Question

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

• A. $99$
• B. $-99$
• C. $96$
• D. $0$

Answer 1

The correct option is A $99$

$m{x}^2+x(2-m)+3=0$
From the equation
$p+q=\frac{m-2}{m},pq=\frac{3}{m}$
Now,
$\frac{p}{q}+\frac{q}{p}=\frac{2}{3}\Rightarrow \frac{p^2+q^2}{pq}=\frac{2}{3}\Rightarrow \frac{(p+q{)}^2-2pq}{pq}=\frac{2}{3}\Rightarrow 3(p+q{)}^2-6pq=2pq\Rightarrow 3(p+q{)}^2=8pq$
On putting $p+q$ and $pq$, we get
$3\times {\left(\frac{m-2}{m}\right)}^2=\frac{24}{m}\Rightarrow m^2-4m+4=8m$
$\Rightarrow m^2-12m+4=0$
So,
$m_1+m_2=12$ and $m_1{m}_2=4$
Now,
$\frac{m_1}{m_2^2}+\frac{m_2}{m_1^2}=\frac{m_1^3+m_2^3}{(m_1{m}_2{)}^2}=\frac{(m_1+m_2{)}^3-3m_1{m}_2(m_1+m_2)}{(m_1{m}_2{)}^2}=\frac{{12}^3-12\times 12}{16}=\frac{{12}^2\times 11}{16}=99$