There is this question in which the real roots of the quadratic equation have to be found:

$x^2 + x + 1 = 0$

To approach this problem, one can see that $x \neq 0$ because:

$(0)^2 + (0) + 1 = 0$

$1 \neq 0$

Therefore, it is legal to divide each term by $x$:

$x + 1 + \frac{1}{x} = 0$

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

Now, substitute $x$ into the original equation and solve:

$x^2 + (-1-\frac{1}{x}) + 1 = 0$

$x^2-\frac{1}{x} = 0$

$x^3 = 1$

$x = 1$

to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.

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    $\begingroup$ Can you describe how you arrived at $x=1$ more precisely? $\endgroup$ – Jakobian Sep 24 '18 at 0:34
  • $\begingroup$ @Jakobian Okay, I have edited it. $\endgroup$ – user595054 Sep 24 '18 at 0:38
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    $\begingroup$ The key point, as in my answer, is that there are three solutions to $x^3 = 1$. $\endgroup$ – Carl Mummert Sep 24 '18 at 0:40
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    $\begingroup$ For similar reasons, if you multiply your original equation by $(x-1)$, you get $x^3-1=0$. $\endgroup$ – Teepeemm Sep 24 '18 at 2:51

For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.

A trivial example of such a case is the equation $\,x=1\,$. We can substitute $\,1 \mapsto x\,$ on the RHS and end up with $\,x=x\,$. Of course that $\,x=1 \implies x=x\,$, but the converse is not true.

In OP's case, the original equation is quadratic in $\,x\,$ which has $2$ roots in $\Bbb C\,$, while the derived equation is a cubic which has $3$ roots in $\Bbb C\,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $\,x=1\,$ as noted already, which does not satisfy the original quadratic.

  • $\begingroup$ Hi. I am confused to why $x = x$ is not valid? Thanks. $\endgroup$ – user595054 Sep 24 '18 at 1:21
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    $\begingroup$ @Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $\,x=x\,$ do indeed include the root $\,x=1\,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $\,x^3=1\,$ are correct, but at this point it is not guaranteed that all roots of $\,x^3=1\,$ also satisfy the original equation $\,x^2+x+1=0\,$, and in fact $\,x=1\,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots. $\endgroup$ – dxiv Sep 24 '18 at 1:25
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    $\begingroup$ @Strikers I proved there is a real root No, you did not, because the steps are not reversible. What you proved is $\,x^2+x+1=0 \implies x^3=1 \implies x=1\,$ (in reals). That's correct, but those are one-way implications, which you cannot "reverse" to prove that $\,x=1 \implies x^2+x+1 = 0\,$ i.e. prove that $\,x=1\,$ is a root. And quite obviously $\,x=1\,$ is not a root to $\,x^2+x+1=0\,$, so the original equation has no real roots. $\endgroup$ – dxiv Sep 24 '18 at 1:52
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    $\begingroup$ @Strikers if I have an equation and I assume there is a real root OK, that's what you did in the original post. but if my assumed real root does not work You did not assume any particular real root. You proved that if such a root existed, then it must be $\,1\,$, which of course turns out not to be a root of the original equation. Sorry, but I don't know how to make that point any more clear as much as I've tried, so I'll just leave it at that. $\endgroup$ – dxiv Sep 24 '18 at 2:47
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    $\begingroup$ Okay, that makes sense. Thank you so much for your time. $\endgroup$ – user595054 Sep 24 '18 at 2:57

You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.

Here is what we get by substituting:

$$ x^2 + x + 1 = 0$$ $$ x^2 + (-1 - 1/x) + 1 = 0$$ $$ x^2 - 1/x = 0 $$ $$ x^2 = 1/x$$ $$ x^3 = 1 $$

There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.

This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.

  • $\begingroup$ Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks. $\endgroup$ – user595054 Sep 24 '18 at 1:18
  • $\begingroup$ In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers. $\endgroup$ – Carl Mummert Sep 24 '18 at 1:39
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    $\begingroup$ Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't. $\endgroup$ – Carl Mummert Sep 24 '18 at 1:41
  • $\begingroup$ Hi. What do you specifically mean by it is not a "reversible operation"? Thanks. $\endgroup$ – user595054 Sep 24 '18 at 1:45
  • $\begingroup$ It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation. $\endgroup$ – Carl Mummert Sep 24 '18 at 1:47

The higher level description of your work is:

  • Assume $x$ is a solution to the original equation.
  • Then $x$ has to be $1$
  • $1$ is not a solution to the original equation.

And therefore we conclude the assumption is false: that is,

  • Therefore, the original equation has no solutions.

Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to

  • Assume $x$ is a solution to the original equation.
  • Then $x$ has to be $1$ or either $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ (because those are the three cube roots of $1$)
  • $1$ is not a solution to the original equation.
  • $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ are solutions to the original equation

and therefore

  • The solutions to the equation are $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$
  • $\begingroup$ Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks. $\endgroup$ – user595054 Sep 24 '18 at 1:20
  • $\begingroup$ @Strikers You assumed that $x$ is a solution because literally, when you write $x^2 + x + 1 = 0", you're saying that $x$ is a value that satisfies that very equation. And later, when you divide by $x$, you're implicitly assuming that such a $x$ exists in the first place (because if the $x$ does not exist, how can you divide an equation by it?) $\endgroup$ – Pedro A Sep 24 '18 at 5:09
  • $\begingroup$ @PedroA Okay that makes sense, thanks. $\endgroup$ – user595054 Sep 24 '18 at 5:43

Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows

$$\begin{align} x_1 &= 1, & \text{or} \\ x_2 &= -\frac{1}{2} + \frac{\sqrt 3}{2} i, & \text{or} \\ x_3 &= -\frac{1}{2} - \frac{\sqrt 3}{2} i \end{align}$$

Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.

This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).

  • $\begingroup$ Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks. $\endgroup$ – user595054 Sep 24 '18 at 1:22
  • $\begingroup$ To answer your first question: $i$ represents the imaginary unit, which has the property $i^2 = -1$. To answer your second question: $x = 1$ is an allowed solution of $x^3 = 1$ (this is why I listed it) but it is not an allowed solution of your original problem $x^2 + x + 1 = 0$ (this is why you must reject it). $\endgroup$ – Trevor Kafka Sep 24 '18 at 3:18

Transformations you apply to an equation may introduce alien solutions.

Taking an extreme example,

$$x=0\implies 0=0$$ which is satisfied by all $x$ !

So you may apply transformations, but validate the solutions using the original equation.

In your example, you establish

$$x^2+x+1=0\implies x^3-1=0.$$

But as $$x^3-1=0=(x-1)(x^2+x+1),$$ nothing is wrong if you ignore the first factor.


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