<s>[INST] Range of $p$ in cubic equation 
If $2x^3+px^2+qx+4=0$ has $3$ real roots, then what is the range of $p$? Here $p,q>0$.

My try: Let $\alpha,\beta,\gamma$ be the roots of $2x^3+px^2+qx+4=0$.$$
\begin{cases}
\alpha+\beta+\gamma=-\dfrac{p}{2}\\
\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{q}{2}\\
\alpha\beta\gamma=-2
\end{cases}
$$
Help me how to solve after that point.
 [/INST] Since $p, q > 0$, the real roots to the equation $2x^3 + px^2 + qx + 4 = 0$ can only be negative. Replacing $x$ by $-x$ and dividing by $x$, the equation is equivalent to $2x^2 - px + q - \dfrac{4}{x} = 0$, where $x > 0$.
Define $f(x) = 2x^2 - px + q - \dfrac{4}{x}\ (x > 0)$, then$$
f'(x) =4x - p + \frac{4}{x^2},\ f''(x) = 4 - \frac{8}{x^3}.
$$
Thus the minimum of $f'(x)$ is $f'(\sqrt[3]{2}) = 6\sqrt[3]{2} - p$.
Case 1: $0 < p \leqslant 6\sqrt[3]{2}$. In this case, because $f'$ is positive on $(0, \sqrt[3]{2})$ and $(\sqrt[3]{2}, +∞)$, respectively, then $f$ is strictly increasing on $(0, \sqrt[3]{2})$ and $(\sqrt[3]{2}, +∞)$, and thus strictly increasing on $(0, +∞)$. Therefore, $f$ has no more than one real root.
Case 2: $p > 6\sqrt[3]{2}$. Now take $q = \sqrt[3]{2} p$, then $f(\sqrt[3]{2}) = 0$. Because $f'(\sqrt[3]{2}) < 0$, then there exist $x_1 < \sqrt[3]{2} < x_2$ such that $f(x_1) > f(\sqrt[3]{2}) = 0 > f(x_2)$. Note that$$
\lim_{x → 0^+} f(x) = -∞,\ \lim_{x →+∞} f(x) = +∞,
$$
then there exist $0 < x_3 < x_1$ and $x_4 > x_2$ such that $f(x_3) < 0 < f(x_4)$. Therefore, $f$ has roots on $(x_3, x_1)$ and $(x_2, x_4)$, respectively. Therefore, $f$ has three real roots.
Hence the range of $p$ is $(6\sqrt[3]{2}, +∞)$.
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