Find all four roots of quartic equation $x^4-x+1=0$

Question

How to solve

$$x^4-x+1=0$$

My attempt:

$$x^4-x+1=0$$

$$\implies x^4-x^3-x+1+x^3=0$$

$$\implies x^3(x-1)-(x-1)+x^3=0$$

$$\implies (x^3-1)(x-1)+x^3=0$$

But, I couldn't find a way to combine $x^3$ into that factorization.

I then looked at Wikipedia to see how to solve a quartic. I'm not sure which method is the best one. The coefficients are pretty simple (in the form $ax^4+bx^3+cx^2+dx+e$, $a=e=1$, $b=c=0$, $d=-1$). Should I just use the general formula for quartic equations, or something else?

Also, I couldn't find a post on here talking about how to solve quartic equations. If someone finds a link to such a post then I might as well just delete this question. The only post I found that might be useful is this question but sadly there are no answers there.

EDIT: I would prefer all four solutions, real or complex.

@Lee How did you get that inequality? Is there some kind of rule that allows you to say that? — Aiden Chow, Jun 04 '20 at 18:38
In general a polynomial equationl with degree 4 cannot solved algebraically. — callculus42, Jun 04 '20 at 18:39
@callculus, that is incorrect. There is a quartic formula, though it is somewhat unwieldy. — Integrand, Jun 04 '20 at 18:40
@AidenChow if we work with real numbers, than if you insert $x>1$ then definitely the equality doesnt hold, same for $x<-1$ — Lee, Jun 04 '20 at 18:41
@EDX Should I use that formula to solve this question or the general formula? — Aiden Chow, Jun 04 '20 at 18:43
@callculus I provided a link to the general formula for quartic equations in my question. — Aiden Chow, Jun 04 '20 at 18:43
Mathematica gives the approximate solutions $-0.727\pm 0.934i$ and $0.727\pm 0.430 i$. I figured you were looking for exact solutions but this gives a place to start. — Aaron Hendrickson, Jun 04 '20 at 18:49
Where is this from? Do you have reason to believe that the solutions can be found via a neat method (i.e. without the monstrous general method for solving quartic polynomials)? — Sebastian Monnet, Jun 04 '20 at 19:03
@Qwertiops I don't really have any reason to believe that this question would give neat answers. I just wanted to see what the best method for solving this would be and then maybe solve it myself. It seems like there isn't any neat method to do this problem based on the answers and comments. — Aiden Chow, Jun 04 '20 at 19:07

score 9 · Answer 1 · answered Jun 04 '20 at 18:46

9

There are no real solutions because $x^4-x+1$ attains a positive minimum at $x=1/\sqrt[3]{4}$.

answered Jun 04 '20 at 18:46

lhf

216,483

Quanto · Answer 2 · 2020-11-17T15:51:50.893

Note that $x^4-x+1=0$ is a deeply-depressed quartic equation, which makes it manageable. In fact, it can be factorized as

$$x^4-x+1= \left( x^2- ax+ \frac{a^3-1}{2a} \right) \left( x^2+ ax+ \frac{a^3+1}{2a} \right) =0\tag1 $$

where $a$ satisfies the cubic equation $(a^2)^3-4a^2-1=0$ and can be obtained analytically $$a = \sqrt{\frac4{\sqrt3} \cos\left( \frac13\cos^{-1}\frac{3\sqrt3}{16}\right)}$$

Then, solve the two quadratic equations in (1) to obtain the four complex roots

$$x = \frac a2 \pm \frac i2\sqrt{a^2-\frac2a},\>\>\> -\frac a2 \pm \frac i2\sqrt{a^2+\frac2a} $$

score 5 · Answer 3 · answered Jun 04 '20 at 19:57

A new method for solving quartics known as the ferrari method which has quite posts on this site so we add a factor of $(ex+f)^2$ on both sides so the equation becomes $$(x^2+ax+b)^2=(ex+f)^2$$ and we have to determine $a,b,e,f$

