In a physics textbook I have encountered a cubic equation of the form:
$$Ax^3-Bx+C=0$$
The book states that there exists a positive solution $x>0$ to this equation if and only if the following inequality holds:
$$\sqrt{A}<\dfrac{2}{3\sqrt{3}}\dfrac{B^{3/2}}{C}$$
I need some help figuring out the mathematical reasoning behind this assumption. Where does such an obscure-looking condition come from and how can I deduce it theoretically?
On
As $f(-\infty)=-\infty$ and $f(0)>0$, the polynomial has at least one negative root. As it is strictly concave on $(-\infty, 0)$, it cannot have three negative roots. Hence it has a positive root if and only if it has three real roots, that is to say if and only if its discriminant $4 A B^3-27 A^2 C^2$ is positive.
Denote the polynomial by $f(x)$ and note that with $A, B, C > 0$, $f(0) = C > 0$ and $f(x) \sim Ax^3 > 0$ for large $x$. Therefore $f$ has a positive root iff it has a local minimum with $x_0 > 0$ and $f(x_0) \le 0$. So $f'(x_0) = 3Ax_0^2 - B = 0 \implies x_0 = \sqrt{B/3A}$. Then $f''(x_0) = 6Ax_0 > 0$ so it is automatically a local minimum, therefore the condition reduces to $f(\sqrt{B/3A}) \le 0$. Plugging in, $$\frac{B^{3/2}}{3\sqrt{3A}} - \frac{B^{3/2}}{\sqrt{3A}} + C \le 0 \iff \sqrt{A} \le \frac{2B^{3/2}}{3\sqrt{3}C}.$$