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So, is it correct that the solution set of $x^2 + 1 = 0$ is [-1;1] ?

Is the error in equation development or in the solution set?

Help me, I don't know how to proceed in this question.

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    $\begingroup$ You can plug each of $1$ and $-1$ into the equation and find it is not satisfied. $\endgroup$ – Ross Millikan Mar 19 at 23:19
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All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.

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    $\begingroup$ (+1) What I would have written if I had more time. $\endgroup$ – José Carlos Santos Mar 19 at 23:21
  • $\begingroup$ The final implication is true only if it's been established that $x$ is restricted to be a real number. $\endgroup$ – Barry Cipra Mar 19 at 23:27
  • $\begingroup$ @BarryCipra Sure. I am basing my answer on the assumption that $x$ is a real number. I will wait for the OP to clarify if he is allowing $x$ to be complex. $\endgroup$ – Kabo Murphy Mar 19 at 23:28
  • $\begingroup$ So, the correct way would be x^2+1=0 implies x=1 or x=−1 and empty implies {-1;1}? $\endgroup$ – Daniel Sehn Colao Mar 19 at 23:48
  • $\begingroup$ @DanielSehnColao Since neither $x=1$ nor $x =-1$ satisfies $x^{2}+1=0$ the conclusion is there is no real number $x$ satisfying the given equation. $\endgroup$ – Kabo Murphy Mar 19 at 23:51
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When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $\pm 1$.

Note that

$$x^4=1$$

has four roots in the complex domain which are $\pm 1$ and also $\pm i$.

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  • $\begingroup$ I'm not sure why the other solutions are not pointing out the erroneous trick that multiplying by $x^2-1$ add extraneous solutions. IMO that is the entire aspect of the problem. So +1 to you. Also note, even if you don't consider complex roots. $x^2 +1=0$ has no real roots. And $x^2 -1=0$ has two. So $x^4 -1$ has $2$ but two came from $x^2 -1$ and zero came from $x^2+1$. $\endgroup$ – fleablood Mar 20 at 0:19
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The final implication should be "$\implies x\in\{1,-1,i,-i\}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.

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No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.

This would b similar to doing this:

Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So

$(x^2 - 5x+6 ) = 0$ so

$(x^2 - 5x + 6)(x+3057) = 0\cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.

The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.

So in this "false proof":

So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).

But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.

By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.

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