on this page a nice guy presents a algorithm for solving cubic formulas.
I have some problems with the formula [13] in the chapter "All Roots Real, and Two Equal"
I can't follow him and don't know where the
-0,5 ± sqrt(3)/2 i
comes from.
Down below in chapter "One Real and Two Complex Roots" he did it again.
I have absolutely no clue where this complex number comes from not even with his written sentence. :D
Thank you a lot for helping me
$$\frac{-1\pm i\sqrt 3}2$$ are cubic roots of unity. As you can check,
$$\frac{(-1\pm i\sqrt 3)^3}8=\frac{-1\pm3i\sqrt3+3\cdot3\mp3i\sqrt3}8=1.$$
In the complex numbers, there are three cube roots, namely
$$\sqrt[3]z,\omega\sqrt[3]z,\omega^2\sqrt[3]z,$$ where $\omega$ is one of the above numbers.
Another method:
Two equal roots (i.e. a double root) occurs when both the function and its first derivative are zero.
$$x^3+px+q=3x^2+p=0,$$
so that
$$x=\pm\sqrt{-\frac p3}$$ with the compatibility condition $$\mp\frac p3\sqrt{-\frac p3}\pm p\sqrt{-\frac p3}+q=0,$$
or
$$4p^3+27q^2=0.$$
Now you can obtain the third root by long division of $x^3+px+q$ by $\left(x\mp\sqrt{-\dfrac p3}\right)^2$.