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Number of roots

  1. Apr 30, 2010 #1
    Why is it said that an equation of nth degree must possess 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.
  2. jcsd
  3. Apr 30, 2010 #2
    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 [itex]-\infty[/itex], 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.
  4. Apr 30, 2010 #3
    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.
  5. Apr 30, 2010 #4
    I guess [itex]x^{1/2} = 4[/itex] has half a solution!
  6. May 1, 2010 #5


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    Homework Helper

    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 [tex]x^{4}-1=0[/tex] to find the fourth roots of unity), or by using polar form ([tex]1 = cos 0 + i sin 0[/tex]) and get the same answers. I usually get a 'wow' moment from my students when I show them this.

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