frac{{\left(m_1+m_2\right)}^3-3m_1{m}_2\left(m_1+m_2\right)}{m_1{m}_2}$
$=\frac{{\left(4\right)}^3-3.\left(-6\right).\left(4\right)}{-6}=-\frac{68}{3}$