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So I found out that some transformations break wolfram alpha's ability to solve polynomials. The simplest case I could find is the polynomial $$2x^{5}+5x^{4}+5x^{2}+1=0$$ for which the solution is $$x=\frac{1}{\sqrt[5]{\sqrt{2}-1}-\sqrt[5]{\sqrt{2}+1}}=\frac{\sqrt[5]{12\sqrt{2}-17}-\sqrt[5]{12\sqrt{2}+17}-\sqrt[5]{2\sqrt{2}+3}+\sqrt[5]{2\sqrt{2}-3}-1}{2}$$ Which derives simply from the solvable Moivre quintic $$x^{5}+5ax^{3}+5a^{2}x+2b=0,\ x=\sqrt[5]{\sqrt{a^{5}+b^{2}}-b}-\sqrt[5]{\sqrt{a^{5}+b^{2}}+b}$$ which wolfram alpha can solve. But for some reason, when I try this polynomial, wolfram alpha can only approximate it. Weirder, wolfram alpha is entirely capable of solving other quintics with a sum of $4$ radicals and a rational number, like $$x^{5}+5x^{4}+10x^{3}+10x^{2}+8=0,\ x=\frac{\sqrt[5]{750\sqrt{25+5\sqrt{5}}+375\sqrt{25-5\sqrt{5}}-1250\sqrt{5}-3125}+\sqrt[5]{375\sqrt{25+5\sqrt{5}}-750\sqrt{25-5\sqrt{5}}+1250\sqrt{5}-3125}-\sqrt[5]{750\sqrt{25+5\sqrt{5}}+375\sqrt{25-5\sqrt{5}}+1250\sqrt{5}+3125}-\sqrt[5]{375\sqrt{25+5\sqrt{5}}-750\sqrt{25-5\sqrt{5}}-1250\sqrt{5}+3125}-5}{5}$$ and that example is considerably more complicated than this $1$. A few more sophisticated examples exist, like $ax^{5}+bx^{2}+c=0$, which is solved with hypergeometric functions. Wolfram alpha does use hypergeometric functions to solve other polynomials, like $ax^{5}+bx+c=0$, so what gives?

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    $\begingroup$ Just because the galois group is solvable this does not mean that it is easy to solve this quintic. Why should Wolfram Alpha be able to calculate formulas so complicated that noone would ever use them ? $\endgroup$
    – Peter
    Commented Oct 16, 2022 at 17:13
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    $\begingroup$ @Peter I mean if it can calculate other formulas which are MORE complicated, then why not? Also, this really is very easy, the solution is just the reciprocal of the solution to a Moivre quintic, which has a nice formula for the roots. $\endgroup$ Commented Oct 16, 2022 at 17:15

1 Answer 1

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Simply put, mathematics programs do not always include the methods we dream up. If we are smart enough and maybe stand on the shoulders of giants, so to speak, we can occasionally beat Wolfram Alpha.

I recall using a TI-89 symbolic calculator to solve the quartic equation

$$x^4+x^3+x^2-x+1=0,$$

knowing that the polynomial has factors $(x^2+ax-1)(x^2+\overline{a}x-1)$ for a pair of complex conjugates $a,\overline{a}$. Thus the roots should be expressible in the form $u+vi$ with $u$ and $v$ containing only rational operations and real square roots:

$x=\frac14[(-1\pm_1\sqrt{2\sqrt5+3})\pm_2i(-\sqrt{11}\pm_1\sqrt{2\sqrt5-3})]$

(Identical subscription on the $\pm$ signs indicate signs that are to be chosen identically.)

The TI-89 bombed out completely, giving only numerical answers. Wolfram Alpha does better; it finds the factorization (or an equivalent result) but then fails to resolve the complex square root radicals from solving the quadratic factors, which "should be" a straightforward process.

The quintic equation

$$x^5+10x^3-5=0$$

has Cayley and Watson resolvents with zero roots, so the equation is solvable and the quintic roots may be rendered from fifth roots of the solution to a pair of quadratic equations (see here). In this case a relatively simple form is given:

$$x=\phi^{1/5}+\phi^{3/5}-\frac{1}{\phi^{1/5}}-\frac{1}{\phi^{3/5}}$$

where $\phi=\frac{1+\sqrt5}2$ is the golden ratio. Complex roots are obtained by rendering the above as

$a+a^3-(1/a)-(1/a^3)$

where $a$ runs through all the fifth roots of $\phi$. But WA misses this and gives only numerical solutions.

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    $\begingroup$ +1 I just love it when the roots of a solvable quintic involve only powers of the golden ratio. $\endgroup$ Commented Jun 19, 2023 at 8:38

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