When does this cubic equation have three distinct real roots?

When does this cubic equation have three distinct real roots?

Question

Prove, by considering the turning points on the graph $y=x^3-3a^2x+b$ or otherwise, that if the equation $x^3-3a^2x+b=0$ has three distinct real roots, then $4a^6>b^2$.

Suggestion

Where are the turning points of $y=x^3-3a^2x+b$, and how many are there?

What is the effect of $a$ being negative rather than positive? What happens when $a=0$?

What happens to the number of roots as we translate the curve in the $y$-direction?

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/dfSZ4gqk/width/600/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Solution

The stationary points of $y=x^3-3a^2x+b$ occur when $\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2-3a^2=0.$ We therefore have two stationary points, at $x=\pm a$, where $a\neq 0$.

If $a=0$, there is a single stationary point at $x=0$, and it’s impossible to get three distinct roots in this case. You could try to sketch some graphs of cubic functions to see this.

Therefore, taking

$a\neq 0$

, we find that the two stationary points are at

$$\begin{align*} x=a, y &= a^3-3a^3 + b = -2a^3+b \\ x=-a, y &= -a^3+3a^3 + b = 2a^3+b. \end{align*}$$

If the curve is to have three distinct roots, then the $y$-values of the stationary points must have opposite signs.

So EITHER $-2a^3+b >0\,\, \text{and} \,\, 2a^3+b < 0, \,\, \text{which gives} \,\, 2a^3 < b < -2a^3,$ OR $-2a^3+b < 0\,\, \text{and} \,\, 2a^3+b > 0, \,\, \text{which gives}\,\, -2a^3 < b < 2a^3.$

The first of these is only possible if $a$ is negative, while the second is only possible if $a$ is positive. In either case, we can combine these inequalities to get $b^2 < 4a^6$.

The graph below shows the situation if $a$ is positive.

Three different possibilities for the number of roots of $y=x^3-3a^2x+b$.