General formulae for the roots of a cubic equation

[Gabriel Maia]

Hi everyone. I'm sorry for the long thread. If you don't want to read all the introductory stuff I will mark the part towards the end where my questions are located. I'm trying to find the general formulae for the roots of the equation

$$ax^3 + bx^2 + cx + d = 0$$

By using some changes of variable (which does not really matter now) I was able to rewrite this equation as

$$z^3 - \frac{\Delta_{_{0}}}{3a^2}z+\frac{\Delta_{_{1}}}{27a^3} = 0$$

I used then the Vieta's substitution

$$z = w + \frac{\Delta_{_{0}}}{9a^2w}$$

to obtain the following:

$$(w^3)^2 + \frac{\Delta_{_{1}}}{27a^3}w^3 + \frac{\Delta_{_{0}}^{3}}{729a^6} = 0$$

This is a quadratic equation on $w^3$ with roots:

$$ w^3 = -\frac{1}{27a^3}\,\left(\frac{\Delta_{_{1}} \mp \sqrt{\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3}}}{2}\right)$$

Now, if $\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3} = 0$, then $w$ has three possible values for each sign and all these six values are the same.

If $\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3} > 0$ there are three distinct values for each sign in $w$, making a total of six distinct roots.

Finally, if $\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3} < 0$ we will have, I think, three different complex roots for each sign in $w$, right?
${\Large AND \,\,\, NOW, \,\,\, FOR \,\,\, THE \,\,\, REAL \,\,\, QUESTIONS:}$
- In the case where $\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3} > 0$ there is one obvious root, but how can I find the others?

- In the case where $\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3} < 0$ there is, again, one obvious root, but how can I find the others? And more... how can I be sure they're all different from each other?- If there are at least two distinct roots for each possible sign of $w$ we have a total of four distinct values. But since each value of $w$ gives one value of $z$, because they are related by

$$z = w + \frac{\Delta_{_{0}}}{9a^2w},$$

I would have four distinct values of $z$ but it is the root of a cubic equation so there should be at most three distinct values. What is going on here? Thank you very much.

 

[SteamKing]

Since the cubic has an odd power, you are going to have one real root and a pair of complex conjugate roots or three real roots. Under some conditions, the three real roots may all be identical.

For more info on the roots of a cubic:

https://en.wikipedia.org/wiki/Cubic_function

 

[Gabriel Maia]

Since the cubic has an odd power, you are going to have one real root and a pair of complex conjugate roots or three real roots. Under some conditions, the three real roots may all be identical.

For more info on the roots of a cubic:

https://en.wikipedia.org/wiki/Cubic_function

I've read the wikipedia article but it's of no help. It mostly throw things at you but doesn't say how to obtain the distinct roots and neither does it helps with the problem of having too many roots.

 

[mathman]

Since the cubic has an odd power, you are going to have one real root and a pair of complex conjugate roots or three real roots. Under some conditions, the three real roots may all be identical.

For more info on the roots of a cubic:

https://en.wikipedia.org/wiki/Cubic_function

This assumes all the coefficients are real.

 

[NascentOxygen]

Where you substitute z with $w+\frac 1 w$ you double the degree of the polynomial and your cubic becomes of degree 6. The equation in w can have a solution that is not shared by the original cubic in z, e.g., being of degree 6 it may have 4 real solutions plus a complex conjugate pair.