x5+ax2+b=0

Simplest examples of irreducible and solvable by radicals quintic trinomials, x⁵+ax²+b, where a and b are integers in range [-1111..1111]. Excluded trivial cases where a or b are equal zero.

Basic notes:

- if x⁵+ax²+b=0 is solvable then x⁵-ax²-b=0 is too. More generally x⁵+ax²+b=0 is solvable if and only if x⁵+ac³x²+bc⁵=0 is solvable (c is non-zero rational number).

- if x⁵+ax²+b=0 is solvable then x⁵+(a/b)x³+(1/b)=0 is too. And in more general: x⁵+(ac²/b)x³+(c⁵/b)=0

Important note:

There are known explicit parameterizations for all solvable equations x⁵+ax²+b=0 - see for example "Spearman, Williams: On solvable quintics..." or wikipedia.

See incredible explicit formulas for the roots of the solvable polynomials.