# Solution to a complex cubic equations

1. Dec 29, 2014

### PcumP_Ravenclaw

1. The problem statement, all variables and given/known data
Solve the equation $z^3 + 6z = 20$ (this was considered by Cardan in Ars
magna).

2. Relevant equations

3. The attempt at a solution

I want to know if my solution is correct because the book (2nd attachment) says that there should only be 3 distinct solutions. I get 4 distinct solutions. My method is also slightly different from the book because I dont understand v and v1 etc...

Please explain to me how to solve cubic equations by the method explained in this attachment only. I don't want to solve this by euler's or any other method.

For my attempt please see the 1st attachment

Danke...

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2. Dec 29, 2014

### Staff: Mentor

A cubic can have at most three distinct solutions. I looked at your work, but it is so small, and the written part is so faint that I couldn't read it. It would be better to post the actual work as text here rather than as an image, especially one that is illegible.

3. Dec 29, 2014

### PcumP_Ravenclaw

I have rewritten the first attachement in latex format below.

$z^3 + 6Z = 20$

Notice carefully! this equation is already in the form $z^3 + 3bZ - C = 0$ so the first substitution of $P(z - a/3)$ as given in the text in the 2nd attachemtn is not necssary.

3bZ = 6Z
b = 2

$(Z - 2/Z)^3 + 6(Z - 2/Z) - 20 = 0$

$(Z^2 - 4 + 4/Z^2)(Z - 2/Z) + 6Z - 12/Z - 20 = 0$

after cancelling some terms

$Z^3 - 8/Z^3 - 20 = 0$

substituting
$y = Z^3$

$y^2 - 8 - 20y = 0$
solving the quadratic equation gives 20.392 or -0.392305
$y = 20.392 or -0.392305$

To find the original Z in the equation $z^3 + 6Z = 20$
we must undo the substitution $p(z - b/z)$

so $20.392 = Z - 2/Z$
solving this quadratic gives Z = 20.4896 or -0.09761044

so $-0.392305 = Z - 2/Z$
solving this quadratic gives Z = -1.62390451 or 1.2316

so there are 4 solutions for Z?

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4. Dec 29, 2014

### Ray Vickson

Theorem (The Fundamental Theorem of Algebra): a non-constant polynomial of degree $n$ has at most $n$ (complex) roots; if we count "multiplicity", it has exactly $n$ roots.

See, eg., http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

The hardest part of the theorem is the proof that a polynomial has at least one root. After that, the rest is easy, if we use a Factorization Lemma:
If $r$ is a root of a polynomial $p(x)$ of degree $n > 1$, then $p(x)$ is of the form $p(x) = (x-r) q(x)$, where $q(x)$ is a polynomial of degree $n-1$.

So, NO: your polynomial does not have 4 distinct roots.

5. Dec 29, 2014

### PcumP_Ravenclaw

So can you point me where I am wrong and give me the solution!!!

6. Dec 29, 2014

### SteamKing

Staff Emeritus
Why haven't you taken your 3 or 4 solutions for Z and plugged them back into the original equation, to see if these are actually solutions?

7. Dec 30, 2014

### Ray Vickson

I will point out your error but will not give you the complete solution.

You want to solve an equation $p(x) = 0$. You set $x = Z - 2/Z$, so your equation becomes $p(Z - 2/Z) = 0$. You find that the function $Q(Z) = p(Z - 2/Z)$ is quadratic in $Z^3$, so you can find its two roots, say $r_1= 10 + 3 \sqrt{3} \doteq 20.39230485$ and $r_2= 10 - 6 \sqrt{3} \doteq -0.39230485$.

After this point you make serious errors: you say that you must solve $Z - 2/Z = r_1$ and $Z - 2/Z = r_2$, but this is not true. Your roots $r_1, r_2$ are values of $Z^3$, so you need to solve $Z^3 = r_1$ and $Z^3 = r_2$. That will give you 6 values of $Z$; two of these are real and the other 4 are complex numbers.

Now find $x$ from $x = Z - 2/Z$. If you proceed carefully you will find that the seemingly 6 separate values of $x$ (corresponding to the 6 values of $Z$) actually collapse into 3 distinct values, each of which is duplicated twice. One of the $x$-values is real and the other two are complex.

Basically, the same type of thing happens for any cubic equation, not just for your special one.

8. Dec 30, 2014

### PcumP_Ravenclaw

Hey Ray Vickson, Thanks a lot man!!!