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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- Oct 22nd 2011, 07:29 AMRaj917deciding 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 - Oct 23rd 2011, 12:40 PMOpalgRe: deciding if roots of a polynomial are real or imaginary
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 $\displaystyle f(x) = (x-41)^{49} + (x-49)^{41} -(x-2009)^{2009}.$ If $\displaystyle -\infty<x\leqslant49$ then the positive term $\displaystyle -(x-2009)^{2009}$ massively outweighs the other two terms, so that $\displaystyle f(x)>0.$ For $\displaystyle 49<x<2009$ all three terms are positive, so again $\displaystyle f(x)>0.$

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 $\displaystyle -\infty,$ 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 $\displaystyle 2009 = 41\times49,$ but I cannot see any way to make use of that fact.