Is $x^5-10x^3+20x= 8.58368$ solvable?

Question

This quintic has 5 real roots, how do we find out if it is solvable and ,in that case, how to solve it? Is there a generally valid numeric approach?

Answer 1

Hint:

Let $x=a\sin t$

$$\dfrac{a^5}{16}=\dfrac{10a^3}{20}=\dfrac{20a}5$$

$\implies a^2=8,$

$$\implies x^5-10x^3+20x=\pm?\sin5t\le?$$ for real $x$

Answer 2

In general, a quintic cannot be solved exatly by radicals. But substiture $x\leftarrow y/10$ to arrive at $$y^5-1000 y^3+200000y = 858368$$ First, we look for rational solutions, which must be among the divisors of $858368$. After finitely many trials, we find $y=28$ ($x=2.8$). The remaining solutions hide in $$ y^4 + 28y^3 - 216y^2 - 6048y + 3065600=0,$$ which unfortunately does not factor so easily. While quartics are solvable by radicals, one would usually still prefer to resort to numerical methods to tackle them (as in, usually a quite exact numerical value helps you more than an extremely convoluted exact expression).