# Error: $x^2 + 1 = 0$ has solution set $\{-1;1\}$

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.

• You can plug each of $1$ and $-1$ into the equation and find it is not satisfied. – Ross Millikan Mar 19 at 23:19

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

• (+1) What I would have written if I had more time. – José Carlos Santos Mar 19 at 23:21
• The final implication is true only if it's been established that $x$ is restricted to be a real number. – Barry Cipra Mar 19 at 23:27
• @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. – Kavi Rama Murthy Mar 19 at 23:28
• So, the correct way would be x^2+1=0 implies x=1 or x=−1 and empty implies {-1;1}? – Daniel Sehn Colao Mar 19 at 23:48
• @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. – Kavi Rama Murthy Mar 19 at 23:51

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 $$\pm i$$.

• 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 $x^2+1$. – fleablood Mar 20 at 0:19

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.

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.