Why must absolute value be used in this equation? I have a simple equation:
$(x-2)^2 < 3$
My first solution was:
$x-2 < \sqrt{3}$
But this gives me only:
$x < \sqrt{3} + 2$
Which is only one solution, so it's not enough. I've figured it out that I must use:
$|x-2| < \sqrt{3}$
Then, solutions are:
$-\sqrt{3} + 2 < x < \sqrt{3} + 2$
My question is how can I know that here:
$(x-2)^2 < 3$
I must use absolute value after square rooting both sides of inequality? Like: $|x-2| < \sqrt{3}$
 A: Because, if $b>0$, then $a^2<b\iff\lvert a\rvert<\sqrt b$. This is so because\begin{align}a^2<b&\iff b-\lvert a\rvert^2>0\\&\iff\left(\sqrt b-\lvert a\rvert\right)\left(\sqrt b+\lvert a\rvert\right)\\&\iff\sqrt b-\lvert a\rvert>0\text{ (since $\sqrt b+\lvert a\rvert>0$)}\\&\iff\lvert a\rvert<\sqrt b.\end{align}
A: The quantity $(x-2)^2$ is derived by the squaring of the quantity $(x-2)$. But, you do not know whether this expression is positive or not. Thus, by square-rooting (as you did), you must take into consideration both cases, which is only yielded by using the absolute value.
Note, that the absolute value represents the essence of distance in the set of reals $\mathbb R$. The expression $|x-2| < \sqrt{3}$ means that the distance of $x$ from $2$ is less than $\sqrt{3}$. But $x$ can be either positive or negative, exactly as we wanted.
Thus, a rigorous solution in such case when you're trying to solve for every $x \in \mathbb R$, shall be as:
$$(x-2)^2 < 3 \Rightarrow |x-2| < \sqrt{3} \Leftrightarrow -\sqrt{3} + 2 < x < \sqrt{3} + 2$$
A simple example to understand the importance of the absolute value: Take into account the equation $x^2 = 4$. If you do not use the absolute value and thus not taking into consideration the negative value case, you will only get $x=2$. But, isn't also $(-2)^2 = 4$ ? Thus, the right way to go is $x^2 = 4 \Rightarrow |x| = 2$.
A: If you are not comfortable using $|x-2|$ then you can divide your derivation into two cases as follows:
Case 1: $x-2\ge 0$ :
$(x-2)^2 < 3 \\ \Rightarrow 0 \le x-2 < \sqrt 3 \\ \Rightarrow 2 \le x < 2+\sqrt{3}$
Case 2: $x-2< 0$ :
$(x-2)^2 < 3 \\ \Rightarrow -\sqrt 3 < x-2 < 0\\ \Rightarrow 2-\sqrt 3 < x < 2$
Taking the union of the intervals from the two cases, we get $2-\sqrt{3}<x<2+\sqrt{3}$. The case by case approach is useful for more complex inequalities e.g. $2 < (x-2)^2<3$.
A: Because $\sqrt{x^2} = |x|$ for $x \in \mathbb{R}$. When $x-2$ is negative, a square root doesn't just negate the exponent.
A: Set $y:=x-2$, then 
$0\le y^2<3$;
$y^2-3\lt0$;
$(y+√3)(y-√3)\lt 0$;
Only viable option:
$y +√3 \gt 0$ and $y -√3 \lt 0$.
$y \gt -√3$, and $y \lt √3$ i.e.
$-√3 <y<√3$, or $-√3 <x-2<√3$.
