<s>[INST] In simplifying $\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}}$ to $\frac{x^2 +x +3}{|1-2q|}$, why use the absolute value? I have seen in a question
$$\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}}$$ was given to be
$$\frac{x^2 +x +3}{|1-2q|}$$
Why was absolute value given to $1-2q$?
 [/INST] Well, by definition you have that:
$$
\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}} = \frac{\sqrt{(x^2 +x +3)^2}}{\sqrt{(1-2q)^2}}
$$
and you got that $\forall x \in \mathbb{R}\  \sqrt{x^2} = |x|$
so you get that:
$$
\frac{\sqrt{(x^2 +x +3)^2}}{\sqrt{(1-2q)^2}} = \frac{|(x^2 +x +3)|}{|(1-2q)|}
$$
Clearly $x^2 \geq 0 \ \forall x \in \mathbb{R}$, and in particular you have that $x^2 \geq x $ while $|x| > 1$, if $x \in [-1,1]$ you can easily see that $ x^2 + x < 3$  so you have that $x^2+x+3$ is always greater than 0 so you can get rid of $|.|$ (as well when you consider $x^2-x+3$)
Now if you have $q > 0$ you can see that $1-2q$ is not necessarily positive, so you need to use your absolute value.
If you still have doubts I suggest you to graph the functions and watch that both $x^2-x+3$ and $x^2+x+3$ are positive in all its codomain! :)
</s>