Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I input:

Solve[x^5 - x - 1 == 0, x]
(* {{x -> Root[-1 - #1 + #1^5 &, 1]}, {x -> 
   Root[-1 - #1 + #1^5 &, 2]}, {x -> Root[-1 - #1 + #1^5 &, 3]}, {x ->
    Root[-1 - #1 + #1^5 &, 4]}, {x -> Root[-1 - #1 + #1^5 &, 5]}} *)

$x^5 - x - 1=0$ is NOT solvable by radicals (according to various sources).

My question is:

When Mathematica does not return an exact solution to a polynomial equation (as in the above example), does this imply that the polynomial is not solvable by radicals?

share|improve this question
1  
But Solve did return an exact solution set. Root objects are exact solutions. – m_goldberg Jan 1 at 14:19
1  
Have a look at Quartics in the documentations. – rewi Jan 1 at 14:28
    
Maybe a better way to state my question is this: Does there exists an algorithm that can determine if a polynomial equation is "solvable by radicals"? – Geoffrey Critzer Jan 1 at 14:58
3  
The direct answer to your question is no, and this is documented in tutorial/AlgebraicNumbers. It is indicated there that Solve[x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 == 0, x] returns a list of Root objects but that 2^(1/3) + 3^(1/2) is a solution. – murray Jan 1 at 18:00
    
@murray. Yes, thank you. This is the answer that I expected, along with the documentation to substantiate it. – Geoffrey Critzer Jan 1 at 18:45
up vote 2 down vote accepted

This is to put murray's comment on record, so it can be accepted by the OP.

The direct answer to your question is no, and this is documented in tutorial/AlgebraicNumbers. It is indicated there that

Solve[x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 == 0, x]

returns a list of root objects, even though 2^(1/3) + 3^(1/2) is a solution

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.