Checking for real roots of an even-th degree polynomial $ (a-x)^{2n+1}+x^{2n+1}-b=0 $

Question

How should one check for real solutions of an even-th degree polynomial similar to this $ (a-x)^{2n+1}+x^{2n+1}-b=0 $ ? a, b and n are constant non-zero natural numbers

Answer 1

In general, the number of distinct real roots of a polynomial can be computed using Sturm's theorem.

In your case, if $P(x) = (a - x)^{2n+1} + x^{2n+1} - b$, it is not hard to show that $P$ is convex and symmetric around $x=a/2$. Thus we have one real root if $b = 2 (a/2)^{2n+1}$, none if $b < 2 (a/2)^{2n+1}$, two if $b > 2 (a/2)^{2n+1}$.