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?

Question

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 $1-2q$?

Answer 1

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! :)