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

*To*: mathgroup at smc.vnet.net*Subject*: [mg54646] Re: finding roots of 1 + 6*x - 8*x^3*From*: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>*Date*: Fri, 25 Feb 2005 01:18:44 -0500 (EST)*Organization*: Uni Leipzig*References*: <cvk4nr$dnj$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, try Roots[1 + 6*x - 8*x^3 == 0, x] /. x == r_ :> x == ComplexExpand[r] Regards Jens "Kennedy" <kennedy at oldnews.org> schrieb im Newsbeitrag news:cvk4nr$dnj$1 at smc.vnet.net... > 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 >