Why nth Degree Equations Have n Roots?

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SUMMARY

An nth degree polynomial, denoted as P(x), possesses exactly n roots in the complex number system, as established by the Fundamental Theorem of Algebra. This theorem asserts that every non-constant polynomial of degree n has n roots, counting multiplicity. For example, the polynomial x^2 + 1 = 0 has two complex roots, i and -i, while having no real roots. Additionally, polynomials of degree 0, such as f(x) = 5, have no roots, highlighting the unique case of degree -∞.

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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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