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

 Messages: [ Previous | Next ]
 Bob Hanlon Posts: 906 Registered: 10/29/11
Re: cubic equation solver
Posted: Mar 28, 2013 5:12 PM

eqn = x^3 + (Sqrt[6] + 2 Sqrt[3] + 2 Sqrt[2] - 9) x +
2 Sqrt[3] - Sqrt[2] - 2 == 0;

sol = Solve[eqn, x] // Simplify

{{x -> (-2*3^(1/3)*(-9 + 2*Sqrt[2] + 2*Sqrt[3] +
Sqrt[6]) + 2^(1/3)*
(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] -
648*Sqrt[3] - 132*Sqrt[6])])^(2/3))/
(6^(2/3)*(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] - 648*Sqrt[3] -
132*Sqrt[6])])^(1/3))},
{x -> (2*3^(1/3)*((-9 - 6*I) + (2 - 3*I)*Sqrt[2] +
(2 + 9*I)*Sqrt[3] + (1 - 2*I)*Sqrt[6]) +
2^(1/3)*(-1 - I*Sqrt[3])*
(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] -
648*Sqrt[3] - 132*Sqrt[6])])^(2/3))/
(2*6^(2/3)*(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] - 648*Sqrt[3] -
132*Sqrt[6])])^(1/3))},
{x -> (2*3^(1/3)*((-9 + 6*I) + (2 + 3*I)*Sqrt[2] +
(2 - 9*I)*Sqrt[3] + (1 + 2*I)*Sqrt[6]) +
I*2^(1/3)*(I + Sqrt[3])*
(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] -
648*Sqrt[3] - 132*Sqrt[6])])^(2/3))/
(2*6^(2/3)*(9*(2 + Sqrt[2] - 2*Sqrt[3]) +
I*Sqrt[6*(2331 - 626*Sqrt[2] - 648*Sqrt[3] -
132*Sqrt[6])])^(1/3))}}

Although not in the simplest form, these are the correct roots

eqn /. sol // Simplify

{True, True, True}

To find simpler forms without using FullSimplify (which seems to go on
indefinitely)

sol2 = ({x -> RootApproximant[#[[-1, -1]]]} & /@ sol) // ToRadicals

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

Verifying that these are exact solutions

eqn /. sol2 // FullSimplify

{True, True, True}

I assume that these are the values in your first sentence which was
garbled (presumably you did not convert to InputForm before copying).

Bob Hanlon

On Thu, Mar 28, 2013 at 4:06 AM, Elim Qiu <elim.qiu@gmail.com> 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))}}
>

Date Subject Author
3/28/13 Bob Hanlon
3/30/13 Dana DeLouis