Square root and principal square root confusion A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the concept with the following two questions...
Question 1:
$\sqrt{12x^2} = \sqrt{4\cdot3x^2} = 2|x|\sqrt{3}$
Why do we only have to keep in mind that $x$ could be negative, yet we just factorize $\sqrt{12}$ to $\sqrt{3\cdot4}$ and put the 2 right in front of the radical sign?
$\sqrt{12}$ could be factorized as $\sqrt{-3 \cdot -4}$ as well.
Question 2:
I've been told to take the square root of both sides in the following equation, but the square root doesn't have its own symbol right? Only the principal root does... So if you'd use the radical sign you're using the principal root and therefore you're missing out on a solution in the following scenario:
$(x + 8)² = 1 \iff \sqrt{(x + 8)^2} = \sqrt1 \iff x + 8 = 1 \iff x = -7$
Does that mean that whenever we want to take the square root of something (not the principal) we just use a $\pm$ symbol instead? For example:
$(x + 8)^2 = 1 \iff x + 8 = \pm 1 \iff x = -8\pm1$

Q1: If $a,b$ are nonnegative, then $\sqrt{ab}=\sqrt a\sqrt b$. If $a<0$ or $b<0$ (and you are not into complex numbers), you better not talk about the square roots on the right. 
Q2:
In general $a=b\implies f(a)=f(b)$.
Since $(x+8)^2$ and $1$ are both nonnegative, we can ltake $\sqrt{\ }$ for the function $f$ and obtain 
$$(x+1)^2=1\implies \sqrt{(x+1)^2}=\sqrt 1.$$
(Actually, since the square root function is injective on $[0,\infty)$, the use of "$\iff$" would be justifed, but itis not for general $f$). 
Now recall that $$\tag0\sqrt {t^2}=|t|$$ for all real $t$ (wheras $\sqrt{t^2}=t$ only for nonnegative $t$, and we don't know if $x+1$ is nonnegative). Hence by simplificatiopn of expressions we find
$$\sqrt{(x+1)^2}=\sqrt 1\iff |x+1|=1$$
Now the appearence of $|\cdot|$ suggests that we investigate two cases:
$$\tag a x+a\ge 0\quad\land\quad x+1=1$$
$$\tag b x+a< 0\quad\land\quad -(x+1)=1$$
So the $\pm$ sign (or any other distinction of two cases, e.g. as I just did) ultimately comes from equation $(0)$.

An alternative way to proceed from the beginning is as follows, using the third binomial formula:
$$ (x+1)^2=1\iff (x+1)^2-1=0\iff((x++1)+1)((x+1)-1)=0$$
and now use that a product is zero if and only if at least one of the factors is zero.
