How to solve $x^4+8x-1=0 $ Any idea how to solve the following equation?
$$x^4+8x-1=0$$
I tried to find some obvious roots of this equation so it could help me to find the other roots (if it has more then $1$) but I had no success, and I would really appreciate some help.
 A: Rewrite our equation in the following form
$(x^2+k)^2-(2kx^2-8x+k^2+1)=0$.
We want that $2kx^2-8x+k^2+1=2k(x+a)^2$ for which we need $16-2k(k^2+1)=0$,
which gives $k^3+k-8=0$.
The last equation has unique positive root.
Thus, we get $(x^2+k)^2-2k\left(x-\frac{2}{k}\right)^2=0$, where $k^3+k-8=0$,
and we can get an exact roots, but it's very ugly.
A: Note
$$x^4+8x-1=\left(x^2-\sqrt a x +\frac a2+\frac4{\sqrt a}\right)\left(x^2+\sqrt a x +\frac a2-\frac4{\sqrt a}\right)=0
$$
where $a^3+4a-64=0$, yielding $a =\frac4{\sqrt3}\sinh\left( \frac13\sinh^{-1}(12\sqrt3)\right)$. Solve to obtain
$$x= -\frac{\sqrt a}2\pm\sqrt{\frac4{\sqrt a}- \frac a4 },\>\>\>
\frac{\sqrt a}2\pm i \sqrt{\frac4{\sqrt a}+\frac a4 }
$$
A: hint see this article from wolramalpha its irreducible over $Q$ http://www.wolframalpha.com/input/?i=is%20x%5E4%2B8%20x-1%20irreducible%3F&lk=2 if you want to find roots they are hard to find . $x\approx -2,0.12,0.95-\pm\sqrt{3}i$
A: The two real roots $r_1, r_2$ are obtainable as follows.
Let
$$ \eqalign{
a_1 &= 2 \sqrt{1299} + 72\cr
a_2 &= 2 \cdot 18^{1/3} a_1^{2/3} - 12\cr
a_3 &= \sqrt{144 \sqrt{a_1} + 6 \sqrt{a_2} - 18^{1/3} a_1^{2/3} \sqrt{a_2}}\cr
r_1 &= \dfrac{18^{1/6}(2 a_3 - \sqrt{2} a_2^{3/4})}{12 a_1^{1/6} a_2^{1/4}}
\cr
r_2&= \dfrac{18^{1/6}(-2 a_3 - \sqrt{2} a_2^{3/4})}{12 a_1^{1/6} a_2^{1/4}}
}$$
I think that way of presenting them, taking advantage of common-subexpression elimination, is much less formidable-looking than writing them out in full.
