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

• To: mathgroup at smc.vnet.net
• Subject: [mg54663] Re: [mg54630] finding roots of 1 + 6*x - 8*x^3
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Fri, 25 Feb 2005 01:19:22 -0500 (EST)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <200502240821.DAA13324@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```A similar question, about a quadratic, just appeared in this list, with
quite a few variants of essentially the same response -- use
ComplexExpand.  One of the simplest, for your example, is:

ComplexExpand[x /. Solve[1 + 6x - 8x^3 == 0, x]]

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)/2)^(1/3)/2 +
>    1/(2^(2/3)*(1 + I*Sqrt)^(1/3)) ||
>  x == -((1 - I*Sqrt)*((1 + I*Sqrt)/2)^(1/3))/4 -
>    ((1 + I*Sqrt)/2)^(2/3)/2 ||
>  x == -(1 - I*Sqrt)/(2*2^(2/3)*(1 + I*Sqrt)^
>       (1/3)) - (1 + I*Sqrt)^(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
>
> Thanks,
> Kennedy
>
> PS.  Mathematica 4.2
>
>

--
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

```

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