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Topic: cubic equation solver
Replies: 2   Last Post: Mar 30, 2013 4:06 AM

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Dana DeLouis

Posts: 24
Registered: 11/18/12
Re: cubic equation solver
Posted: Mar 30, 2013 4:06 AM
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Hello. For this particular problem, perhaps

equ=x^3+(Sqrt[6]+2 Sqrt[3]+2 Sqrt[2]-9) x+2 Sqrt[3]-Sqrt[2]-2 ;

Solve[equ==0, x, Reals]

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

%[[-1]] //ToRadicals
{x->Sqrt[5-2 Sqrt[6]]}

'// Or done together:

Solve[equ==0,x,Reals] //ToRadicals

{
{x->-2+Sqrt[2]},
{x->2-Sqrt[3]},
{x->Sqrt[5-2 Sqrt[6]]}
}

= = = = = = = = = =
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[6] + 2 Sqrt[3] + 2 Sqrt[2] - 9) x + 2 Sqrt[3] -
>
> Sqrt[2] - 2 == 0, x]
>
>
>
> {{x -> (1/
>
> 2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(1/3)/3^(
>
> 2/3) - (-9 + 2 Sqrt[2] + 2 Sqrt[3] + Sqrt[
>
> 6])/(3/2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(
>
> 1/3)}, {x -> -(((1 + I Sqrt[3]) (1/
>
> 2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(1/3))/(
>
> 2 3^(2/3))) + ((1 - I Sqrt[3]) (-9 + 2 Sqrt[2] + 2 Sqrt[3] +
>
> Sqrt[6]))/(
>
> 2^(2/3) (3 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(
>
> 1/3))}, {x -> -(((1 - I Sqrt[3]) (1/
>
> 2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(1/3))/(
>
> 2 3^(2/3))) + ((1 + I Sqrt[3]) (-9 + 2 Sqrt[2] + 2 Sqrt[3] +
>
> Sqrt[6]))/(
>
> 2^(2/3) (3 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>
> I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>
> 264 Sqrt[6])]))^(1/3))}}







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