My question is in determining when the equation $x^3 = px+q$ has only one real solution.
For this question we are using the restriction $p,q \gt 0$
The trouble is that I am getting what at first appear as rather opposite answers from two different approaches, and I want to make sure the algebraic details are being followed correctly and are simplifying correctly so that they prove the same thing.
The first approach is a calculus based-approach.
Let us fix the value of the slope $p$. Then we know that we need to make sure our line only crosses the curve $y=x^3$ once: in the first quadrant. Thus, our line must not touch the curve in the third quadrant. So to start my goal is to find the line with slope $p$ that is tangent to $y=x^3$ in the third quadrant. Thus, we have
\begin{array}{rrr} &\frac{d}{dx} x^3 = &p \\ \implies &3x^2 = &p \\ \implies &x = &-\sqrt{\frac{p}{3}} \end{array}
As we want it to be tangent in the third quadrant, we throw in the negative sign.
Plugging in this particular $x$ into $y=x^3$ yields the tangent point to be $$\left( -\sqrt{\frac{p}{3}} ,-\frac{p^{3/2}}{3\sqrt{3}}\right)$$
Our line must also pass through this point, so we have that $$-\frac{p^{3/2}}{3\sqrt{3}} = p\left( -\sqrt{\frac{p}{3}} \right) + q$$ $$\implies q = \frac{2\sqrt{p^3}}{3\sqrt{3}}$$
So if we pick a value of $p$, the equation $x^3 = px+q$ will only have one real solution if we have $$q \gt \frac{2\sqrt{p^3}}{3\sqrt{3}}$$
The second method uses Cardano's formula for solving the equation $x^3 + px + q = 0$.
Because we brought the $px+q$ across the equal sign (as opposed to what I did for my first method) we now have that $p,q <0$
From Cardano's formula, we get that $$\frac{q^2}{4} + \frac{p^3}{27} \gt 0$$ $$\implies 27q^2 + 4p^3 \gt 0$$ $$\implies q^2 \gt \frac{-4p^3}{27}$$ $$\implies \left| q\right| \gt \frac{2\sqrt{-p^3}}{3\sqrt{3}}$$ $$\implies -q > \frac{2\sqrt{-p^3}}{3\sqrt{3}}$$ And since we had originally that $p,q < 0$ in this case, both methods are proving the same result, yes?
Thanks for any info or comments given
I found the several points of terminology in the question to be confusing and so won't comment on those. We can, however, solve this problem directly, and without invoking radicals, by appealing to the discriminant.
Specializing the formula for the discriminant of a general cubic to depressed cubics -- those of the form $x^3 - p x - q$ -- gives $$\Delta = 4 p^3 - 27 q^2,$$ and a real cubic has precisely one real root iff $\Delta < 0$.
In particular, after rearranging (and after accounting for the relabeling $(p, q) \mapsto (-p, -q)$ implicit in the second method in the original question), the answer this method produces agrees with the answer given by both methods there.