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.
 A: Let $\beta = \sqrt[3]{2}$. Change variable to $y = \beta x$, $p = 6a\beta$ and $q = 6b\beta^2$. The problem is equivalent to:

For what $a > 0$ we can find a $b > 0$ so that the cubic equation
  $$y^3 + 3a y^2 + 3b y + 1 = 0\tag{*1a}$$
  has $3$ real roots.

Let $z = y+a$, $(*1a)$ is equivalent to
$$z^3 -3(a^2-b) z + (1 + 2a^3 - 3ab) = 0\tag{*1b}$$
In order for $(*1b)$ and hence $(*1a)$ to have $3$ real roots (counting multiplicity), the discriminant of $(*1b)$ has to be non-negative. More precisely,
$$-4(-3(a^2-b))^3 - 27(1 + 2a^3 - 3ab)^2 \ge 0
\quad\iff\quad
 4(a^2-b)^3 \ge (1 + 2a^3- 3ab)^2\tag{*2}
$$
Notice RHS$(*2)$ $\ge 0$, this imposes $a^2 \ge b$ as a necessary condition for $(*1b)$ to have $3$ real roots. 


*

*Case I ( $a < 1$ )
We have $1 - a^3 > 0$. This implies
$${\rm RHS} = (1 + 2a^3 - 3ab)^2 = ((1 - a^3) + 3a(a^2-b))^2 \ge (1-a^3)^2 > 0$$
This leads to $a^2 - b > 0$ and
$$4 (a^2-b)^3 = {\rm LHS} \ge {\rm RHS} \ge (3a(a^2-b))^2 \implies 4 (a^2-b) \ge 9a^2$$
which is clearly impossible. This means there is no $b$ which can make
$(*1b)$ having $3$ real roots.

*Case II ( $a = 1$ )
$(*2)$ reduces to $4(1-b)^3 \ge 9(1-b)^2$ and it has a unique solution $b = 1$.

*Case III ( $a > 1$ )
If we take $b = \frac{1+2a^3}{3a}$, we have RHS$(*2) = 0$ and 
$${\rm LHS}(*2) = 4\left(a^2 - \frac{1+2a^3}{3a}\right)^3 = 4\left(\frac{a^3-1}{3a}\right)^3 > 0$$
For such a $b$, $(*2)$ is satisfied and $(*1b)$ has $3$ distinct real roots.
Combine these, we can conclude:


*

*When $a < 1$, there is no $b$ which makes $(*1b)$ has $3$ real roots.

*When $a = 1$, there is a unique $b$ where $(*1b)$ has $3$ real roots
(counting multiplicity).
Namely $b = 1$ and $(*1a)$ has a triple root at $y = -1$.

*When $a > 1$, there is at least one $b$ which makes $(*1b)$ has $3$ distinct real roots.


Translate this to $p$, the desired range of $p$ is $[6\sqrt[3]{2},\infty)$ or 
$(6\sqrt[3]{2},\infty)$ depends on whether having $3$ real roots means
$3$ real roots (counting multiplicity) or $3$ distinct real roots.
A: 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 $(6\sqrt[3]{2}, +∞)$.
