<s>[INST] 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?
 [/INST] 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.
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