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Solution to a complex cubic equations

  • #1

Homework Statement


Solve the equation ## z^3 + 6z = 20 ## (this was considered by Cardan in Ars
magna).

Homework Equations



Please see the 2nd attachement.

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...
 

Attachments

Answers and Replies

  • #2
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Homework Statement


Solve the equation ## z^3 + 6z = 20 ## (this was considered by Cardan in Ars
magna).

Homework Equations



Please see the 2nd attachement.

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...
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
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?
 

Attachments

  • #4
Ray Vickson
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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?
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
So can you point me where I am wrong and give me the solution!!!
 
  • #6
SteamKing
Staff Emeritus
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So can you point me where I am wrong and give me the solution!!!
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
Ray Vickson
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So can you point me where I am wrong and give me the solution!!!
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
I will point out your error but will not give you the complete solution.
Hey Ray Vickson, Thanks a lot man!!!
 

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