How to prove $-\sqrt a < x<\sqrt a$ from $x^2I want to prove the following:

\begin{align}
x^2<a \iff \tag 1 \\
-\sqrt a < x<\sqrt a \tag 2
\end{align}
  Update:
Do I need some  restriction on $a$ or is $a\in\mathbb R$?
Between $(1)$ and $(2)$, is it correct to use "equivalence" or is it just "implies"?

My attempt:
Case 1: $x>0$: 
We have 
\begin{align}
x^2&<a\iff \tag 3\\
\sqrt{ x^2} &<\pm \sqrt a \tag 4
\end{align}
But $\sqrt{ x^2}=\lvert x \rvert$? This looks weird?
Case 2: $x<0$. 
If $x<0$ then $-x>0$ is positive, so
\begin{align}
(-x)^2=x^2&<a \iff \tag 5\\
x&<\pm \sqrt a \tag 6
\end{align}
 A: For $a\le0$ the inequality has not solutions then for $a>0$ we have that
$$x^2<a \iff x^2-a<0 \iff (x-\sqrt a)(x+\sqrt a)<0$$
that is


*

*$x<-\sqrt a\quad \land \quad x>\sqrt a$
which is impossible or


*

*$x>-\sqrt a\quad \land \quad x<\sqrt a$
that is indeed 
$$-\sqrt a<x<\sqrt a$$
A: Write $$x^2-\sqrt{a}^2<0$$ and the use the formula $$a^2-b^2=(a-b)(a+b)$$
A: From
$$x^2<a,$$ by taking the square root, a monotonic function, you get
$$\sqrt{x^2}=|x|<\sqrt a.$$
Writing $|x|<\pm a$ just doesn't make sense.
Now, if $x\ge 0$, then $$|x|=x<\sqrt a,$$
and if $x\le 0$, then
$$|x|=-x<\sqrt a.$$ 
Grouping these results,
$$(x\ge0\land x<\sqrt a)\lor(x\le 0\land -\sqrt a<x)$$ can be simplified to $$-\sqrt a<x<\sqrt a.$$
A: Take square roots of both sides in the inequality $$x^2< a$$ to get $$|x|<\sqrt a.$$ This is permissible provided the original inequality is true, which would automatically imply that $0\le a.$
Now $|x|=\pm x,$ depending on the sign of $x.$ Thus, if $x$ is positive, then the last inequality becomes $x<\sqrt a.$ And if $x\le 0,$ then the inequality is $-x<\sqrt a,$ or that $x>-\sqrt a.$ Combining these inequalities gives the result.
A: First of all, do you know that $0 < m < n$ means $\sqrt{m} < \sqrt{n}$?  Can you take that as a given.  If so, can you take for $m>0, n> 0$ then $m< n \iff \sqrt{m} < \sqrt{n}$ is a given?
In general:  You have an axiom that if $a < b$ and $c > 0$ then $ac < bc$. (as well as if $a < b$ then $a+d < b+d$
From that you can prove many propositions including i) if $a > 0$ then $-a < 0$; ii) if $a<b$ and $c < 0$ then $ac > bc$ and that iii) $x^2 = 0$ if $x=0$ and $x^2 > 0$ if $x \ne 0$, iv) if $a > 0$ there are two $x_0, x_1$ so that $x_i^2 = a$.  $x_0 > 0$ and $x_1 = -x_0 <0$; we call $x_0 := \sqrt{x_0}$ and $x_1=-\sqrt a$.  v) $\sqrt{x^2} = |x|$.  $|x| =x$ if $x\ge 0$ and $|x|=-x$ if $x \le 0$. etc. These are basic I won't go over them but I think it worth pointing out that 
For $a > 0; b> 0$ then 1)  $a < b \iff a^n < b^n\iff \sqrt[k]a < \sqrt[k]b$ for all $n,k \in \mathbb N$, is worth noting why it is true.
$a < b \implies a^2 = a*a < a*b < b*b < b^2$ and so by induction if $a^{n-1} < b^{n-1}$ then $a^n = a^{n-1}*a < b^{n-1}*a < b^{n-1}*b = b^n$.  And if $a \ge b$ then the same argument shows $a^n \ge b^n$ and so $a<b \iff a^n < b^n$.  And therefore $\sqrt[k]a < \sqrt[k]b \iff a=(\sqrt[k]a)^k< (\sqrt[k]b)^k = b \iff a^n < b^n$.
...
Okay, now we can begin:
To prove $x^2 < a \implies -\sqrt a < x <\sqrt a$.
$x^2 \ge 0$ so $a > 0$.  So $\sqrt a$ exists and $\sqrt a > 0$.
$x^2 \ge a$ so $\sqrt{x^2}=|x| > \sqrt a \ge 0$
If $x \ge 0$ we have $0 \le x=|x| = \sqrt{x^2} < \sqrt a$.
... and so $-\sqrt a < 0 \le x < \sqrt a$.
If $x < 0$ we have $0 \le -x=|x| =\sqrt{x^2} < \sqrt a$ and therefore:
$-\sqrt a < -\sqrt{x^2} = -|x| = x < 0$.
.... and so $-\sqrt a < x < 0 < \sqrt a$.
In either case we have $-\sqrt a < x < \sqrt a$.
...
to prove $-\sqrt a < x < \sqrt a\implies x^2 < a$.
(Note: if $\sqrt a$ exists at all, that means $a \ge 0$.)
If $x \ge 0$ then $0 \le x < \sqrt a$ and so $x^2 < (\sqrt a)^2 = a$
If $x < 0$ then $-\sqrt a < x < 0$ means $0 < -x < \sqrt a$.  And so $x^2 = (-x)^2 < (\sqrt a)^2 = a$.
Either way, $x^2 < a$.
