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Re: cubic equation solver
Posted:
Mar 30, 2013 4:06 AM


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>2Sqrt[3]}, {x>Root[110 #1^2+#1^4&,3]} }
%[[1]] //ToRadicals {x>Sqrt[52 Sqrt[6]]}
'// Or done together:
Solve[equ==0,x,Reals] //ToRadicals
{ {x>2+Sqrt[2]}, {x>2Sqrt[3]}, {x>Sqrt[52 Sqrt[6]]} }
= = = = = = = = = = Good Luck. :>) Dana DeLouis Mac & Math 9 = = = = = = = = = =
On Thursday, March 28, 2013 4:05:44 AM UTC4, 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=9A22, =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))}}



