# Extraneous solution from substituting in equations

I came across this example of how you could end up with an extraneous solution but I was wondering how it arose. We have the equation: $$x^2+x+1=0$$ Since x=0 does not satisfy the equation, you can divide by x on both sides which yields: $$x+1+\frac{1}{x}=0$$ which is equivalent to our first equation. From our first eqution we can conclude that: $$-x^2=x+1$$ We now substitute that into the second equation to get:$$x^2=\frac{1}{x}$$ which results in $$x^3=1$$ which is equivalent to our previous equation since x cannot be 0. However, one solution from our last equation is x=1, which is not a solution to our original equation. I have a vague idea that it may have to do with the fact that you get a cubic equation and you began with a quadratic, and that steps imply the following and not vice versa, but can you provide a very detailed answer as to why it arises? Can you please provide more examples?

• Take a look this: mathforum.org/library/drmath/view/69477.html Aug 16 '20 at 9:10
• You should make than an answer if you can. Aug 16 '20 at 9:11
• It's a duplicate but i am not able to find the older post Aug 16 '20 at 9:14
• If $A(x)=B(x)$ has the same solutions as $A(x)=C(x)$, it does not follow that $B(x)=C(x)$ also has exactly those solutions. E.g., $x=2$ has the same solution as $x=2$, but $2=2$ is satisfied by all values of $x$. Aug 16 '20 at 9:17

If we call $$A(x)=x^2+x+1$$ and $$B(x)=x+1+\frac1x$$, we can schematize your passages as such: $$A(x)=0\Leftrightarrow \begin{cases}x\ne 0\\ B(x)=0\end{cases}\Leftrightarrow \begin{cases}x\ne 0\\ A(x)=0 \\B(x)=0\end{cases}\stackrel{!!!}\Rightarrow \begin{cases}x\ne 0\\ B(x)-A(x)=0\end{cases}$$

Whereas to preserve equivalence you should have kept $$A(x)=0$$ in $$\begin{cases}x\ne0\\ B(x)-A(x)=0\\ A(x)=0\end{cases}$$

• When you say x is not equal to zero, is it just saying if x is not 0 then blablabla... is it just a remainder? Could we avoid it since they are equal for all x in the domain of A? (of course you have to consider if it is 0 before dividing but I'm thinking whether you can omit that). Aug 16 '20 at 9:40
• I am saying that the condition $[A(x)=0]$ is equivalent to the condition $[x\ne 0\land B(x)=0]$. What I have written is a schematization of the relevant passages of your work (explicit or implicit). How you prove (co-)implications is mostly a communicative aspect: for instance, strictly speaking I do not need someone to explain why $x^2+x+1=0$ and $\begin{cases} x\ne0\\ x+\frac1x+1=0\end{cases}$ are equivalent, because I can figure out by myself the few standard passages of interest.
– user239203
Aug 16 '20 at 11:25
• When I'm doing a system of inequations, I like to keep existence conditions of the terms inside the system itself as much as I can (it becomes somehow cumbersome with trigonometric functions, though). I have my reasons to believe this is a good course of action.
– user239203
Aug 16 '20 at 11:30

This substitution ($$x+1=-x^2$$) expands a set of roots of the equation

because $$-x^2$$ also depends on $$x$$.

You can substitute $$x+1=y$$, for example.

More example, when a similar substitution gives similar problems.

Let we need to solve $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1}.$$

We obtain: $$\left(\sqrt[3]{2x+1}+\sqrt[3]{x+1}\right)^3=x-1$$ or $$2x+1+x+1+3\sqrt[3]{2x+1}\cdot\sqrt[3]{x+1}\left(\sqrt[3]{2x+1}+\sqrt[3]{x+1}\right)=x-1.$$ Now, since $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1},$$ which can get something bad, we obtain: $$3\sqrt[3]{(2x+1)(x+1)(x-1)}=-3-2x$$ or $$x(440x^2+630x+189)=0$$ and we got as one of options $$x=0$$.

Easy to see that $$0$$ is not a root of the starting equation and it happened

because we used a not correct substitution $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1}.$$

Now, we need to check that all roots of the equation $$440x^2+630x+189=0$$ are roots of the starting equation, which is not so easy.

If we want to avoid from these problems, so we need to use the following identity. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$

All transformations of an equation must be reversible. With $$x=0$$,

$$x^2+x+1=0\leftrightarrow x+1+\frac1x=0$$ is fine.

But combining two equations in one $$\begin{cases}x+1=-\dfrac1x\\x+1=-x^2\end{cases}\leftrightarrow x^2=\frac1x$$ is not.