Why are the solutions of the equation different? : $x=2 => x^2=4 => x=±2$

Question

If I define the variable $x$ as $x=2$, then $x^2=4$. But the solutions of $x^2=4$ are $±2$(two solutions). I defined what the variable $x$ is, then why are the solutions for the equation $x^2=4$ two, not only one?

Answer 1

You are confusing neccesary and sufficient conditions. It is true that

1. if $x=2$, then $x$ is a solution to the equation $x^2=4$
2. if $x^2=4$, then $x$ is either equal to $2$ or to $-2$.

There is no logical mistake here, it's just that $x^2=4$ means that $x$ can be one of two things, and $x=2$ means it is one precise thing.

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In a way, you are losing information when you go from $x=2$ to $x^2=4$. For example, you can conclude, from $x=2$, that $x>0$, but you cannot conclude that from $x^2=4$.

Logically, there is nothing wrong with that. If I tell you I weigh 70kg, I have also told you that I weigh more than $50$kg, right? But if I tell you I weigh over 50kg, you cannot then conclude that I weigh 70kg...

Or, if I tell you I have a male cat, you can conclude I have a cat. But if I tell you I have a cat, you cannot conclude I have a male cat.

Answer 2

The implication $$x^2 = 4 \implies x = \pm 2$$ says that if $x^2 = 4$ then $x=2$ or $x=-2$, which is true. So the implication $$x = 2 \implies x = \pm 2$$ is correct.

Answer 3

Because $f(x) = x^2$ is not injective (i.e. not one to one).

$x=2$ implies $x^2 =4$ but $x^2 =4$ does not necessarily imply $x=2$.

More generally, raising both sides of an equation to an even power can introduce extraneous or "redundant" roots that satisfy the transformed equation but not the original. If you do this manipulation, you must substitute any solutions you find back into the original equation and verify them.