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Math Help - deciding if roots of a polynomial are real or imaginary

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    deciding if roots of a polynomial are real or imaginary

    the roots of (x-41)^49 + (x-49)^41 -(x-2009)^2009 = 0 are

    A. All necessarily real
    B. non real except three positive real roots
    C. non real except one positive real roots
    D. non real except for three real roots of which exactly one is positive
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  2. #2
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    Re: deciding if roots of a polynomial are real or imaginary

    Quote Originally Posted by Raj917 View Post
    the roots of (x-41)^49 + (x-49)^41 -(x-2009)^2009 = 0 are

    A. All necessarily real
    B. non real except three positive real roots
    C. non real except one positive real roots
    D. non real except for three real roots of which exactly one is positive
    My guess is that the answer is C (only one real root, which is positive), but I don't have a proof of that.

    Let f(x) = (x-41)^{49} + (x-49)^{41} -(x-2009)^{2009}. If -\infty<x\leqslant49 then the positive term -(x-2009)^{2009} massively outweighs the other two terms, so that f(x)>0. For 49<x<2009 all three terms are positive, so again f(x)>0.

    At x=2009 the function is increasing. It looks very much as though it then has a single maximum as x increases, after which it decreases monotonically to -\infty, thereby crossing the x-axis just once and giving rise to the single real root of the function.

    I am sure that there must be a more convincing way of attacking this problem. It should presumably take advantage of the fact that 2009 = 41\times49, but I cannot see any way to make use of that fact.
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