Question

Let $p\&q$ be the two roots of the equation, $m{x}^2+x(2-m)+3=0$. Let $m_1,m_2$ be the two values of $m$ satisfying $\frac{p}{q}+\frac{q}{p}=\frac{2}{3}$. Determine the numerical value of $\frac{m_1}{m_2^2}+\frac{m_2}{m_1^2}$.

Answer 1

Answer: 99

$p+q=\frac{m-2}{m};pq=\frac{3}{m}$
now $m_1$ and $m_2$ satisfies $\frac{p}{q}+\frac{q}{p}=\frac{2}{3}\Rightarrow \frac{p^2+q^2}{pq}=\frac{2}{3}$
$\frac{(p+q{)}^2-2pq}{pq}=\frac{2}{3}$
${\left(\frac{m-2}{m}\right)}^2-\frac{6}{m}=\frac{2}{3}\cdot \frac{3}{m}=\frac{2}{m}\Rightarrow \frac{(m-2{)}^2}{m^2}=\frac{8}{m}$
$m^2-4m+4=8m\Rightarrow m^2-12m+4=0$
$\therefore m_1+m_2=12\text{ and }{m}_1{m}_2=4$
now $\frac{m_1}{m_2^2}+\frac{m_2}{m_1^2}=\frac{m_1^2+m_2^3}{{\left(m_1{m}_2\right)}^2}=\frac{{\left(m_1+m_2\right)}^3-3m_1{m}_2\left(m_1+m_2\right)}{{\left(m_1{m}_2\right)}^2}$
$=\frac{12^3-12\cdot 12}{16}=\frac{12^2\cdot 11}{16}=99$