# 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?

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.
It is because $$|2a|$$ is either $$+2a$$ or $$-2a$$. Whichever sign it is, it gets "absorbed," effectively, by the $$\pm$$ sign from the square root from that very same term in the expression.