$\frac{p^3}{27}<-\frac{q^2}{4}$ implies $x^3+px+q$ has 3 distinct real roots

Question

This is my first homework in Galois Theory, and my professor seems to be going for a style that is pretty concrete; for example on the first day he derived Cardano's Formula for cubics, he spent a couple hours justifying Viete's Formula, and he only has just now mentioned fields.

I've been asked to show that $x^3+px+q\in\mathbb R[x]$ has three distinct real roots if $\frac{p^3}{27}+\frac{q^2}{4}$ is negative. After fiddling around with it for a while, I am just generally confused by this question. Is this supposed to be a lot of computation? The discriminant looks pretty bad, Viete only seems to be marginally useful, so I guess I'm supposed to use the Cardano derivation?

Answer 1

Note that the derivative $3x^2+p$ is always non-negative if $p\ge 0$, so if $p\ge 0$ there is exactly one real root (possibly a multiple root).

So we want $p$ negative, say $p=-a$. Our function $x^3-ax+q$ is increasing up to $-(a/3)^{1/2}$, then decreasing up to $(a/3)^{1/2}$, then increasing.

There can be at most one real root in each of these intervals. So there are three distinct real roots precisely if our function is positive at $-(a/3)^{1/2}$ and negative at $(a/3)^{1/2}$. That happens if the product of the function values at these two points is negative.

Now substitute. The function value at $-(a/3)^{1/2}$ is $\frac{2a^{3/2}}{3\sqrt{3}}+q$, and the function value at $(a/3)^{1/2}$ is $-\frac{2a^{3/2}}{3\sqrt{3}}+q$.

The product of the function values is $-\frac{4a^3}{27} +q$. The condition that the product is $\lt 0$ is equivalent to $\frac{p^3}{27} +\frac{p^2}{4}\lt 0$.