How to solve $x^4+8x-1=0 $

Question

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.

Answer 1

There will be a real root a little above zero, as your polynomial has $f(0)=-1, f(1)=8$ You can use any root finding technique from here. Bisection is easy-try $f(\frac 12)$ and replace the endpoint of the interval $(0,1)$ that has the same sign as $f(\frac 12)$. This isn't the fastest, but it will work. There are other techniques in any numerical analysis text. The Newton-Raphson method will have an easy time here. There is another root near $x=-2$. Graphing the function will show that, then you can use the same technique. The other two roots are complex.

Here is the bisection $$\begin {array} {c c c c c c}a&b&(a+b)/2&f(a)&f(b)&f((a+b)/2)\\0&1&0.5&-1&8&3.0625\\0&0.5&0.25&-1&3.0625&1.003906\\0&0.25&0.125&-1&1.003906&0.000244\\0&0.125&0.0625&-1&0.000244&-0.49998\\0.0625&0.125&0.09375&-0.49998&0.000244&-0.24992\\0.09375&0.125&0.109375&-0.24992&0.000244&-0.12486\\0.109375&0.125&0.117188&-0.12486&0.000244&-0.06231\\0.117188&0.125&0.121094&-0.06231&0.000244&-0.03103\\0.121094&0.125&0.123047&-0.03103&0.000244&-0.0154\\0.123047&0.125&0.124023&-0.0154&0.000244&-0.00758\\0.124023&0.125&0.124512&-0.00758&0.000244&-0.00367\\0.124512&0.125&0.124756&-0.00367&0.000244&-0.00171\\0.124756&0.125&0.124878&-0.00171&0.000244&-0.00073\\0.124878&0.125&0.124939&-0.00073&0.000244&-0.00024 \end {array}$$

The root is always between $a$ and $b$. The length of the interval gets cut in half each step. You can see the root is just below $x=1/8$, which you might guess as $x^4$ is rather small, so you want $8x-1$ to be a tiny amount positive.

Answer 2

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 . $x\approx -2,0.12,0.95-\pm\sqrt{3}i$

Answer 3

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.

Answer 4

All equations of degree $\le 4$ can be solved by general formulae. The ones for $4$-th degree equations may be found on Wikipedia. Unfortunately, they are so long that, even in your easy case when the coefficients of $x^3$ and $x^2$ are zero, they are too long to present here. Anyway, computer algebra systems can solve this explicitly, instantly; Mathematica outputs:

$$\frac{1}{\sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}}-\frac{1}{2} \surd \left(-\frac{2 \left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}+\frac{2}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}-8 \sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}\right), \\ \frac{1}{\sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}}+\frac{1}{2} \surd \left(-\frac{2 \left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}+\frac{2}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}-8 \sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}\right), \\ -\frac{1}{\sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}}-\frac{1}{2} \surd \left(-\frac{2 \left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}+\frac{2}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}+8 \sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}\right), \\ -\frac{1}{\sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}}+\frac{1}{2} \surd \left(-\frac{2 \left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}+\frac{2}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}+8 \sqrt{\frac{2}{\frac{\left(36+\sqrt{1299}\right)^{1/3}}{3^{2/3}}-\frac{1}{\left(3 \left(36+\sqrt{1299}\right)\right)^{1/3}}}}\right)$$

Answer 5

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.

Answer 6

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

$$x= -\frac{\sqrt a}2\pm\sqrt{\frac4{\sqrt a}- \frac a4 },\>\>\> \frac{\sqrt a}2\pm i \sqrt{\frac4{\sqrt a}+\frac a4 } $$