I remember that there should be a formula for computing the root of $x^n+ax+b=0$. But I can't find it online. Could anybody point me the solutions? Thanks.
Asked
Active
Viewed 1,779 times
6
-
1Seems likely that there would be $n$ roots? – copper.hat Jul 02 '13 at 06:40
-
3By the way if you just put your math between dollars signs it usually looks much better. x^n+ax+b=0 becomes $x^n+ax+b=0$ just by putting it in between two dollar signs – Ovi Jul 02 '13 at 06:41
-
To add to Ovi's comment: For some basic information about writing maths at this site see e.g. here, here, here and here. – Lord_Farin Jul 02 '13 at 06:44
-
You can have a look to Glasser's derivation in the link http://en.wikipedia.org/wiki/Bring_radical – Mathlover Jul 02 '13 at 06:51
1 Answers
2
If by a "formula" you mean an expression in radicals, then you remember incorrectly. The polynomial $x^5-x+1$ has no solutions in radicals. See also http://en.wikipedia.org/wiki/Bring%E2%80%93Jerrard_normal_form#Bring.E2.80.93Jerrard_normal_form
Mikhail Katz
- 42,112
- 3
- 66
- 131
-
https://math.stackexchange.com/questions/2507093/why-is-there-no-solution-with-radicals-for-the-quintic-equation-x5-x1-0 exchange post that explains why x^5 - x + 1 is not solveable by radicals using galois theory – jsmith Jun 28 '23 at 10:36