Question

Let $p \& q$ be the two roots of the equation, $mx ^2+ x (2- m )+3=0$. 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

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