- We have to keep in mind that $x$ might be negative because we don't know what $x$ is and it might by negative.
But we don't have to keep anything in mind about $12$ because we know exactly what $12$ is.
But perhaps we need to understand why we are allowed to split $\sqrt {ab}=\sqrt{a}\sqrt{b}$ in the first place.
Real Number Answer (Assumes Since $m^2 \ge 0$ for all $m \in \mathbb R$ then $x^2 = $a negative number is impossible)
If $k=\sqrt{a}\sqrt{b}>0$ (presuming $a,b> 0$) then $k^2 = (\sqrt{a}\sqrt{b})^2 = \sqrt{a}^2 \sqrt{b}^2 = ab$ so if we ask "What is $\sqrt {ab}$, that is what is the unique non-negative number so $x^2=ab$". Well, we just demonstrated that $k = \sqrt{a}\sqrt{b}$ is such a number.
So if we have $M > 0$ and $M$ could be written as $M = a*b$ where $a,b\ge 0$ we would have $\sqrt M = \sqrt{a}\sqrt {b}$.
Now you are correct that $M = (-a)*(-b)$. But how does that help us in anyway. We explained why $M =a*b$ works but $M=(-a)*(-b)$ is as useful for find square roots as noting that $M = j+k$ for some $j$ and $k$. It's just not relevant.
We need $\sqrt{a}$ so that $\sqrt{a}^2 =a$ as well as a $\sqrt{b}$ so that $\sqrt{b}^2 = b$ for this to work and if we tried it with $-a*-b$ we'd need $\sqrt{-a}$ so that $\sqrt{-a}^2 = -a$ and there's just no such thing.
Unless we use complex numbers...
Complex number: There exist $i$ and $-i = (-1)*i$ so that $i^2=-1$. (And $(-i)^2 = i^2 = -1$ as well.
So if $k^2 = 12$ we could do $(\sqrt{3}2)^2 = 3*4 = 12$ so $k=\sqrt 3\sqrt 4=2$ is an acceptable answer. So is $(\sqrt{3}i*2i)^2 = (\sqrt\3i)^2*(2i)^2 = (-3)*(-4) = 12$ but $\sqrt{3}i*2i$ is just another way of writing $\sqrt{3}i*2i = (\sqrt{3}2)*(i^2)= -2\sqrt 3$.
So the two values of $k$ are $2\sqrt 3$ and $-2\sqrt 3$ just as we wanted.
But we can do $\sqrt{12} = \sqrt{-3}\sqrt{-4} = \sqrt 3i* 2i=-2\sqrt 3$.
(I suppose I should point out we tend not to worry about primary square roots in complex numbers because it just opens a can of worms.)
- "I've been told to take the square root of both sides in the following equation"
Well, short answer, since we do not know whether $x+8$ is positive or negative we don't know that $\sqrt{(x+8)^2} = x+8$. If $x = -17$ that is false.
$\sqrt{(-17+8)^2} = \sqrt{(-9)^2} =\sqrt{81}= 9\ne -17+8$.
Instead all we know is that $\sqrt{(x+8^2} = |x+8|$
$\sqrt{(-17+8)^2}=\sqrt{81} = 9 $ which does $|-9| = |-17+8|$.
So to solve
$(x + 8)^2 = 1 \implies \sqrt{(x+8)^2} = \sqrt 1\implies |x+8| = 1$
$\implies -(x+8) = 1$ OR $x+8 = 1$ $\implies$
$x+8 = -1$ ** OR ** $x+8 = 1$
And we can write that last line as
$x+8 = \pm 1$.
Or we can take a short cut and go directly to:
$(x+8)^2 = 1$
$\sqrt{(8+x)^2} = \sqrt 1$
$8+ x = \pm 1$.
.... and from here we can go
$x = \pm 1 -8$ so $x= -1 -8=-9$ OR $x=1-8=-7$.