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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.

ryang
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Ransplito
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  • To summarize the already given responses, let Eq. 2 denote $x^2 - \frac{1}{x} = 0.$ Then Eq. 1 is a 2nd degree equation and Eq. 2 is a disguised form of a 3rd degree equation. Every root of Eq. 1 is a root of Eq. 2, but not vice versa. In effect, it is as if you took the equation $(x - r_1)(x - r_2) = 0$ and transformed it into the equation $(x - r_1)(x - r_2)(x - r_3) = 0$. – user2661923 Nov 01 '21 at 21:28
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    I find it funny though that none of the answers or the comment addresses the actual question. OP seems very aware that Eq.2 is implied by but does not imply Eq.1, but they wonder what exact step they made is not reversible. Nobody has identified that clearly (although Emilio's hint goes in the right direction). – Torsten Schoeneberg Nov 01 '21 at 21:30
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    ... well, as I was typing that, Arturo Magidin spelled it out: It's the step that turns one equation into two equations and then drops the first of those two, leaving us with the more general second one ("first" and "second" in the order Arturo writes them underneath each other). You cannot reverse that "dropping of the first equation". – Torsten Schoeneberg Nov 01 '21 at 21:34
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    By the way: the easiest way to pinpoint the step in which the extraneous solution is introduced is simply to go step-by-step and see where you suddenly get that $1$ is a solution. Doing that it becomes immediately apparent that it's the step where you go form $x^2+x+1+\frac{1}{x}=x^2$ (in which $1$ is not a solution) to the step $\frac{1}{x}=x^2$ (where $1$ is a solution). Once you pinpoint the step in which this occurs, it becomes a question of figuring out why it occurs. – Arturo Magidin Nov 02 '21 at 01:38
  • By the way, the step where you multiply by $1/x$ needs more justification than just saying $1/x$ exists (and is $\neq 0$). Actually, if you are precise you have to write two equations exactly like in Arturo's answer there as well, it just happens that in this specific case, those two equations are equivalent and you can drop either of them without changing the solutions. For comparison, you can also tell that $x=7$ is not a solution for Eq.1, so $x-7 \neq 0$ exists, but when you multiply both sides through with $(x-7)$, you do introduce another solution. – Torsten Schoeneberg Nov 02 '21 at 04:32
  • You might want to check Where is the error in this "proof" that 3=0? and the linked posts. – Sil Nov 02 '21 at 11:10
  • To be extremely clear, the error only occurs in the final line "The real root of the eqn is..." immediately after the system of two equations had been created; your mistake was simply in ignoring the first of the two equations that you had just created! – ryang Nov 02 '21 at 21:28
  • @TorstenSchoeneberg It always preserves equivalence to divide both sides of an eqn by a nonzero variable; OTOH, multiplying both sides by any variable is always valid $({\Rightarrow})$ but doesn't preserve equivalence $({\not\Leftarrow})$ (so, multiplying both sides by $(x-7)$ creates an extraneous root $x=7$ unless it's already a root of the original eqn). – ryang Nov 02 '21 at 21:34

3 Answers3

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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 $x^2+x+1=0$).

Arturo Magidin
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    Thanks @Arturo. You are the only one who has grasped the essential logical issue here. – Cheerful Parsnip Nov 01 '21 at 21:44
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    @CheerfulParsnip: Exactly. People who do not fully grasp the logical reasoning involved (which is actually quite basic) should not be posting 'answers'. To be specific, finding real solutions to this equation means finding a property $Q$ such that $∀x{∈}ℝ\ ( \ x^2+x+1 = 0 ⇔ Q(x) \ )$. Anyone who actually grasps that and can perform proper logical deduction can easily see that the attempt in the question can only yield $∀x{∈}ℝ\ ( \ x^2+x+1 = 0 ⇒ x^2 = 1/x \ )$, and not an equivalence. – user21820 Nov 02 '21 at 06:23
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Hint:

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

Emilio Novati
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The first step when we got an extraneous root is the equation $x^2 - \frac{1}{x} = 0$. It is a corollary of equation(1). But it's only corollary: it's not equivalent to it.

Botnakov N.
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  • I think OP knows that. That's their entire question: Which one of the steps to get from Eq.1 to $x^2-\frac{1}{x}$ is not reversible? – Torsten Schoeneberg Nov 01 '21 at 21:36
  • @TorstenSchoeneberg, the step when we write $x^2 - \frac{1}{x} = 0$: it's a corollary and it's the first time when we have extraneous root. I made an addition to emphasize this. – Botnakov N. Nov 01 '21 at 21:38