Re: finding roots of 1 + 6*x - 8*x^3

*To*: mathgroup at smc.vnet.net*Subject*: [mg54661] Re: [mg54630] finding roots of 1 + 6*x - 8*x^3*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 25 Feb 2005 01:19:19 -0500 (EST)*References*: <200502240821.DAA13324@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Kennedy wrote: >Hello All, > >I am trying to find the roots of >1 + 6*x - 8*x^3. > >Roots[1+6*x-8*x^3==0,x] yields this ugly thing: >(made uglier by my converting to InputForm) > >x == ((1 + I*Sqrt[3])/2)^(1/3)/2 + > 1/(2^(2/3)*(1 + I*Sqrt[3])^(1/3)) || > x == -((1 - I*Sqrt[3])*((1 + I*Sqrt[3])/2)^(1/3))/4 - > ((1 + I*Sqrt[3])/2)^(2/3)/2 || > x == -(1 - I*Sqrt[3])/(2*2^(2/3)*(1 + I*Sqrt[3])^ > (1/3)) - (1 + I*Sqrt[3])^(4/3)/(4*2^(1/3)) > >This monstrosity is chock full of imaginaries, >even though I know all three roots are real. > >I tried Solve too but got the same thing, except >in the form of a set of replacements. My guess is >that Solve just calls Roots when handed a poly- >nomial. > >When I ran the above through FullSimplify, I >got three "Root" objects, the upshot of which is >that the roots of the polynomial are indeed the >Roots of said polynomial. Huh. > >What command can I use to get the roots into >a form that are >(a) purely real, and >(b) in radical form? > >Thanks, >Kennedy > >PS. Mathematica 4.2 > > It is a classical result of mathematics that this cannot be done: http://mathworld.wolfram.com/CasusIrreducibilis.html That "monstrosity" you saw is simply the Cardano (or Tartaglia, I suppose) formulation for roots of a cubic. The Root[] form is often more convenient for various purposes than is the radical form. To get it directly one can do SetOptions[Roots, Cubics->False, Quartics->False] This will have the effect of disabling use of the Cardano formulas. Then real roots will indeed be explicitly real for purposes of e.g. the funtions N, Re, and Im. Daniel Lichtblau Wolfram Research

**References**:**finding roots of 1 + 6*x - 8*x^3***From:*"Kennedy" <kennedy@oldnews.org>