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Topic: Rearranging a cubic equation
Replies: 6   Last Post: Aug 11, 2013 8:31 PM

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umumenu@gmail.com

Posts: 910
Registered: 3/9/08
Re: Rearranging a cubic equation
Posted: Aug 7, 2013 3:51 PM
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> Am 07.08.2013 08:09, schrieb dwyerdan@gmail.com:
> > Hi.
> >
> > Can anyone re-arrange the following equation to

> make x the subject of the equation (i.e. x = ...)?
> >
> > y = (x^3 * a) + (x * b * [x * c + 91]) + (x * d) +

> (x * e)
> >
> > It is a real world problem, not a theoretical

> exercise. I have had a look at the wikipedia page on
> cubic functions, but I don't understand enough of it
> to be helpful.

> >
> > I am struggling to get it into the basic form of a

> cubic function (f(x) = ax^3 + bx^2 + cx + d).
> >
> > Any takers?

>
>
> See
>
> http://www.wolframalpha.com/
>
> The three soltions are
>
> {{x -> -((b c)/(
> 3 a)) + (2^(
> 1/3) (273 a b - b^2 c^2 + 3 a d + 3 a e))/(3
> a e))/(3 a (-819 a b^2 c +
> 2 b^3 c^3 - 9 a b c d - 9 a b c e -
> 27 a^2 y + \[Sqrt](4 (273 a b - b^2 c^2 + 3
> ^2 c^2 + 3 a d +
> 3 a e)^3 + (-819 a b^2 c + 2 b^3 c^3
> ^2 c + 2 b^3 c^3 - 9 a b c d -
> 9 a b c e - 27 a^2 y)^2))^(1/3)) -
> y)^2))^(1/3)) - (1/(
> 3 2^(1/3)
> a))((-819 a b^2 c + 2 b^3 c^3 - 9 a b c d - 9
> c d - 9 a b c e -
> 27 a^2 y + Sqrt[
> 4 (273 a b - b^2 c^2 + 3 a d + 3 a e)^3 +
> e)^3 + (-819 a b^2 c +
> 2 b^3 c^3 - 9 a b c d - 9 a b c e - 27 a^2
> e - 27 a^2 y)^2])^(
> 1/3))}, {x -> -((b c)/(
> 3 a)) - ((1 + I Sqrt[3]) (273 a b - b^2 c^2 + 3
> ^2 + 3 a d +
> 3 a e))/(3 2^(2/3)
> a (-819 a b^2 c + 2 b^3 c^3 - 9 a b c d - 9
> b c d - 9 a b c e -
> 27 a^2 y + \[Sqrt](4 (273 a b - b^2 c^2 + 3
> ^2 c^2 + 3 a d +
> 3 a e)^3 + (-819 a b^2 c + 2 b^3 c^3
> ^2 c + 2 b^3 c^3 - 9 a b c d -
> 9 a b c e - 27 a^2 y)^2))^(1/3)) +
> y)^2))^(1/3)) + (1/(
> 6 2^(1/3)
> a))(1 - I Sqrt[3]) (-819 a b^2 c + 2 b^3 c^3 -
> 3 c^3 - 9 a b c d -
> 9 a b c e -
> 27 a^2 y + \[Sqrt](4 (273 a b - b^2 c^2 + 3 a
> ^2 + 3 a d +
> 3 a e)^3 + (-819 a b^2 c + 2 b^3 c^3 -
> + 2 b^3 c^3 - 9 a b c d -
> 9 a b c e - 27 a^2 y)^2))^(1/3)}, {x ->
> (1/3)}, {x -> -((b c)/(
> 3 a)) - ((1 - I Sqrt[3]) (273 a b - b^2 c^2 + 3
> ^2 + 3 a d +
> 3 a e))/(3 2^(2/3)
> a (-819 a b^2 c + 2 b^3 c^3 - 9 a b c d - 9
> b c d - 9 a b c e -
> 27 a^2 y + \[Sqrt](4 (273 a b - b^2 c^2 + 3
> ^2 c^2 + 3 a d +
> 3 a e)^3 + (-819 a b^2 c + 2 b^3 c^3
> ^2 c + 2 b^3 c^3 - 9 a b c d -
> 9 a b c e - 27 a^2 y)^2))^(1/3)) +
> y)^2))^(1/3)) + (1/(
> 6 2^(1/3)
> a))(1 + I Sqrt[3]) (-819 a b^2 c + 2 b^3 c^3 -
> 3 c^3 - 9 a b c d -
> 9 a b c e -
> 27 a^2 y + \[Sqrt](4 (273 a b - b^2 c^2 + 3 a
> ^2 + 3 a d +
> 3 a e)^3 + (-819 a b^2 c + 2 b^3 c^3 -
> + 2 b^3 c^3 - 9 a b c d -
> 9 a b c e - 27 a^2 y)^2))^(1/3)}}
>
> --
>
> Roland Franzius


Weird .. Just thought that x = 0 must be a solution.

Han de Bruijn



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