       Re: cubic equation solver

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• Subject: [mg130255] Re: cubic equation solver
• From: Dana DeLouis <dana01 at icloud.com>
• Date: Sat, 30 Mar 2013 04:06:36 -0400 (EDT)
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```Hello.   For this particular problem, perhaps

equ=x^3+(Sqrt+2 Sqrt+2 Sqrt-9) x+2 Sqrt-Sqrt-2 ;

Solve[equ==0, x, Reals]

{
{x->-2+Sqrt},
{x->2-Sqrt},
{x->Root[1-10 #1^2+#1^4&,3]}
}

{x->Sqrt[5-2 Sqrt]}

'// Or done together:

{
{x->-2+Sqrt},
{x->2-Sqrt},
{x->Sqrt[5-2 Sqrt]}
}

= = = = = = = = = =
Good Luck.  :>)
Dana DeLouis
Mac & Math 9
= = = = = = = = = =

On Thursday, March 28, 2013 4:05:44 AM UTC-4, Elim Qiu wrote:
> x^3 + (=E2=88=9A6 + 2=E2=88=9A3 + 2=E2=88=9A2 -9)x + 2=E2=88=9A3 -=E2=88=9A2 -2 = 0
>
> has exact roots =E2=88=9A2-2, =E2=88=9A3-=E2=88=9A2, 2-=E2=88=9A3
>
>
>
> But Mathematica says:
>
>
>
> Solve[x^3 + (Sqrt + 2 Sqrt + 2 Sqrt - 9) x + 2 Sqrt -
>
>    Sqrt - 2 == 0, x]
>
>
>
> {{x -> (1/
>
>       2 (18 + 9 Sqrt - 18 Sqrt +
>
>         I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>             264 Sqrt)]))^(1/3)/3^(
>
>     2/3) - (-9 + 2 Sqrt + 2 Sqrt + Sqrt[
>
>      6])/(3/2 (18 + 9 Sqrt - 18 Sqrt +
>
>         I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>             264 Sqrt)]))^(
>
>     1/3)}, {x -> -(((1 + I Sqrt) (1/
>
>         2 (18 + 9 Sqrt - 18 Sqrt +
>
>           I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>               264 Sqrt)]))^(1/3))/(
>
>      2 3^(2/3))) + ((1 - I Sqrt) (-9 + 2 Sqrt + 2 Sqrt +
>
>        Sqrt))/(
>
>     2^(2/3) (3 (18 + 9 Sqrt - 18 Sqrt +
>
>          I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>              264 Sqrt)]))^(
>
>      1/3))}, {x -> -(((1 - I Sqrt) (1/
>
>         2 (18 + 9 Sqrt - 18 Sqrt +
>
>           I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>               264 Sqrt)]))^(1/3))/(
>
>      2 3^(2/3))) + ((1 + I Sqrt) (-9 + 2 Sqrt + 2 Sqrt +
>
>        Sqrt))/(
>
>     2^(2/3) (3 (18 + 9 Sqrt - 18 Sqrt +
>
>          I Sqrt[3 (4662 - 1252 Sqrt - 1296 Sqrt -
>
>              264 Sqrt)]))^(1/3))}}

```

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