# What is the correct way to solve the equation: $x^4-x^3+x^2-x+1=0$

Given the equation: $$x^4-x^3+x^2-x+1=0$$ we need to find both its real and complex roots. What is the easiest and correct method for solving the equation?

Here is my approach, but it gives wrong result on the end. Since the equation is symmetric we can group the terms.

$$x^4+1 - (x^3+x)+x^2=0 \text{ Divide everything by } x^2 \\(x^2+\frac{1}{x^2})-(x+\frac{1}{x})+1=0\\ \text{ Let } t = x + \frac{1}{x}, \text{ we can see that } x^2 + \frac{1}{x^2} = t^2 - 2 \\ \text{back in our equation: } t^2 - 2 - t + 1 = 0 \\ t_{12} = \frac{1 \pm \sqrt{5}}{2} \\ \text{however if we go back in } x+\frac{1}{x} = t_{12} \text{ we don't get the correct result }$$.

As given in the textbook the solutions are: $$x_{12}=\frac{1+\sqrt5 \pm \sqrt{10-2\sqrt5}}{4}, x_{34}=\frac{1-\sqrt5 \pm \sqrt{10-2\sqrt5}}{4}$$ Can someone say if those are the correct solutions or not?

• math.stackexchange.com/questions/403025/… – lab bhattacharjee Apr 24 at 5:58
• Multiply the equation by $(x+1)$ and find $x^5=-1$. – Empy2 Apr 24 at 6:04
• Your textbook's solutions are not the solutions to the equation you give. They are the solutions to $x^4+1=0$. The "answers in the back of the book" are not always right. – Lord Shark the Unknown Apr 24 at 6:09
• It turned out that I was looking at the solution of another task all the time, the given solutions are updated, can someone verify if they are correct or not? – someone123123 Apr 24 at 6:20
• As you've written them, all four of the new "solution" have the same imaginary part. That is not the case for the actual solutions. – Lord Shark the Unknown Apr 24 at 6:51

The easiest way to solve is to note that if $$x\neq-1$$ then $$\frac{x^5+1}{x+1}=x^4-x^3+x^2-x+1,$$ and so writing $$x=re^{\theta i}$$ quickly yields $$r=1$$ and $$\theta=\tfrac k5\pi$$ with $$k$$ odd.
Your approach is also fine; you get the same solutions by solving the two quadratics $$x+\frac1x=\frac12\pm\frac{\sqrt{5}}{2}.$$
• @someone123123 It seems you've posted the pair of solutions $x_{12}$ twice, and haven't posted $x_{34}$. And the pair you give is not a solution (but close); check your calculations again. – Inactive - Objecting Extremism Apr 24 at 6:22