Why nth Degree Equations Have n Roots?

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Discussion Overview

The discussion revolves around the assertion that an nth degree polynomial equation must possess n roots, exploring the implications of this statement in the context of complex and real numbers, as well as addressing exceptions and nuances related to polynomial degrees.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the statement about nth degree equations having n roots, citing examples where lower degree equations yield fewer solutions.
  • Another participant clarifies that a polynomial of degree n has n complex roots, noting that the case of x^0 = 1 does not fit the standard degree classification.
  • A third participant references the Fundamental Theorem of Algebra, stating that every polynomial of degree n has exactly n roots when counting multiplicity over the complex numbers, while acknowledging that real roots may be fewer.
  • One participant humorously suggests that an equation like x^{1/2} = 4 has "half a solution," which is challenged as not being a polynomial.
  • A later reply expresses enthusiasm for the topic of nth roots of complex numbers, highlighting the methods of finding these roots and the educational impact it has on students.

Areas of Agreement / Disagreement

Participants express differing views on the nature of roots for polynomials of various degrees, with some agreeing on the Fundamental Theorem of Algebra while others raise exceptions and question the applicability of the degree concept in certain cases.

Contextual Notes

There are unresolved nuances regarding the classification of polynomials of degree 0 and the implications of complex versus real roots. The discussion also touches on the informal treatment of non-polynomial equations.

johncena
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Why is it said that an equation of nth degree must possesses n roots ?
if x^1 = y, x has only 1 value
x^2 = y, x has 2 values (the 2 values may be equal)
x^3 = y, x has 3 values
going on like this, we have, x^0 = 1 , implies x has no solutions. but x has infinite number of solutions.
 
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What people say when they say that an equation of degree n has n roots is that given a polynomial P(x) of degree n, it has n complex roots (where n is a non-negative integer). However x^0=1 so the polynomial in your last example is:
P(x) = x^0 - 1 = 1-1 = 0
This does not have degree 0. We often say that it has degree -\infty, but since it's the only polynomial with degree not a non-negative integer this is the single polynomial to which the rule does not apply.
 
Fundamental Theorem of Algebra: Every polynomial of degree n (n=/=0) has exactly n roots counting multiplicity over the Complex numbers. In the case of the real numbers, it has d roots where d is less than or equal to n.
For instance, the easiest example x^2+1=0 only has complex solutions, namely i and -i, but over the Reals, it has no roots.
Now to finish your question consider nonzero polynomials of degree 0, suggesting they are nonzero, they have no solutions. E.g., f(x)=5 is a polynomial of degree 0 and has 0 roots.
 
I guess x^{1/2} = 4 has half a solution!
 
g_edgar said:
I guess x^{1/2} = 4 has half a solution!
I'm assuming that you're joking, as this would not be a polynomial in the first place. The Fundamental Theorem of Algebra applies to non-constant single-variable polynomials with complex coefficients.

I don't know why, but finding nth roots of complex numbers is one of my favorite topics in teaching Pre-Calculus. I find it fascinating to see that you can find nth roots algebraically (like, for instance solving the equation x^{4}-1=0 to find the fourth roots of unity), or by using polar form (1 = cos 0 + i sin 0) and get the same answers. I usually get a 'wow' moment from my students when I show them this.


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