Solving the cubic without complex numbers

cubicsreal numbersroots

I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.

The general equation can in all cases be reduced to the depressed form,

$$x^3+px+q=0.$$

By a suitable scaling of the variable, this can be further reduced to

$$x^3-\frac34x-\frac r4=0$$ where $r>0$, or

$$4x^3-3x=r.$$

Then depending on the magnitude of $r$, we write

$$4\cos^3t-3\cos t=\cos3t=r,$$ $$x=\cos\frac{\arccos(r)+2k\pi}3$$

or
$$4\cosh^3t-3\cosh t=\cosh3t=r,$$ $$x=\cosh\frac{\text{arcosh}(r)}3.$$

This correctly handles the cases of $1$ and $3$ real roots.

Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?

Best Answer

When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.

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