Simplifying inequality contradicts actual inequality I am doing a problem and am not sure why the final step is the following
$x^{2}<4 <=> -2<x<2$

When simplifying
$x^{2}<4$ it becomes $x<±2$ which is the same as $x<2$, $x<-2$
but this does not match the actual inequality of $-2<x<2$, since $x>-2$ not $x<-2$ which is one of the inequalities
obtained in the line above
My question:
Inequalities are swapped when multiplying & dividing by negative numbers, so is the reason why $x^{2}<4 <=> -2<x<2$
because square rooting both sides of $x^{2} <4$ give $x<2$ AND $x >-2$ since $4$ became $-2$ so it is like the inequality rule of dividing by a negative number?
Hence the sign is flipped when considering  $√(4)=-2$? $<=> 4/-2=-2$?

What I think:
$x^{2}<4$ gives four possible inequalities:
(1)$x<2$
(2)$-x<-2=>x>2$
(3)$-x<2=>x>-2$
(4)$x<-2$.
By inspection, (1) & (3) are the actual inequalities and (2) & (4) are false solutions.
(1) & (3) = $x<2$, $x>-2$ $<=>$ $-2<x<2$
Is this the reason why??
 A: Care must be exercised when considering negative square roots. For example, we have $9\gt4$, but not $-3\gt-2$.
So we can only take positive square roots.
From $x^2\lt4$ and $(-x)^2\lt4$ this gives two equations:

*

*$x\lt2$

*$-x\lt2\implies x\gt-2$
A: Your thinking is correct but it is quite tedious to consider the four possible inequalities.
A faster method would be simply observing the fact that,
$$|x|<2$$ where $x^{2}<4$ is true simply when the magnitude of $x$ is less than $2$ (e.g. $x=0,1,1.3,1.999$)
Note that |x| is a piecewise function
$$|x| = \begin{cases}
                -x\\
                 x
                    \end{cases}$$
Therefore, we obtain

*

*$x<2$

*$-x<2 \Rightarrow x>-2$
$$-2<x<2$$
Also if you were to simplify the inequality using algebra like you have shown in your question, you would have to look through the four inequalities and see which are valid. As mentioned that is tedious, so you should just consider the magnitude of $x$ and make use of $|x|$
A: You have to deal with inequalities carefully:
$x^2 < 4 \iff x^2-4<0 \iff(x-2)(x+2)<0 \iff -2< x < 2$
Now can you see why?
A: You must take care when simplifying the inequality:
You say that $(1):x<2 , (2): x<−2$
Squaring $(2)$:  $x^2>4$. The reason this is, to square $-2$ is to multiply by a negative on the RHS (-2). And on the LHS, you are also multiplying by a negative since $x$ is less than  -2 (hence negative).
Squaring  $(1)$: $x^2<4$.  Well that contradicts $(2)$. So you have made an error.
Also: $x<2$ and $x<-2$ implies $x<-2$, and I showed previously that this will imply that $x^2>4$, which contradicts your premise.
