I'm doing calculations with a lot of symbols, which don't always gives nice expressions. Now I have a solution with a Root-expression, where the coefficients are symbolic, e.g.

Root[Z + Y#1 + X#1^2 - Conjugate[Y]#1^3 + Conjugate[Z]#1^4,1]

Is there a way to simplify this with Mathematica? How would you treat these kind of expressions?

  • $\begingroup$ Are y and z complex or real-valued? $\endgroup$ Apr 24, 2015 at 14:01
  • $\begingroup$ Your Root object is missing the &. If you use ToRadicals you can convert the Root object to the equivalent algebraic expression which is much more complicated. Adding any appropriate assumptions to Simplify may help but the Root object is probably the simplest form. Read the documentation for examples on how to use the Root objects. $\endgroup$
    – Bob Hanlon
    Apr 24, 2015 at 14:05
  • $\begingroup$ @2012rcampion every variable is complex valued... $\endgroup$ Apr 24, 2015 at 18:14
  • $\begingroup$ @BobHanlon OK, I hoped there was a simpler way, but thanks! $\endgroup$ Apr 24, 2015 at 18:14

1 Answer 1


If you look at the output of

Solve[Z + Y #1 + X #1^2 - Conjugate[Y] #1^3 + Conjugate[Z] #1^4 &[x] == 0, x, Quartics -> True]

you'll see that the solutions in radicals are far more complicated.

How do I treat solutions in terms of Root? I leave them that way. Root is nice. It's handier to have solutions ordered by magnitude than by principal values of radicals. When you need numerical answers, real solutions in terms of radicals tend to end up with spurious imaginary parts, but the methods behind Root avoid this. And, of course, beyond quartics you can't generally have radicals as a solution.

  • $\begingroup$ Ok, I didn't realise this. It's just that I want to understand the output from for example Integrate[-((2 I \[Pi])/Sqrt[(r + A xiR - rb xiR^2)^2 - 4 (l xiR + w xiR^2 + Z) (wb - lb xiR + xiR^2 Zb)]),xiR] which gives an EllipticF[ArcSin[...]] from a gigantic expressions with a lot of Roots... I want to find the an expression for A wherefore this integral becomes zero but it's too complicated to understand so I wanted to simplify this integral... $\endgroup$ Apr 28, 2015 at 10:34

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