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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?

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    $\begingroup$ Take a look this: mathforum.org/library/drmath/view/69477.html $\endgroup$ – Ryan Soh Aug 16 at 9:10
  • $\begingroup$ You should make than an answer if you can. $\endgroup$ – Toby Mak Aug 16 at 9:11
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    $\begingroup$ It's a duplicate but i am not able to find the older post $\endgroup$ – Baba Yaga Aug 16 at 9:14
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    $\begingroup$ 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 $x$. $\endgroup$ – Gerry Myerson Aug 16 at 9:17
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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 $\begin{cases}x\ne0\\ B(x)-A(x)=0\\ A(x)=0\end{cases}$

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  • $\begingroup$ 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). $\endgroup$ – Curiousmind Aug 16 at 9:40
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    $\begingroup$ 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. $\endgroup$ – Gae. S. Aug 16 at 11:25
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    $\begingroup$ 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. $\endgroup$ – Gae. S. Aug 16 at 11:30
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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. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$

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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.

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