I am following a method to solve a cubic equation, which is given by:
$x^3 + px + c = 0$
For the equation above they set:
$\epsilon = (\frac{c}{2})^2 + (\frac{p}{3})^3 $
and state that if $\epsilon > 0$, there is one real root and a pair of complex conjugate roots.
I don't understand where this $\epsilon$ has come from and how they classify the solutions based on it - can someone please explain?
Let $f(x)=x^3+px+c$. If $p\geqslant0$, then $f'(x)=3x^2+p$. Therefore, $f$ is strictly increasing and so the equation $f(x)=0$ has, at most, one real solution. But it has to have some solution, since $\lim_{x\to\pm\infty}f(x)=\pm\infty$. So, in this case, $x^3+px+c$ has one and only one real root, no matter the value of $\varepsilon$.
Now, suppose that $p<0$; then $f$ is strictly decreasing on$\left[-\sqrt{-\frac p3},\sqrt{\frac p3}\,\right]$ and strictly increasing on $\left(-\infty,-\sqrt{-\frac p3}\right]$ and on $\left[\sqrt{-\frac p3},\infty\right)$. It is easy to deduce that $f$ has one and only one real root if and only if $f\left(-\sqrt{-\frac p3}\right)<0$ or $f\left(\sqrt{-\frac p3}\right)>0$.
Now, note that\begin{align}f\left(\sqrt{-\frac p3}\right)>0&\iff-\frac p3\sqrt{-\frac p3}+p\sqrt{-\frac p3}+c>0\\&\iff\frac{2p}3\sqrt{-\frac p3}+c>0\\&\iff-\frac p3\sqrt{-\frac p3}<\frac c2\end{align}and that\begin{align}f\left(-\sqrt{-\frac p3}\right)<0&\iff\frac p3\sqrt{-\frac p3}-p\sqrt{-\frac p3}+c<0\\&\iff-\frac{2p}3\sqrt{-\frac p3}+c<0\\&\iff-\frac p3\sqrt{-\frac p3}<-\frac c2.\end{align}If $c\geqslant0$, then $-\frac p3\sqrt{-\frac p3}<-\frac c2$ cannot hold (recall that $p<0$) and$$-\frac p3\sqrt{-\frac p3}<\frac c2\iff-\left(\frac p3\right)^3<\left(\frac c2\right)^2\iff\varepsilon>0.$$And if $c\leqslant0$, then $-\frac p3\sqrt{-\frac p3}<\frac c2$ cannot possibly hold and$$-\frac p3\sqrt{-\frac p3}<-\frac c2\iff-\left(\frac p3\right)^3<\left(\frac c2\right)^2\iff\varepsilon>0.$$