<s>[INST] Why is the $2a$ term in the quadratic formula $|2a|$? I am reading through Algebra by Gelfand/Shen. There was a construction of the quadratic formula as follows: 
$ax^2 + bx + c = 0$. Dividing by $a$ gives us $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ and we can apply the formula of the equation $x^2 + px + q = 0$ with $p = \frac{b}{a}, q = \frac{c}{a}$, we get:
$x_{1,2} = -\frac{b}{2a} \pm \sqrt{(\frac{b}{2a})^2 - 4\frac{c}{a}} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
My question is if we take $\sqrt{4a^2}$ why do we assume it is always equal to $2a$ instead of $|2a|$? Why does the negative part not matter?
 [/INST] Because there is a $\pm$ in front of the fraction. Since you are considering either of the signs, both $\frac{\sqrt{b^2-4ac}}{\lvert 2a\rvert}$ and $\frac{\sqrt{b^2-4ac}}{-\lvert 2a\rvert}$ will have to be considered, which is the same as considering both $\frac{\sqrt{b^2-4ac}}{2a}$ and $\frac{\sqrt{b^2-4ac}}{-2a}$. This, as far as real values of $a$ are considered. If $a\in\Bbb C\setminus\Bbb R$, then $\sqrt{a^2}=\lvert a\rvert$ doesn't hold in the first place.
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