Extra solutions appearing like magic

Question

Take $(x+1)^2 = x^2 + 2x +1$ which is true for all $x$. Then we have $x = 2x^2-5x$ which is true for $x = 0 $ and $x = 3$. If we have any linear combination of these two, I think you can only use $x$ which is valid in both, which is just $0$ and $3$. Then writing $(2x^2 - 5x + 1)^2 = x^2 + 2x +1$, should only be satisifed by $0$ and $3$, however this is not the case ( we have an extra solution $x=1$). Now I would like to ask, does this extra solution affect what we already did? What I mean by this is that just because we could only be sure that the equation held for some $x$, does that interfece with the logic that it holds for other $x$ which we did not expect? Or is it a case of, all we can say is that it holds for $0 $ and $3$. (Not that they are the only solutions, but that it definitely holds for those $x$)

Answer 1

Say you have two equations $(E_1)$ and $(E_2)$, and let $a,b$ be numbers. If $x$ is a solution to both $(E_1)$ and $(E_2)$, then it is indeed a solution to the linear combination $a(E_1)+b(E_2)$, however there is no reason why the converse should hold in general.