Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $x \in \mathbb R$ and $a \neq 0$. Throughout this derivation, I will use the definition: $\sqrt{x^2}=|x|$.
Here is my derivation, and I have placed a $\color{red}{\dagger}$ next to the part that I would like some clarificaiton about:
$ax^2+bx+c =0 \iff x^2+\frac{bx}{a}+\frac{c}{a}=0 \iff (x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})=0 \iff (x+\frac{b}{2a})^2+(\frac{4ac-b^2}{4a^2})=0 $
Bringing the right summand over to the right side of the equation:
$(x+\frac{b}{2a})^2=(\frac{b^2-4ac}{4a^2}) \iff \left| x+\frac{b}{2a}\right|=\sqrt{b^2-4ac}\cdot\sqrt{\frac{1}{4a^2}} \iff \left| x+\frac{b}{2a}\right|=\left|\frac{1}{2a} \right| \cdot \sqrt{b^2-4ac} \quad \quad \color{red}{\dagger}$
My confusion stems from how the final expression above is equivalent to the syntax "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" .
For $\color{red}{\dagger}$, we have 4 total cases:

*

*$ a \lt 0$ and $x+\frac{b}{a} \lt 0$


*$ a \lt 0$ and $x+\frac{b}{a} \geq 0$


*$ a \gt 0$ and $x+\frac{b}{a} \lt 0$


*$ a \gt 0$ and $x+\frac{b}{a} \geq 0$
Case 1: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
Case 2: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 3: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 4: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
From the four scenarios, is the way we get to "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" simply by noting that for a fixed $a$, we have $x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ or } x=\frac{-b-\sqrt{b^2-4ac}}{2a}$? ...where the '$\text{ or }$' here is denoting the logical or.
At this point, we define "$x=\pm \alpha$" as meaning $x = \alpha \text { or } x=-\alpha$...therefore meaning that $x=\beta\pm \alpha$ is equivalent to $x=\beta+\alpha \text{ or } x=\beta - \alpha$.
Is that the proper understanding?
 A: Your analysis is spot on, up until the point where you concluded that
$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \tag1 $$
I didn't bother reading after that, because there is no point.  All that is necessary from the point of (1) above is to consider that any equation of the form
$$[f(x)]^2 = C$$
may be dissected as follows:

*

*If $C < 0$ then there can't be any real roots.  So, without loss of generality (assuming that you are only concerned with real roots), assume that $C \geq 0.$


*Then, you must have that $f(x) = \pm \sqrt{C}$. 
Bang, that makes it: game over right there. 
So, no further analysis is needed.

This begs the question: what happens if $C < 0$.
Then, if you let $D = \sqrt{-C}$, then the equation becomes
$$[f(x)] = \pm (iD) ~: ~i = \sqrt{-1}.$$
A: From your working step $\left(x+\frac{b}{2a}\right)^2=\left(\frac{b^2-4ac}{4a^2}\right),$
take square root (permissible because non-negative);
then apply your definition $\sqrt{x^2}=|x|,$ which implies that $\sqrt{x^2}=\pm x$ (what does $\pm x$ mean);
then simplify;
finally, verify (by substitution) that your two solutions indeed satisfy $ax^2+bx+c=0.$
This considers two instead of four separate cases.
A: An appealing way of understanding why $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be written as $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ is to take a set perspective.
The statement $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be thought of as saying:

$x \in S$ where $S:=\left\{x\in \mathbb R:\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}  \right\}$

The statement $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ can be though of as saying:

$ x \in T$ where $T:=\left\{x\in \mathbb R:x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}\right\}$

The objective is to show that $x \in S \rightarrow x \in T$ and $x \in T \rightarrow x \in S$.
There is a lemma that follows directly from the definition of $|\cdot|$, which reads as:

$|x|=C \iff x=C \text{ OR } x= -C \quad (*_1)$

To prove the $\rightarrow$ direction, simply exhaust all possible values of $x \in \mathbb R$ by splitting it into the two cases of $x \geq 0$ and $x \lt 0$. The $\leftarrow$ direction is trivial.

Prove:$\quad x \in S \rightarrow x \in T$
By assumption, we have that $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}  $. Taking the square root of both sides, and applying the definition of $\sqrt{\cdot}$, we have:
$$\left |x + \frac{b}{2a} \right|=\sqrt{\frac{b^2 - 4ac}{4a^2}}$$
Applying our lemma $(*_1)$, we have the logical statement:
$$x+\frac{b}{2a}=\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad\text { OR }\quad x+\frac{b}{2a}=-\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad (*_2)$$
The symbol $\pm$ is defined to capture the meaning of $(*_2)$ and is equivalently written as:

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

Subtraction gives us: $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$, which means that $x \in T$.

Prove:$\quad x \in T \rightarrow x \in S$
By assumption, we have that $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$. This means that $x= \sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}\text { OR } x=-\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$.
In the first case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$.
In the second case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(-\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$. Therefore, in all cases, we have that $x \in T$.

In conclusion, we have: $x \in S \iff x \in T$, which means that the two statements:
(1) $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
(2) $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$
are equivalent.
