# Can Mathematica solve $x^{b^2} + ax^b -1 = 0$ for $x$ with $a>0$, $b>1$?

I have a problem that I can't seem to solve neither analytically nor with Mathematica. Namely, I have the expression $$x^{b^2} + ax^b - 1= 0,$$ which I want to solve for $x$ with $a>0$, $b>1$.

But I can't find a method in Mathematica that would provide an answer to this without specifying $a$ and $b$ (for example, Solve, Root, FindRoot and Reduce don't work). I am quite new to the software, however, so I was wondering whether I am just missing something or can this equation truly not be solved in general using it?

According to Abel-Ruffini theorem, there is no general solution in radicals to polynomial equations of degree five or higher, except for those with solvable Galois group. You did not specify whether the parameter b is integer. But even if it is, it is not possible to construct a general analytical solution of this equation with Mathematica.

• Thank you, and I figured this would be the case. I guess I was just being naively hopeful Mathematica would nevertheless spit something out :) And no, $b \in \mathbb{R}$ with $b>1$, so not necessarily an integer. I assume there is no theorem that is more general than the Abel-Ruffini one you mention and that would apply here, is there? That is, for each particular $b$, the given equation is a beast that needs to be tackled individually and you cannot a priori say whether you'll be successful in finding a root or not without knowing the value of $b$, right? – Ryker May 18 '14 at 17:03
• Let's take x>0 to deal with real-valued functions. The LHS of your equation is a monotonically increasing function of x (within the specified range of parameters), which at x=0 is equal to -1. Thus, it turns to zero at a single point x_1>0, but, generally speaking, x_1 can not be expressed via b and a using radicals. You can rely on a numerical solution, however, or construct an approximate analytic solution, e.g. via series expansion. – Yasha Gindikin May 18 '14 at 17:55
• Yeah, I've already shown that there exists a unique real root, but was curious as to whether I could find an explicit expression for it in general. Too bad this isn't possible, but I probably don't need anything but existence anyway. Thanks again. – Ryker May 18 '14 at 18:23

I would leave a unassigned, set b to whatever you like, and solve for x. Then you can pick the solution you like from the list:

With[{b = 2}, Solve[x^b^2 + a x^b == 0, x]]


With b=2, there are four solutions:

{{x -> 0}, {x -> 0}, {x -> -I Sqrt[a]}, {x -> I Sqrt[a]}}

• I know, that would be trivial to solve, but I'm really interested in the general solution. – Ryker May 18 '14 at 6:09
• In that case, don't assign anything to b: Solve[x^b^2 + a x^b == 0, x] – phosgene May 18 '14 at 6:13
• Yeah, you're right, I messed up on my question, sorry about that. I've edited the original post and the title. What you're saying works, but it doesn't work with the edited equation. – Ryker May 18 '14 at 6:16
• With the extra -1 in the equation, I would recommend my previous method (using With). If Reduce can't solve the general case, you may be out of luck. – phosgene May 18 '14 at 6:21
• By With you mean choosing $b$ again, right? Yeah, as I said, I'm not really interested in such a simplification, and Reduce can't solve the general case. I get the "This system cannot be solved with the methods available to Reduce." message :/ – Ryker May 18 '14 at 6:35

Multiply by x^(-b), take the log, and the Exp you will get:

x=Exp[Log[-a]/(b^2-b)]

• You can try this also: – Basheer Algohi May 18 '14 at 5:43
• Solve[FullSimplify[(x^b^2 + a x^b ) x^(-b)] == 0, x] – Basheer Algohi May 18 '14 at 5:44
• Thanks, although as mentioned above, I messed up on my question. There should be an additional -1 in the equation, and I've now edited the original post. So I guess what you suggested would work, but not on the amended problem. – Ryker May 18 '14 at 6:17