How can we solve the particular equation $16x^5-200x^3-200x^2+25x+30=0$ in closed form?

Question

How can we solve the following quintic in closed form ? $$16x^5-200x^3-200x^2+25x+30=0$$

What is special about this equation? What can we say about solvability? I'm not particularly into math. I saw this equation in a discussion group. Here are some ideas about the equation from non-mathematicians .

1. There is no closed-form formula for $5$-degree equations. Therefore there is no solution. In my opinion this is wrong. For example, $x^5=1$ can be easily solved. The true version if that idea would be there is no general solution formula . (by radicals)

2. The equation is not factored. Therefore, there is no closed-form solution. This idea is also wrong in my opinion. Because the factors do not have to be rational numbers.

3. Wolfram Alpha does not give closed form for the equation. Therefore, it cannot be solved. I cannot comment on this idea.

I found that the quintic is irreducible over $\Bbb Q$.

I've heard a little bit about Galois theory. But I don't have the mathematical knowledge. If possible, could you help to find out if this equation is solvable? The problem author argues that the equation has a solution.

I tried WA software several times. But, it does not succeed.

Answer 1

This polynomial $f$ has Galois group $F_5$, the Frobenius group of order 20. Since this group is solvable, then $f$ is solvable by radicals. Here are Magma commands showing this.

R<x> := PolynomialRing(Rationals());
f := 16*x^5-200*x^3-200*x^2+25*x+30;
G, data := GaloisGroup(f);
TransitiveGroupDescription(G);
IsSolvable(G);

which give as output

F(5) = 5:4
true

The docstring for the command SolveByRadicals describes how it gives solutions to a solvable polynomial:

Compute a splitting field as a radical tower and the roots of $f$ as elements in this tower. Return also the non-trivial roots of unity used.

(See here for more details.) We enter

K, roots := SolveByRadicals(f : Name := "K.");
K:Maximal;

which yields output

   K<K.1>
     |
     |
  $1<K.2>
     |
     |
  $2<K.3>
     |
     |
  $3<K.4>
     |
     |
     Q

K  : K.1^5 + 1/512*(-1875*K.4 - 3125)*K.3*K.2 - 2560000000
$1 : K.2^2 + 15118284881920000*K.4 - 34359738368000000
$2 : K.3^2 + 32*K.4 + 160
$3 : K.4^2 - 5

Denote by $\alpha_i$ the elements K.i for $i = 1, \ldots, 4$. Then this output is saying that we have a chain of simple radical extensions $$ K = K_1 \supseteq K_2 \supseteq K_3 \supseteq K_4 \supseteq \mathbb{Q} $$ where $K_i = K_{i+1}(\alpha_i)$ and the $\alpha_i$ satisfy the equations \begin{align*} 0&={\alpha_1}^5+1/512(-1875\alpha_4-3125)\alpha_3\alpha_2-2560000000\\ 0&={\alpha_2}^2+15118284881920000\alpha_4-34359738368000000\\ 0&={\alpha_3}^2+32\alpha_4+160\\ 0&={\alpha_4}^2-5\,.\end{align*} Or in other words, \begin{align*} \alpha_4 &= \sqrt{5}\\ \alpha_3 &= \sqrt{-32 \alpha_4 - 160}\\ \alpha_2 &= \sqrt{-15118284881920000 \alpha_4 + 34359738368000000}\\ \alpha_1 &= \sqrt[5]{1/512 (1875 \alpha_4 + 3125) \alpha_3 \alpha_2 + 2560000000} \end{align*} up to choosing the correct roots of unity when taking these roots.

To express the roots of $f$ in terms of these generators, we enter [K!r : r in roots];, which yields

