Extraneous solution when solving $x^2+x+1=0$ by getting $x^2=1/x$

Let's assume that we have$$x^2+x+1=0.\tag1$$ Substituting $x=0$, we get $1=0$, so $0$ is not a root for the quadratic equation and thus, $x\neq0$. Therefore, there exists $\frac{1}{x}$, which we'll multiply by both sides of $(1)$, giving us $$x+1+\frac{1}{x}=0.$$ We will, then, move $\frac{1}{x}$ to the other side and get $$x+1=-\frac{1}{x}.$$ If we add $x^2$ to both sides and note that $x^2+x+1=0$, we will have $x^2-\frac{1}{x}=0$. The real root of this equation is $x=1$, which is not a root of $(1)$.

I was wondering at which step did I do something that was incorrect and resulted in this supposed root.

Solution 1:

To go from $x^2+x+1+\frac{1}{x}=x^2$ to $\frac{1}{x}=x^2$ you are also assuming $x^2+x+1=0$. So you are changing the single equation $x^2+x+1+\frac{1}{x}=x^2$ to the system of equations $$\begin{align*} x^2+x+1&=0\\ x^2&=\frac{1}{x}. \end{align*}$$

This is equivalent to the system $$\begin{align*} x^2+x+1&=0\\ x^3&=1. \end{align*}$$

This is, in turn, equivalent to the system $$\begin{align*} x^2+x+1&= 0\\ (x-1)(x^2+x+1)&=0. \end{align*}$$ But this system is equivalent to $x^2+x+1=0$... that is, your original equation.

The error (or rather, the non-reversible step) lies in "forgetting" about the condition $x^2+x+1=0$ which you are assuming to get to $x^2=\frac{1}{x}$; by dropping it explicitly, you end up with the equation $x^3-1=0$, or $(x-1)(x^2+x+1)=0$; this has the solutions $x^2+x+1=0$ (the original equation) plus the solution $x-1=0$ (the extraneous solution which was introduced when you forgot to keep the global condition that $x^2+x+1=0$).

Solution 2:

Hint:

$x^2+x+1=0$ is true only for some value of $x$ not $\forall x$.