so expand $(x^2+ax+b)^2$ and you will get $$x^4+a^2x^2+b^2+2bx^2+2ax^3+2abx=x^4-x+1+e^2x^2+f^2+2efx$$ on comparing coefficients we get $$\begin{align} a =0 \rightarrow (1) & \\2ef = 1 \ \ \ \rightarrow (2) \\1+f^2=b^2\rightarrow (3) \\e^2 = 2b\rightarrow (4) \end{align}$$ now square the $2^{nd}$ equation to get $$f^2 = \frac{1}{8b}$$ put this result in $(3)$ and form a cubic polynomial in $b$ which is $$8b^3-1-8b=0$$ after this I think you can proceed

score 4 · Answer 4 · answered Jun 04 '20 at 18:46

It's sufficent to show that it has no roots in $\mathbb{R}$:
Let $f(x)=x^4-x+1$, then $f'(x)=4x^3-1$, $x_0=\sqrt[3]{\frac{1}{4}}$,
$f(x)$ decreases on $(-\infty,x_0)$ and increases on $(x_0,\infty)$ so it's sufficent to find $f(x_0)$. $$f(x_0)=\frac{1}{8}\left(8-3\sqrt[3]{2}\right)>0\hbox{ as } 8^3>3^3\cdot 2$$ For complex roots one can try Ferrari method. Encyclopedia of Mathematics.

score 3 · Answer 5 · answered Jun 04 '20 at 19:03

Before diving into any details, I consulted Wolfram Alpha and noted that the roots are non-real complex conjugate pairs. Results from Wolfram Alpha for $x^4-x=1=0$.

From the section on the nature of solutions, I cite: The possible cases for the nature of the roots are as follows: [...] If $P > 0$ or $D > 0$ then there are two pairs of non-real complex conjugate roots. [...]

We calculate some of the related coefficients. We find that $$P=8ac-3b^2=0$$ and $$R=b^3+8da^2-4abc=-8<0$$ and $$D=64a^3e-16a^2c^2+16ab^2c-16a^bd-3b^4=64>0$$ and $\Delta_0=12>0$.

The case $P=0$ and $D>0$ does not seem to be listed. But actually, I should have started with the discriminant $\Delta$ (which has only two non-zero terms, subject to human error) and I calculate that $\Delta=229>0$.

Whenever $\Delta>0$, all four roots are real or none of them are. I do not see a reason why $P=0$ is not listed.

Sebastiano · Answer 6 · 2020-06-05T12:26:37.067

The Newton-Raphson method uses an iterative process to approach one root of a any function: $$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$$ This could be a method to see that the equation $$x^4-x+1=0$$ not have any real solutions.

Indeed if you separated the fourth degree equation into two functions, the first $f(x)=x^4$ and the second $g(x)=x-1$, starting from the equation $x^4-x+1=0$, using Desmos to draw one function, for example, you can observe that there is no intersections beetween $f$ and $g$. I have chosen the graphic way.

score 0 · Answer 7 · answered Jun 12 '20 at 04:58

Equation $\displaystyle p x + x^4 = t$

Solution:

$\displaystyle Q = ((-(27 p^4 + 128 t^3) + 3 (3 p^4 (27 p^4 + 256 t^3))^{1/2})/2)^{1/3}$

$\displaystyle A = (Q + 4 t (4 t/Q - 1))/(6 p)$

$\displaystyle B = (32 (3 p A + t))^{-1/6}$

$\displaystyle F = 256 B^{12} t (16 A^4 + 2 A p - t)$

$\displaystyle R_2 = cos((arccos(1 + 8 F) + 2 \pi j)/4)$

$\displaystyle j=0,1,2,3$

$\displaystyle R = 4 B^3 (1 - A^2)$

$\displaystyle y = (R_2 - R)/(4 B^4)$

$\displaystyle x= A \pm (1 + B y)^{1/2}$

score 0 · Answer 8 · answered Jun 12 '20 at 20:48

0

The polynomial is irreducible but solvable.

$$x≈-0.72714 \pm 0.93410 i\qquad \land\qquad x≈0.72714 \pm 0.43001 i$$

answered Jun 12 '20 at 20:48

poetasis

6,338

Find all four roots of quartic equation $x^4-x+1=0$

8 Answers8