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Real roots?

  1. Feb 18, 2005 #1
    In math class today, we were discussing quadratic residues, and one of the things that came up was the fact that

    [tex]
    x^n-a=0
    [/tex]

    has n roots.

    This just made me start thinking about real and complex roots. A question that I had was whether, given n>2, is it possible to have n real roots in the above eqn? If not, is there a simple proof to show why it is not possible?
     
  2. jcsd
  3. Feb 18, 2005 #2
    Take a = 0. One might count that as a trivial special case though...
     
  4. Feb 18, 2005 #3
  5. Feb 18, 2005 #4

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    Well, for
    [tex]x^n=a[/tex]
    and
    [tex]a>0[/tex]
    The roots will be of the form:
    [tex]\sqrt[n]{a} (\cos{\frac{k2\pi}{n} + i \sin\frac{k2\pi}{n}})[/tex]
    with [itex]k[/itex] ranging from [itex]1[/itex] to [itex]n[/itex].
    (You can check this for yourself by, for example, multiplying, if you like).
    Now, it's easy to see that this is real only if:
    [tex]\sqrt[n]{a}=0[/tex]
    or
    [tex]sin\frac{2k\pi}{n}=0 \Rightarrow \frac{k}{n} \in \{1,\frac{1}{2}\}[/tex]
    (There are other values, but they aren't possible for our range of k.)

    From there it's easy to see that for non-zero [itex]a[/itex] if [itex]n[/itex] is even, there will be 2 real roots, and when [itex]n[/itex] is odd, there will be one.
     
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