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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
>
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