[
(1/268435456000000000*(-3*K.4 - 5)*K.3*K.2 + 1/128000000)*K.1^4 + 1/640000*K.1^3 + (1/33554432000000*(-3*K.4 - 5)*K.3*K.2 + 1/16000)*K.1^2 + 1/80*K.1,
((1/268435456000000000*(2*K.4 + 5)*K.3 + 1/13421772800000000*(K.4 + 3))*K.2 + (1/4096000000*(K.4 - 1)*K.3 + 1/512000000*(-K.4 - 1)))*K.1^4 + (-1/10240000*K.3 + 1/2560000*(K.4 - 1))*K.1^3 + ((1/67108864000000*(-K.4 - 5)*K.3 + 1/838860800000*(K.4 + 2))*K.2 + (1/256000*K.3 + 1/64000*(K.4 - 1)))*K.1^2 + (1/2560*(-K.4 + 1)*K.3 + 1/320*(-K.4 - 1))*K.1,
((1/536870912000000000*(-K.4 - 5)*K.3 + 1/6710886400000000*(K.4 + 2))*K.2 + (1/2048000000*K.3 + 1/512000000*(K.4 - 1)))*K.1^4 + (1/20480000*(K.4 - 1)*K.3 + 1/2560000*(-K.4 - 1))*K.1^3 + ((1/33554432000000*(2*K.4 + 5)*K.3 + 1/1677721600000*(-K.4 - 3))*K.2 + (1/512000*(-K.4 + 1)*K.3 + 1/64000*(-K.4 - 1)))*K.1^2 + (-1/1280*K.3 + 1/320*(K.4 - 1))*K.1,
((1/268435456000000000*(2*K.4 + 5)*K.3 + 1/13421772800000000*(-K.4 - 3))*K.2 + (1/4096000000*(-K.4 + 1)*K.3 + 1/512000000*(-K.4 - 1)))*K.1^4 + (1/10240000*K.3 + 1/2560000*(K.4 - 1))*K.1^3 + ((1/67108864000000*(-K.4 - 5)*K.3 + 1/838860800000*(-K.4 - 2))*K.2 + (-1/256000*K.3 + 1/64000*(K.4 - 1)))*K.1^2 + (1/2560*(K.4 - 1)*K.3 + 1/320*(-K.4 - 1))*K.1,
((1/536870912000000000*(-K.4 - 5)*K.3 + 1/6710886400000000*(-K.4 - 2))*K.2 + (-1/2048000000*K.3 + 1/512000000*(K.4 - 1)))*K.1^4 + (1/20480000*(-K.4 + 1)*K.3 + 1/2560000*(-K.4 - 1))*K.1^3 + ((1/33554432000000*(2*K.4 + 5)*K.3 + 1/1677721600000*(K.4 + 3))*K.2 + (1/512000*(K.4 - 1)*K.3 + 1/64000*(-K.4 - 1)))*K.1^2 + (1/1280*K.3 + 1/320*(K.4 - 1))*K.1
]

You can try this code for yourself using the Magma online calculator.

Answer 2

If we are content to evaluate the solvability, we can use the Cayley resolvent described by Wikipedia. This gives an unexpected result.

First we note that the given equation has a leading coefficient of $2^4$. If we multiply by $2$ and define $y=2x$, we get an equation with smaller integer coefficients:

$ y^5-50y^3-100y^2+25y+60=0$

The Cayley resolvent method is applied to this equation. The quintic is solvable if the combination

$[P_3(z)]^2-1024z\Delta$

has a rational zero, where $\Delta$ and the coefficients of the cubic polynomial are evaluated from the coefficients of the reduced quintic. Formulas are given in the referenced Wikipedia article.

For the case at hand, the relevant outcome is that $P_3$ happens to have a zero constant term:

$P_3(z)=z^3-3000z^2+12800000z+\color{blue}{0}.$

This forces the complete sextic to also have a zero constant term, and so the sought rational root of the Cayley resolvent is simply zero!

Thus this particular quintic is solvable. But actually obtaining the solution requires a complex formula which, as reported by Wikipedia, runs to three pages (with the Cayley resolvent root already in hand). We might be content just to use the Magma online calculator from here on out, as Viktor Vaughn does, or use the trigonometric trick by lone student.

Answer 3

I have no idea how to solve the equation by radicals. However, I wanted to write it assuming that this answer will somehow work for us as well .

---

$$ \begin{align}16x^5-200x^3-200x^2&+25x+30=0\end{align} $$

Using the linear substitution $\thinspace x:=u-\dfrac 12\thinspace $, we have :

$$ \begin{align}2u^5-5u^4-20u^3+10u^2&+10u-1=0\end{align} $$

Then, recall the following trigonometric angle formula :

$$ \begin{align}\tan (5\alpha)&=\frac{\tan^5\alpha-10\tan^3\alpha+5\tan \alpha}{5\tan^4\alpha-10\tan^2\alpha+1}\end{align} $$

Let $\thinspace\tan\alpha=u\thinspace$ and $\thinspace\tan (5\alpha)=a\thinspace$, then :

$$ \begin{align}u^5-5ay^4-10u^3+10au^2&+5u-a=0\end{align} $$

which is equivalent to :

$$ \begin{align}2u^5-10au^4-20u^3+20au^2&+10u-2a=0\end{align} $$

Therefore, taking $\thinspace a=\dfrac 12\thinspace $ then reduces the original equation to :

$$ \begin{align}&\tan (5\alpha)=\frac 12\\ \implies &\alpha=\frac 15\left(\arctan \left(\frac 12\right)+\pi k\right)\end{align} $$

---

Thus, the final answer becomes :

$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{x=-\frac 12+\tan\left(\frac 15\left(\arctan \left(\frac 12\right)+\pi k\right)\right)}\end{align} $$

where $\thinspace k\in\{0,1,2,3,4\}\thinspace .$