I am trying to solve the quadratic equation $x^2 + x + 1 = 0$. $x^2 = -1 - x $ $\iff x = -\frac{1}{x} - 1$, assuming $x\neq 0$.

Substituting that into the original equation gives $x^2 + (-\frac{1}{x} -1) + 1 = 0$

$x=1$ is a solution to this second equation, but it isn't a solution to the first. How did this extra solution arise? Which step wasn't reversible and why wasn't it?

  • 4
    $\begingroup$ In essence you are just multiplying the equation by $x-1$. $\endgroup$ – njguliyev Oct 19 '13 at 21:45
  • 1
    $\begingroup$ here is another way of asking a similar question: $(x-1)(x^2+x+1)=0(x-1)$, then $x^3-1=0$, hence $x=1$ $\endgroup$ – newzad Oct 19 '13 at 21:48
  • $\begingroup$ Cool, I am going to try this out at work... $\endgroup$ – imranfat Oct 19 '13 at 23:19

The source of this extraneous solution is the same as that of those encountered while completing the square: mistaking $ P \implies Q $ to mean the converse $ Q \implies P $, where $P$ is $ x^2 + x + 1 = 0 $ and $Q$ is $ x^2 + (-\frac{1}{x} - 1) + 1 = 0 $.

Try going backwards from $Q$ to $P$ yourself: you can't, at least not without the proposition $R$ $ x = -\frac{1}{x} - 1 $ (which is rightly derivable from $P$, but not from $Q$).

A more rigorous way to go about solving for $x$ would be to assume $ P \iff Q \ \& \ R $. So now you don't have to go back to $P$ to check if your solution is extraneous, but you do have the added restriction of $R$ which you have to keep in mind.

Since $R$ eliminates $ x = 0 $, and there are no other solutions for $Q$; you are forced to conclude that $P$ does not have any solutions either, and indeed your quadratic equation has no solutions in $\mathbb{R}$ (the two roots are complex: $-\sqrt[3]{-1} = -i^{\frac{2}{3}} $ and $ (-1)^\left(2/3\right) = i^{\frac{4}{3}}$).

That would mean $x = undef$ (or rather $\not\exists x \in \mathbb{R}$), I'm not sure what what the mathematical meaning or validity of $(x−1)(x^2+x+1)=0(x−1)$ would be other than just be saying $undef = undef$. (Could someone formalize what's happening here?)

Also, if we're trying to go backward from $x^2 + \frac{1}{x} = 0$ (which does have the solution 1), we can indeed multiply both sides by the identity $\frac{(x - y)(x^2 + xy + y^2}{(x^3 - y^3)}$ as it is equal to $1$; but we must ensure that the denominator we're dividing by is not 0 (as if we break the rules, we're kicked out of the nice world of our number theory), which in the case of $x = 1$ is indeed that exactly.


You have $x=-x^2-1$ and $x=-\frac1x-1$,

Let $A=x$, $B=-x^2-1$ and $C=-\frac1x-1$. You start with $A=B$ and $A=C$. You conclude that $B=C$. The trouble is going backwards from $B=C$ to $A=B$ and $A=C$.

  • $\begingroup$ Could you clarify what A, B and C are in this case to save poor readers of some extra mental effort? $\endgroup$ – Yatharth Agarwal Jul 19 '14 at 17:19

Your reasoning is:

$\hspace{57pt} x^{2}+x+1\overset{1}{=}0 \ \left(x\in\mathbb{R}\right)$

$\overset{1}{\iff} \hspace{30pt} \text{Rearrange:} \ x^{2}\overset{2}{=}-x-1$

$\overset{2}{\iff} \hspace{30pt} \text{Since }x=0\text{ doesn't solve } \overset{1}{=} \text{, we have } x\neq0. \text{ So, divide } \overset{2}{=} \text{ by } x \text{ to get}$:

$$ x\overset{3}{=}-1-\frac{1}{x}$$

$\overset{3}{\iff} \hspace{30pt} \text{Now plug } \overset{3}{=} \text{ into } \overset{1}{=} \text{to get:}$

$$x^{2}+\left(-1-\frac{1}{x} \right)+1\overset{4}{=}0.$$

And now we see that $x=1$ solves $\overset{4}{=}$ but not $\overset{1}{=}$. Why?

The error in the above chain of reasoning is in $\overset{2}{\iff}$.

The operation of "dividing by $x$" presumes that a (real) solution to $\overset{1}{=}$ exists. Thus, $\overset{2}{\iff}$ should not be an $\iff$ step but a $\implies$ step. It is thus at this step that extraneous solutions may be introduced.

Here there is indeed no (real) solution to $\overset{1}{=}$ and an extraneous solution was indeed thereby introduced.


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