# Cubic Discriminant Uses

The discriminant for the cubic equation $$ax^3+bx^2+cx+d=0$$ is

$$Δ​\:=b^2c^2−4ac^3−4b^3d−27a^2d^2+18abcd$$

And I am aware that you can determine the number of roots a cubic has using method shown below -

$$Δ​\:>0$$: the equation has three distinct real roots

$$Δ​\:=0$$: the equation has a repeated root and all its roots are real

$$Δ​\:<0$$: the equation has one real root and two non-real complex conjugate roots

But I was wondering if one could determine whether a cubic has rational or integer roots, as you can do with the discriminant for quadratics, and if so what the method would be.

I have noticed that with the cubics I have checked: if the discriminant is a perfect square there are 3 integer solutions, although I have not checked many and I am not sure of the reasoning behind it.

Any help would be greatly appreciated.

• This is a good question, but I think you meant something else when you wrote “cubic for quadratics” Commented May 19, 2020 at 17:49
• Even if you use the discriminant for quadratics, you cannot say wether the roots are integer or rational.. You also need to know the coefficients Commented May 19, 2020 at 18:01
• Yes sorry you are right I meant to say discriminant for quadratics Commented May 19, 2020 at 20:12
• And I meant just rational not integer or rational. Commented May 19, 2020 at 20:13

$$x^3 + x^2 - 2x - 1$$ has $$2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \; 2 \cos \frac{8 \pi}{7} \; , \; \;$$
more in a minute $$x^3 - 3x + 1$$ has $$2 \cos \frac{2 \pi}{9} \; , \; \; 2 \cos \frac{4 \pi}{9} \; , \; \; 2 \cos \frac{8 \pi}{9} \; , \; \;$$  $$x^3 + x^2 - 4x + 1$$ has $$2 \cos \frac{2 \pi}{13} + 2 \cos \frac{10 \pi}{13}\; , \; \; 2 \cos \frac{4 \pi}{13} + 2 \cos \frac{6 \pi}{13} \; , \; \; 2 \cos \frac{8 \pi}{13} +2 \cos \frac{12 \pi}{13} \; , \; \;$$