x^1/3 +x^1/2 +2=0

By Brad Rodgers (P1930) on Tuesday, January 23, 2001 - 12:46 am :

How do I solve an equation such as

x^1/3 +x^1/2 +2=0

for x?

Thanks,

Brad

By Dan Goodman (Dfmg2) on Tuesday, January 23, 2001 - 01:18 am :

The problem is, you have to specify a branch of the functions

x^{1 / 3}

and

x^{1 / 2}

. There are three roots of

x^{1 / 3}

, if

y^{3} = x

, then

(y ρ)^{3} = x

and

(y ρ^{2})^{3} = x

where

ρ = \exp (2 π i / 3)

. Similarly, if

y^{2} = x

, then

(- y)^{2} = x

. However, suppose you have a number

y

and a number

z

such that

y^{3} = x

and

z^{2} = x

and

y + z + 2 = 0

, what can we say about

y

and

z

? Well,

z = - 2 - y

, so

(- 2 - y)^{2} = x

, i.e. (1)

y^{2} + 4 y + 4 = x

. Also, (2)

y^{3} = x

This turns out to be trickier than I thought, and it's quite late and I have a lecture in a few hours, must get some sleep. I'll write more tomorrow if nobody has answered by then. Hopefully the beginnings of an analysis above will give you some ideas.

Don't be too hopeful for a nice solution though, I just plugged in

''Solve[{Y^3==X,Z^2==X,Y+Z+2==0},{X,Y,Z}]'' into Mathematica and got a page long answer back, involving all sorts of yucky stuff.

By Dan Goodman (Dfmg2) on Wednesday, January 24, 2001 - 12:53 am :

OK, update on this analysis. To solve x^1/3 +x^1/2 +2=0, you need to specify branches of the functions involved, if you want solutions for arbitrary branches, you need to simultaneously solve the equations:

(1) y³ =x
(2) z² =x
(3) y+z+2=0

Solving these is pretty tricky, as mentioned above, Mathematica can do it, but the answer is disgusting. If you're really interested, the solutions are included at the end of this post.

One way to go about solving it is to introduce w such that w⁶ =x. Let w_j =we^{2pi ij/6} where j=0,1,2,3,4 or 5, then w_j ⁶ =x. Therefore, any solution to (1) is of the form y=w_j ² for some j, and a solution to (2) is of the form z=w_k ³ for some k. So, equation (3) becomes (3') w_j ² +w_k ³ +2=0. Expanding this we get (3'') w² e^{2pi ij/3} + (-1)^k w³ + 2 = 0. Thus, if we vary k=0 or 1 and j=0,1 or 2, we get 6 cubic polynomial equations in w. Each of these can be solved for w, so we will have at most 18 solutions for w. Now we just work out the corresponding values of y, z and x (for which we need the values of j and k which each solution for w represents), and we have solved the original problem. As an example, let j=k=0, we get the equation w³ +w² +2=0. This has 3 (unfortunately nasty) solutions, call them w₀ , w₁ and w₂ . Now, set y_p =w_p ² , z_p =w_p ³ and x_p =w_p ⁶ , and this will be a solution to the original equation. I just checked one in Mathematica, and it works, phew. In this example, the solutions aren't nice, but this is the best method I could come up with. Sorry it's so complicated!

{{x -> I/18 x (-98 x I +
14 x I x (73 - 6 x Sqrt[87])^1/3 -
14 x Sqrt[3] x (73 - 6 xSqrt[87])^1/3 +
I x (73 - 6 x Sqrt[87])^2/3 +
Sqrt[3] x (73 - 6 x Sqrt[87])^2/3 +
14 x I x (73 + 6 x Sqrt[87])^1/3 +
14 x Sqrt[3] x (73 + 6 x Sqrt[87])^1/3 +
I x (73 + 6 x Sqrt[87])^2/3 -
Sqrt[3] x (73 + 6 x Sqrt[87])^2/3 -
4 x I x ((73 - 6 x Sqrt[87]) x (73 + 6 x Sqrt[87]))^1/3 ),

z -> (-14 + (73 - 6 x Sqrt[87])^1/3 +
I x Sqrt[3] x (73 - 6 x Sqrt[87])^1/3 +
(73 + 6 x Sqrt[87])^1/3 -
I x Sqrt[3] x (73 + 6 x Sqrt[87])^1/3 )/6,

y -> 1/3 - ((1 + I x Sqrt[3]) x
(73 - 6 x Sqrt[87])^1/3 )/6 -
((1 - I x Sqrt[3]) x (73 + 6 x Sqrt[87])^1/3 )/6},

{x -> -I/18 x (98 x I -
14 x I x (73 - 6 x Sqrt[87])^1/3 -
14 x Sqrt[3] x (73 - 6 x Sqrt[87])^1/3 -
I x (73 - 6 x Sqrt[87])^2/3 +
Sqrt[3] x (73 - 6 x Sqrt[87])^2/3 -
14 x I x (73 + 6 x Sqrt[87])^1/3 +
14 x Sqrt[3] x (73 + 6 x Sqrt[87])^1/3 -
I x (73 + 6 x Sqrt[87])^2/3 -
Sqrt[3] x (73 + 6 x Sqrt[87])^2/3 +
4 x I x ((73 - 6 x Sqrt[87]) x (73 + 6 x Sqrt[87]))^1/3 ),

z -> (-14 + (73 - 6 x Sqrt[87])^1/3 -
I x Sqrt[3] x (73 - 6 x Sqrt[87])^1/3 +
(73 + 6 x Sqrt[87])^1/3 +
I x Sqrt[3] x (73 + 6 x Sqrt[87])^1/3 )/6,

y -> 1/3 - ((1 - I x Sqrt[3]) x
(73 - 6 x Sqrt[87])^1/3 )/6 -
((1 + I x Sqrt[3]) x (73 + 6 x Sqrt[87])^1/3 )/6},

{x -> (49 + 14 x (73 - 6 x Sqrt[87])^1/3 +
(73 - 6 x Sqrt[87])^2/3 +
14 x (73 + 6 x Sqrt[87])^1/3 +
(73 + 6 x Sqrt[87])^2/3 +
2 x ((73 - 6 x Sqrt[87]) x (73 + 6 x Sqrt[87]))^1/3 )/9,

z -> (-7 - (73 - 6 x Sqrt[87])^1/3 -
(73 + 6 x Sqrt[87])^1/3 )/3,

y -> 1/3 + (73 - 6 x Sqrt[87])^1/3 /3 +
(73 + 6 x Sqrt[87])^1/3 /3}}