Re: cubic equation solver
- To: mathgroup at smc.vnet.net
- 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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- Delivered-to: l-mathgroup@wolfram.com
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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))}}