I have the following quartic polynomial equation, where $U$ and $V$ are positive real numbers:
$$ x^4 + \left(\frac{13}{6} × U\right) x^3 + \left(\frac{3}{2} × U^2\right) x^2 + \left(\frac{1}{3} × U^3\right) x - V = 0 $$
It has one positive real solution. When I go through the process for solving a quartic given on Wikipedia, or ask WolframAlpha to solve it, the result I get uses only square roots and cube roots.
Is there a way to solve this quartic that uses quartic roots?
The solution to a linear polynomial (degree 1) uses linear roots (degree $\frac{1}{1}$), the solution to a quadratic polynomial (degree 2) uses square roots (degree $\frac{1}{2}$), and the solution to a cubic polynomial (degree 3) uses cube roots (degree $\frac{1}{3}$). I want to be able to continue this pattern and find a way to write the solution to my quartic polynomial (degree 4) using quartic roots (degree $\frac{1}{4}$).
For example, the above quartic is part of the following sequence of polynomial equations:
- $x − V = 0$
- $x × \left(x + \frac{1}{1} × U\right) − V = 0$
- $x × \left(x + \frac{1}{1} × U\right) × \left(x + \frac{1}{2} × U\right) − V = 0$
- $x × \left(x + \frac{1}{1} × U\right) × \left(x + \frac{1}{2} × U\right) × \left(x + \frac{2}{3} × U\right) − V = 0$
They all have one positive real solution, the first three of which are as follows:
- $x = V$
- $x = \sqrt[2]{V + \left(\frac{U}{2}\right)^2} − \frac{U}{2}$
- $x = \sqrt[3]{\frac{V}{2} + \sqrt[2]{\left(\frac{V}{2}\right)^2 − \left(\frac{\left(\frac{U}{2}\right)^2}{3}\right)^3}} + \sqrt[3]{\frac{V}{2} − \sqrt[2]{\left(\frac{V}{2}\right)^2 − \left(\frac{\left(\frac{U}{2}\right)^2}{3}\right)^3}} − \frac{U}{2}$
For these first three solutions, they can be written in such a way that there is a clear progression in the structure. The first has an invisible first-order root; the second has a square root, with the invisible first-order root inside it; and the third has two cube roots, each of which contains a square root.
The fourth equation's solution as calculated by WolframAlpha is quite messy, as is to be expected for a quartic polynomial. It also uses only square roots and cube roots, with no quartic roots. I've tried rearranging it in a few different ways to see if I can get it to fit with the progression of the previous three, but I haven't had any success.
Can the positive real solution to the quartic equation be written as a sum of three or four quartic roots, each containing one or two cube roots which each contain a square root, to fit the pattern of the solutions to the other three equations?
The pattern you observed is valid for prime degrees $p$. Hence,
$$x_2 = \sqrt{u_1}$$ $$x_3 = \sqrt[3]{u_1}+\sqrt[3]{u_2}$$ $$x_5 = \sqrt[5]{u_1}+\sqrt[5]{u_2}+\sqrt[5]{u_3}+\sqrt[5]{u_4}$$
etc, where $u_k$ are the roots of its Lagrange resolvent of degree $p-1$. But for solvable equations of composite degree $pq$, the first method is to decompose it into equations of smaller degree with coefficients that are no longer rational.
However, since a quartic is also a power of 2, there is a second method and Euler found the alternative way to solve it (which can be extrapolated to certain $8$-degs, $16$-degs, etc). These are Method 1 and Method 2 in this old answer. Method 2 is given below.
I. Square roots
Given the depressed quartic,
$$x^4+px^2+qx+r = 0$$
and its resolvent cubic in $u$,
$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$
then, with the appropriate signs,
$$x_k = \frac{\pm\sqrt{u_1}\pm\sqrt{u_2}\pm\sqrt{u_3}}2$$
II. Fourth roots
But we can "cheat" by squaring the roots of the cubic $u^2=v$ using resultants. The WA command is,
which eliminates $u$ and collects the new variable $v$ to get,
$$v^3 - 2(p^2 + 4r)v^2 + \big((p^2 - 4r)^2 + 4p q^2\big)v - q^4 = 0$$
and we get the alternative solution of a quartic root as a sum of three $4$th roots,
$$x_k = \frac{\pm\sqrt[4]{v_1}\pm\sqrt[4]{v_2}\pm\sqrt[4]{v_3}}2$$
as requested.
III. Examples
Using the minpoly of the tetranacci constant $x \approx 1.92756$,
$$x^4-x^3-x^2-x-1 = 0$$
then,
\begin{align}x &=\frac14\left(1+\frac{\sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}}2\right)\approx 1.92756\\ &=\frac14\left(1+\frac{\sqrt[4]{v_1}\,+\sqrt[4]{v_2}\,+\sqrt[4]{v_3}}2\right)\approx 1.92756 \end{align}
where $u_k$ and $v_k$ are roots of the cubics,
$$u^3 - 44u^2 + 1840u - 104^2 = 0$$ $$v^3 + 1744v^2 + 2433792v - 104^4 = 0$$
P.S. This is just proof of principle it can be done, but may be a wee bit complicated in practice, especially when dealing with complex roots.