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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

Hans Lundmark
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thale00
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  • @DietrichBurde: this is not a case of "All Roots Real, and Two Equal", is it ? –  Dec 17 '19 at 10:08

1 Answers1

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$$\frac{-1\pm i\sqrt 3}2$$ 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 $\left(x\mp\sqrt{-\dfrac p3}\right)^2$.

  • Thank you Yves. But I don't got the mathematical intention to multiply with the cubic roots of unity? – thale00 Dec 17 '19 at 12:30
  • @thale00: $\sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}$. –  Dec 17 '19 at 12:38
  • And when I am doing that I will get the other two roots? Why? And how can be the first one real? there is a (-q/2)^(1/3) ... – thale00 Dec 17 '19 at 13:13