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Root question

  1. Nov 30, 2011 #1
    My friends teacher posted a sample problem saying find the real roots of x^6 + 1. Is this a trick question? All roots for this function are imaginary right?
     
  2. jcsd
  3. Nov 30, 2011 #2
    Yep!

    Consider the fact that [itex]x^6 + 1 = 0 \Rightarrow x^6 = -1 \Rightarrow x = (-1)^{1/6}[/itex]. If you write [itex]-1 = e^{i(\pi + 2\pi n)}[/itex], then this is simply [itex]e^{i(\pi/6 + n\pi/3)} = \cos(\pi/6 + n\pi/3) + i\sin(\pi/6 + n\pi/3)[/itex] for n = 0, 1, ..., 5. If there were any real roots, then the imaginary part of this would be zero; i.e., we would have [itex]\pi/6 + n\pi/3 = m\pi[/itex] for some integer m. But since there are only finitely many values of n, you can just do this manually. We have:

    [itex]\pi/6, \pi/2, 5\pi/6, 7\pi/6, 3\pi/2, 11\pi/6[/itex]

    none of which would cancel the imaginary part.
     
  4. Nov 30, 2011 #3
    If you've taken calculus (and I assume you have if you're on this board), another way to see it would be to note that since this is an even degree polynomial, the limit as [itex]x\rightarrow \pm\infty[/itex] will be the same. In this case, it is positive infinity. Now find the minimum of this function by taking the derivative ([itex]6x^5[/itex]) and setting it equal to zero. Then you can see that the only critical point is at x = 0, which corresponds to f(0) = 1 in the original function. You can perform a derivative test if you want to verify that it's a minimum, but graphically you can see that it is. Since the minimum is y = 1, there are no roots.
     
  5. Nov 30, 2011 #4

    HallsofIvy

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    Well, not "imaginary", but "complex". the only "imaginary" roots of [itex]x^6+ 1= 0[/itex] is are i and -i. The other for are non-real complex numbers.
     
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