Is this valid when deriving quadratic equation? When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres the quadratic formula. I just don't understand why its not absolute value of $(x+\frac{b}{2a})$. For example the square root of $x^2$ is the absolute value of $x$, which is equal to $\pm{x}$. Sorry if this is confusing its more of a conceptual thing. Thank you in advance.

philalethesnew
You seem to be asking why after the step $\;\left(x+\cfrac b{2a}\right)^2=\cfrac{b^2-4ac}{4a^2}\;$ , they don't go to

$$\left|x+\cfrac b{2a}\right|=\sqrt{\frac{b^2-4ac}{4a^2}}$$

But we actually do! To write the above is exactly the same as to write

$$x+\cfrac b{2a}=\pm\sqrt\frac{b^2-4ac}{4a^2}$$

just as $\;x^2=a\implies |x|=a\;\; (\,a\ge 0)\;$ is exactly the same as $\;x^2=a\implies x=\pm a\;\;(a\ge0)\;$ , under the assumption, of course, that once we take the + sign and the second time we take the - sign.
When they get to point of having
$$\bigl{(}x+\frac{b}{2a}\bigr{)}^2 =-\frac{c}{a}+\bigl{(}\frac{b}{2a}\bigr{)}$$ the square root of both sides has no absolute value sign on the left because the right side can be $\pm$.
$$\biggl{|}x+\frac{b}{2a}\biggr{|}\ne\sqrt{-\frac{c}{a}+\biggl{(}\frac{b}{2a}\biggr{)}}$$

One complete derivation explained simply is shown here.