# How to solve in radicals this family of equations for any degree $k$?

Part I. Given any constant $a,b$, the equation in $x$,

$$\left(\frac{x+\sqrt{x^2+4a}}{2}\right)^{k}+\left(\frac{x-\sqrt{x^2+4a}}{2}\right)^{k}=b\tag1$$

is solvable in radicals for any degree $k$. The general solution is,

$$x = \frac{-a}{\beta^{1/k}}+\beta^{1/k},\quad\text{where}\;\beta = \frac{b+\sqrt{b^2+4a^k}}{2}$$

For example, expanding at $k=5$, we get the DeMoivre quintic,

$$x^5+5ax^3+5a^2x=b$$

Part II. Given any constant $a,b$, the equation in $x$,

$$\frac{\big(x+\sqrt{b}\big)^k+\big(x-\sqrt{b}\big)^k}{2}+a\frac{\big(x+\sqrt{b}\big)^k-\big(x-\sqrt{b}\big)^k}{2\sqrt{b}}=0\tag2$$

is also solvable in radicals for any $k$. For example, for $k=5$, we get,

$$x^5+5ax^4+10bx^3+10abx^2+5b^2x+ab^2=0$$

Question: What is the general solution of $(2)$?

• I have a doubt now since my answer is rather trivial... Did you want the polynomial or did you hope an explicit solution $x=\cdots$ ? (I should probably erase my answer!) Dec 23 '15 at 14:10
• @RaymondManzoni: The polynomials are easy to get by expansion. I was hoping for the explicit solution $x$. But as I look at your answer, it seems you are on the right track. Dec 23 '15 at 14:14
• In the mean time I had reverted $(2)$ in my solution. Now are there others? Dec 23 '15 at 14:18
• @RaymondManzoni: I think all $k$ roots can be found by using something similar to this. (P.S. Maybe you can delete everything after $(3)$? The solution $x$ is what I'm after.) Dec 23 '15 at 14:22
• @RaymondManzoni: I figured out the expression for all $k$ roots. Do you want me to do the edit, or did you figure it out also? :) Dec 23 '15 at 14:29

Let's rewrite your equation $\,(2)\,$ for $\;r:=\dfrac {\sqrt{b}}x\,$ as : $$\tag{1}\dfrac{(1+r)^k-(1-r)^k}{(1+r)^k+(1-r)^k}=-\frac {\sqrt{b}}a$$ then from Ron Gordon's recent answer : $$\tag{2}\tanh(k\;\operatorname{tanh^{-1}}r)=\frac{e^{(2 k \tanh^{-1} r)}-1}{e^{(2 k\tanh^{-1} r)}+1}=-\frac {\sqrt{b}}a$$

We may revert this to get one real solution : $$\tag{3}\dfrac {\sqrt{b}}x=-\tanh\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a\right)$$

Other solutions may be obtained by noticing that $\,e^{\large{(2 k \tanh^{-1} r+2\pi in)}}\,$ will give the same expansion in $(2)$ so that we have too : $$\tag{4}\tanh(k\;\operatorname{tanh^{-1}}r+\pi i n)=-\frac {\sqrt{b}}a$$ and the $k$ roots $x_n$ (for $n=0..k-1$) given by : $$\tag{5}\boxed{\displaystyle x_n=-\frac 1{\sqrt{b}}\;\operatorname{cotanh}\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a+\frac{\pi in}k\right)}$$ We may too revert the passage from $(1)$ to $(2)$ (with $k$ replaced by $\dfrac 1k$ and for $\alpha:=\dfrac {\sqrt{b}}a$ )
and rewrite $(5)$ as :
$$\tag{6}\frac {\sqrt{b}}x=-\dfrac{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}-1}{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}+1}$$ that is : $$\tag{7}x=\sqrt{b}\;\dfrac{1+R^{1/k}}{1-R^{1/k}}$$

where $\;R^{1/k}\;$ is one of the $k\;$ $k$-th roots of $\;R:=\dfrac{1+\alpha}{1-\alpha}=\dfrac{a+\sqrt{b}}{a-\sqrt{b}}$.

• I'm glad you noticed that post with R. Gordon's answer. Dec 23 '15 at 14:53
• @TitoPiezasIII: so am I. Excellent continuation! Dec 23 '15 at 14:55
• One thing I would like though. This family has a solvable Galois group. Do you know a way to express your $x_n$ in terms of plain trigonometric functions? (Hence in radicals.) Dec 23 '15 at 14:56
• I'll try to edit (and convert the log in powers if possible...) Dec 23 '15 at 15:03
• Oh, I already tested it with Mathematica. Besides, the form is verification enough for me. If it had been ugly and complicated, I would be doubtful. But since it is beautiful and simple, then it has to be true. (Admittedly not the most rigorous test.) :) Dec 23 '15 at 16:03