From equation (i)
$$
l=-(m+n)
$$
Putting in equation (ii), we get
$$
\begin{aligned}
& & m n+2(m+n) n-(m+n) m & =0 \\
\Rightarrow & & m n+2 m n+2 n^2-m^2-n m & =0 \\
\Rightarrow & & 2 n^2-m^2+2 m n & =0 \\
\Rightarrow & & 2\left(\frac{n}{m}\right)^2+\frac{2 n}{m}-1 & =0
\end{aligned}
$$

[where $\frac{n_1}{m_1}, \frac{n_2}{m_2}$ are the roots of the equation]
From equaiton (i)
$$
m=-(n+l)
$$
Putting in equation (ii), we get
$$
\begin{aligned}
& & -(n+l) n-2 \ln -l(n+l) & =0 \\
\Rightarrow & & n^2+\ln +2 \ln +\ln +l^2 & =0 \\
\Rightarrow & & l^2+3 \ln +n^2 & =0 \\
\Rightarrow & & \left(\frac{l}{n}\right)^2+\frac{3 l}{n}+1 & =0
\end{aligned}
$$

[where $\frac{l_1}{n_1}, \frac{l_2}{n_2}$ are the roots of the equation] From equations (iii) and (iv), we get