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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?

Blue
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Lawton
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  • Have you tried any variations in the form of your quartic? The $\frac23$ you have in there is not the only way to continue the pattern. – Chris Lewis Jul 11 '23 at 19:10
  • @ChrisLewis The value of $\frac{2}{3}$ is what works in the context of the project this is a part of, not an arbitrarily chosen value, so it's rather inflexible. Unless you just mean changing the format? – Lawton Jul 11 '23 at 20:14
  • I just meant that you're expecting a pattern in the form of the roots, but that will be entirely dependent on how the pattern in the coefficients is continued. To me, neither pattern is obvious. What specific form do you think the next term in the pattern of the roots might take? (My point is, maybe this is part of a sequence, but it's possible you're looking at the wrong next term; perhaps there's a "nice" root if you have, say $\frac13$ in place of $\frac23$, or $\frac14$. You really only have a sequence that starts $1,\frac12$, and there are lots of those.) – Chris Lewis Jul 12 '23 at 08:46
  • @ChrisLewis Unfortunately, I don't know what the pattern in the polynomial coefficients is, nor do I know the exact form to expect for the quartic root. As I say in the bold text in the question, I expect it to be a sum of three or four quartic roots, each containing one or two cube roots which each contain a square root; however, that expectation is based only on guesswork based on the first three roots. I don't have any mathematical logic to go on. – Lawton Jul 12 '23 at 13:11
  • You can turn the outer square roots from $\sqrt{y}$ into $\sqrt[4]{y^2}$. – L. F. Jul 13 '23 at 14:00
  • @L.F. Just squaring the terms in the output from WolframAlpha results in incredibly messy expressions. Is there a neater way to go about it? – Lawton Jul 13 '23 at 14:53

1 Answers1

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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,

Collect[Resultant[u^3 + 2p u^2 + (p^2 - 4r)u - q^2, u^2-v , u], v]

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.

Tito Piezas III
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  • I'm trying to follow along with your answer, but the results I'm getting are extremely messy and aren't reducing. Is that expected, or am I doing something wrong? – Lawton Aug 02 '23 at 16:13
  • @Lawton I assume you remember how to reduce any equation to depressed form? Also, using the example, did you get $x = \approx 1.92756$ for both formulas? – Tito Piezas III Aug 03 '23 at 10:16
  • I haven't worked on the tetranacci examples, I've been using the equation from my question. I reduced it to depressed quartic form, wrote out the resolvent cubic in $u$ and resultant form in $v$, and tried to solve for $v$, but the solutions I'm getting are extremely messy. – Lawton Aug 03 '23 at 14:28
  • Given the equation $x^4 + \left(\frac{13}{6} U\right) x^3 + \left(\frac{3}{2} U^2\right) x^2 + \left(\frac{1}{3} U^3\right) x - V = 0$, which is in the form $a x^4 + b x^3 + c x^2 + d x + e = 0$, we have $a = 1$, $b = \frac{13}{6} U$, $c = \frac{3}{2} U^2$, $d = \frac{1}{3} U^3$, and $e = -V$. Given the definitions $p = \frac{8 c - 3 b^2}{8}$, $q = \frac{b^3 - 4 b c + 8 d}{8}$, and $r = \frac{-3 b^4 + 256 e - 64 b d + 16 b^2 c}{256}$, I get that $p = -\frac{25}{96} U^2$, $q = -\frac{35}{1728} U^3$, and $r = \frac{143}{110592} U^4 - V$. – Lawton Aug 03 '23 at 14:38
  • Then with the resolvent cubic cheated using resultants to get $v^3 - 2 (p^2 + 4 r) v^2 + ((p^2 - 4 r)^2 + 4 p q^2) v - q^4 = 0$, substituting in $p$, $q$, and $r$ and expanding I get $v^3 - \left(\frac{1009}{6912} U^4 - 8 V\right) v^2 + \left(\frac{501217}{143327232} U^8 + \frac{433}{864} U^4 V + 16 V^2\right) v - \left(\frac{1500625}{8916100448256} U^{12}\right) = 0$. – Lawton Aug 03 '23 at 14:45
  • @Lawton Simplifying equations is all about substitutions. I've verified your steps and they are correct. However, you can reduce the size of the coefficients by doing the substitution $U = 12W$ to get the simpler, $$v^3 + (8 V - 3027 W^4) v^2 + (16 V^2 + 10392 V W^4 + 1503651 W^8) v - (35 W^3)^4 = 0$$ Of course, since you have TWO parameters ($V,W$), then solving this cubic in Wolfram Alpha will naturally give long expressions for the roots. But, in principle, you got what you wanted: a quartic root expressed as a sum of 4th roots. – Tito Piezas III Aug 03 '23 at 18:53
  • @Lawton Also, since $$\frac{\sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}}2 = \frac{\sqrt[4]{v_1}+\sqrt[4]{v_2}+\sqrt[4]{v_3}}2$$ where you are getting the 4th roots of the RHS, then you can expect the cubic roots $v_k$ to be larger and more complicated than the $u_k$. Price to pay, I'm afraid. – Tito Piezas III Aug 03 '23 at 19:20