Why isn't this approach in solving $x^2+x+1=0$ valid?

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

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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 $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

Answer 4

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).

Answer 5

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.

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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.