Why are there 3 roots to a cubic equation?

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Discussion Overview

The discussion centers on the nature of the roots of cubic equations, specifically addressing why a cubic equation has three roots instead of six. Participants explore concepts related to complex numbers, polynomial roots, and the implications of conjugate pairs in the context of real cubic equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about why cubic equations have three roots instead of six, referencing the three cube roots of a complex number and the potential for positive and negative imaginary parts.
  • Another participant challenges the assumption that complex conjugates would lead to six roots, stating that the cube roots of a complex number do not imply that both a+ib and a-ib yield the same result when cubed.
  • A later reply clarifies that any nth degree polynomial can be expressed as the product of n linear terms, thus a cubic polynomial has at most three distinct solutions.
  • One participant notes that while a real cubic equation has one real root and two conjugate roots, this does not increase the total number of distinct roots beyond three.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial assumptions about the number of roots. There are multiple competing views regarding the interpretation of roots in the context of complex numbers and conjugates.

Contextual Notes

Some participants' arguments depend on specific interpretations of polynomial roots and the behavior of complex conjugates, which remain unresolved. The discussion also touches on the relevance of Galois theory, but its application is contested.

okkvlt
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I find complex numbers very fascinating. But i don't understand something.
Why does a cubic equation have 3 answers instead of 6?

I know that there are 3 cube roots of a complex number, and the imaginary part of the complex number can be either positive or negative, so there should be 6 answers. Actually, if each side of the formula was independent of the other, there would be 3*4=12 answers.

I tried reading an article on the internet about galois theory, but it used a lot of jargon about fields and groups that i didnt understand.

Could somebody explain to me why there are only 3 roots, without all the jargon?
 
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It is nought to do with Galois theory.

What you've written implies that you think that a+ib and a-ib cube to give the same number. That is clearly not true: just try it.

All you need to use is the euclidean algorithm for polynomials: if I have a polynomial f(x) and f(a)=0, then I can write f(x)=(x-a)g(x) where the degree of g(x) is one less than the degree of f(x).

Finally, why have you chosen cubic equations? Do you accept that quadratics have two roots, and if so why have you accepted that without question?
 
okkvlt said:
I find complex numbers very fascinating. But i don't understand something.
Why does a cubic equation have 3 answers instead of 6?

I know that there are 3 cube roots of a complex number, and the imaginary part of the complex number can be either positive or negative, so there should be 6 answers.
NO. "the imaginary part of the complex number can be either positive or negative" is true only for square roots of real numbers.

Actually, if each side of the formula was independent of the other, there would be 3*4=12 answers.

I tried reading an article on the internet about galois theory, but it used a lot of jargon about fields and groups that i didnt understand.

Could somebody explain to me why there are only 3 roots, without all the jargon?
In the complex numbers any nth degree polynomial can be written as the product of exactly n linear terms. In particular, any cubic polynomial can be written as the product of 3 linear terms. Setting each term equal to 0 will give at most 3 distince solutions.
 
conjugates of roots to a real cubic

okkvlt said:
I know that there are 3 cube roots of a complex number, and the imaginary part of the complex number can be either positive or negative, so there should be 6 answers.

Hi okkvlt! :smile:

You're obviously thinking that the conjugate of a real cubic equation is itself, and therefore the conjugates of its roots must also be roots.

That's correct! :smile:

But that doesn't mean that there are 6 answers … one root of a real cubic equation will always be real, and the other two will be conjugates of each other (whether real or not) …

so including the conjugates still leaves the total as three! :smile:
 

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