Only five solvable quintic equations of the form $x^5+ax^2+b=0$? What are their solutions?

According to Wikipedia there are only five solvable quintic equations of the form $x^5+ax^2+b=0,~~a,b \in \mathbb{Q}$ (up to a scaling constant $s$).

$$x^5-2s^3x^2-\frac{s^5}{5}=0 $$ $$ x^5-100s^3x^2-1000s^5=0 $$ $$x^5-5s^3x^2-3s^5=0 $$ $$x^5-5s^3x^2+15s^5=0 $$ $$ x^5-25s^3x^2-300s^5=0 $$

But the source of this claim is a web-page (even if it's Harvard), not an article.

Thus, my questions are:

Why is this true, and what is the original source of this knowledge?

What are the solutions to these equations (in radical form, or possibly trigonometric/hyperbolic form)?

Edit

I found the proper citation on the linked Web-Page (for which I sincerely thank the author, if they ever visit this post).

The paper is On Solvable Quintics $X^5+aX+b$ and $X^5+aX^2+b$ by Blair K. Spearman and Kenneth S. Williams and the full text is available with open-access. I will see if my questions are answered by this paper and update the post.

Solution 1:

According to the linked paper On Solvable Quintics $X^5+aX+b$ and $X^5+aX^2+b$ by Blair K. Spearman and Kenneth S. Williams, the solutions are as follows (we take $s=5$ in the first equation and $s=1$ for the rest):

$$x=\omega^{j}u_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4$$

$\omega$ - the fifth root of unity.

$$x^5-5x^2-3=0 $$

$$u_1=\left(-\frac{1}{4}+\frac{\sqrt{5}}{20}-\frac{1}{100} \sqrt{150+30\sqrt{5}}+\frac{1}{50} \sqrt{150-30\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(-\frac{1}{4}-\frac{\sqrt{5}}{20}-\frac{1}{100} \sqrt{150+30\sqrt{5}}-\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$u_3=\left(-\frac{1}{4}-\frac{\sqrt{5}}{20}+\frac{1}{100} \sqrt{150+30\sqrt{5}}+\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$u_4=\left(-\frac{1}{4}+\frac{\sqrt{5}}{20}+\frac{1}{100} \sqrt{150+30\sqrt{5}}-\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$x^5-5x^2+15=0 $$

$$u_1=\left(\frac{5}{4} + \frac{13\sqrt{5}}{20} - \frac{7}{100} \sqrt{750+330\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(\frac{5}{4} - \frac{13\sqrt{5}}{20} - \frac{7}{100} \sqrt{750-330\sqrt{5}} \right)^{1/5}$$

$$u_3=\left(\frac{5}{4} - \frac{13\sqrt{5}}{20} + \frac{7}{100} \sqrt{750-330\sqrt{5}} \right)^{1/5}$$

$$u_4=\left(\frac{5}{4} + \frac{13\sqrt{5}}{20} + \frac{7}{100} \sqrt{750+330\sqrt{5}} \right)^{1/5}$$

$$ x^5-25x^2-300=0 $$

$$u_1=\left(-\frac{25}{2} - \frac{5\sqrt{5}}{2} - \frac{5}{2} \sqrt{30+6\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(-\frac{25}{2} + \frac{5\sqrt{5}}{2} - \frac{5}{2} \sqrt{30-6\sqrt{5}} \right)^{1/5}$$

$$u_3=\left(-\frac{25}{2} + \frac{5\sqrt{5}}{2} + \frac{5}{2} \sqrt{30-6\sqrt{5}} \right)^{1/5}$$

$$u_4=\left(-\frac{25}{2} - \frac{5\sqrt{5}}{2} + \frac{5}{2} \sqrt{30+6\sqrt{5}} \right)^{1/5}$$

$$ x^5-100x^2-1000=0 $$

$$u_1=-2^{6/5},~~~~u_2=-2^{7/5},~~~~u_3=2^{3/5},~~~~u_4=-2^{4/5}$$

$$x^5-250x^2-625=0 $$

$$u_1=(-125+50 \sqrt{5})^{1/5}$$

$$u_2=\frac{(-375-175 \sqrt{5})^{1/5}}{2^{1/5}}$$

$$u_3=\frac{(-375+175 \sqrt{5})^{1/5}}{2^{1/5}}$$

$$u_4=(-125-50 \sqrt{5})^{1/5}$$

Solution 2:

Like all solvable quintics, these can be expressed in terms of trigonometric and inverse trigonometric functions, and two of them have nice representations using the golden ratio,

$$\phi = \tfrac{1+\sqrt{5}}{2}$$

The unique real roots are given by,

I. $\quad x^5-5x^2-3=0 $

$$x=\frac{-2}{5^{1/4}}\left(\cos\Big(\tfrac{1}{5}\arccos\big(\frac{5^{3/4}}{2\,\phi}\big)\Big)-i\,\sin\Big(\tfrac{1}{5}\arcsin\big(\frac{5^{3/4}\phi\;i}{2}\big)\Big)\right)=-1.8087\dots$$

III. $\quad x^5-25x^2-300=0 $

$$x=2\cdot 5^{1/4}\left(\cos\Big(\tfrac{1}{5}\arccos\big(5^{1/4}\phi\big)\Big)-i\,\sin\Big(\tfrac{1}{5}\arcsin\big(\frac{5^{1/4}\;i}{\phi}\big)\Big)\right)=3.6287\dots $$