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
 A: 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.
A: Well actually it is done. 
$$ax^2 + bx + c = 0$$
$$\implies x^2+\frac{bx}{a} +\frac{c}{a} = 0$$
Now completing the square: 
$$\left(x+\frac{b}{2a}\right)^2= \frac{b^2-4ac}{4a^2}$$
Now since we know that, 
$$x^2 = a, \quad a > 0$$ 
has two solutions, $x=\pm a$. 
Now if $b^2 - 4ac > 0$, there are two solutions indeed! 
$$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$. 
Does that help?
A: You can look at it differently:
$$ax^2+bx+c=0 \iff \\
x^2+\frac bax+\frac ca=0\iff \\
\left(x+\frac b{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}=0 \iff \\
\left(x+\frac b{2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \iff \\
\left[\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}\right]\cdot \left[\left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}\right]=0 \iff \\
\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}=0 \ \ \text{or} \ \ \left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}=0 \iff \\
x=\frac{-b+\sqrt{b^2-4ac}}{2a} \ \ \text{or} \ \ x=\frac{-b-\sqrt{b^2-4ac}}{2a} \iff \\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
A: 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.
