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I was told the discriminant of the cubic is

$$\Delta=-27q^2-4p^3$$

and that $\Delta>0$ means that there are three real roots. Simply put, why is this the discriminant? I ask this because, looking at Cardano's formula, I thought that we want everything inside the square root to be positive to get real roots(just as in quadratic cases).

Namely, $\frac{q^2}{4}+\frac{p^3}{27}>0$. Which is essentially, $27q^2+4p^3>0$. But the discriminant has a minus on it, and I don't see why. Does the derivation involve resultants and whatnot? Will it be rather complex?

I am wondering if there is a simple explanation as to why this is the case. Similarly, for the quartic, quintic discriminants...will I need to go through resultants for them? Or is there a simpler faster way to determine them?

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  • $\begingroup$ I suggest reading this page. $\endgroup$ – user307169 Apr 7 '16 at 15:44
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Your discriminant refers to a cubic equation reduced to the form $$ y^3+py+q=0 $$ and the Cardano formula says that the solutions are: $$ y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} $$

Note that if $$ \Delta=\frac{q^2}{4}+\frac{p^3}{27}>0 $$ the arguments of the cubic roots are real numbers, so the roots give only one real value ( and two complex values), so the other solutions of the equation must be complex numbers and we cannot have three real roots.

The proof that for $\Delta \le 0$ we find three real roots is not so simple, you can see here.

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