how one can prove that real roots, which look non-real when using Cardano's formula.

Question

how one can prove that real roots, which look non-real when using Cardano's formula, can be shown to be real.

Answer 1

You show that when the two complex quantities you add up are rendered into polar form, they are complex conjugates.

As an example consider the equation $x^3-7x+6=0$. Of course the roots are $1,2,-3$ but what do they look like with the Cardano method?

$x=w-(-7)/3w=w+(7/3w), 27w^6+162w^3+343=0$

$w^3=-3+(\sqrt{100/27})i$

Either root for the quadratic equation in $w^3$ will ultimately yield the same sums.

Convert to polar form:

$w^3=\sqrt{343/27}(\cos\theta+i\sin\theta)$

$\theta\in\{\cos^{-1}(-3\sqrt{27/343}),\cos^{-1}(-3\sqrt{27/343})+2\pi,\cos^{-1}(-3\sqrt{27/343})+4\pi\}$

The three values of $\theta$ are identical for evaluating $w^3$, but each one will give different cube roots for $w$ itselfcwhencee dividevthe argument by $3$. In what follows we shall assume that $\theta$ takes on each value from the above set in turn.

So we have

$w=\sqrt{7/3}(\cos(\theta/3)+i\sin(\theta/3))$

$7/3w=(7/3\sqrt{7/3})(\cos(-\theta/3)+i\sin(-\theta/3))=\sqrt{7/3}(\cos(\theta/3)-i\sin(\theta/3))$

and thus $x=w+(7/3w)$ is the sum of two complex conjugates, therefore real as claimed.

This trigonometric method was cutting edge ... in 1600, when Vieta published it.