Answer
$\pm \sqrt{\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m}}$
Work Step by Step
$m{{n}^{4}}-{{r}^{2}}p{{m}^{3}}-{{r}^{2}}{{n}^{2}}+p=0$
Solving the equation for $n$,
Let $u={{n}^{2}}$
$m{{u}^{2}}-{{r}^{2}}p{{m}^{2}}+p=0$
Now, using the quadratic formula, for $n$,
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute the values in the equation,
$a=m,b=-{{r}^{2}},c=-{{r}^{2}}p{{m}^{3}}+p$
Then, the equation is,
$\begin{align}
& u=\frac{-\left( -{{r}^{2}} \right)\pm \sqrt{{{\left( -{{r}^{2}} \right)}^{2}}-4\times m\left( -{{r}^{2}}p{{m}^{3}}+p \right)}}{2\times m} \\
& =\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m}
\end{align}$
Where $u={{n}^{2}}$
${{n}^{2}}=\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m}$
Squaring both sides,
$\begin{align}
& {{n}^{2}}=\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m} \\
& n=\pm \sqrt{\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m}} \\
\end{align}$
Thus, the value is $n=\pm \sqrt{\frac{{{r}^{2}}\pm \sqrt{{{r}^{4}}+4{{m}^{4}}{{r}^{2}}p-4mp}}{2m}}$.
