6
$\begingroup$

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.

$\endgroup$
15
  • 2
    $\begingroup$ Is $x\in \mathbb R$? $\endgroup$ – callculus Jun 4 '20 at 18:34
  • $\begingroup$ @callculus check my edit. $\endgroup$ – Aiden Chow Jun 4 '20 at 18:36
  • 2
    $\begingroup$ @callculus, that is incorrect. There is a quartic formula, though it is somewhat unwieldy. $\endgroup$ – FearfulSymmetry Jun 4 '20 at 18:40
  • 1
    $\begingroup$ @callculis Look at Ferrari Formula. $\endgroup$ – EDX Jun 4 '20 at 18:41
  • 1
    $\begingroup$ @EDX Should I use that formula to solve this question or the general formula? $\endgroup$ – Aiden Chow Jun 4 '20 at 18:43
9
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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

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

$\endgroup$
6
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

enter image description here

$\endgroup$
0
$\begingroup$

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

Solution:

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

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

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

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

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

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

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

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

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

$\endgroup$
0
$\begingroup$

The polynomial is irreducible but solvable.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.