Why doesn't the quadratic equation contain $2|a|$ in the denominator? When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step:
$$
\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}
$$
the square root is taken from both sides, so why is 
$$\sqrt{4a^2} = 2a$$ 
in the denominator and not
$$ \sqrt{4a^2} = 2\left |a  \right | $$
Could somebody explain this to me? Thank you very much
 A: The two square roots of $a^2$ are $a$ and $-a$, sometimes written together as $\pm a$.
For real numbers $\pm a$ is equivalent to $\pm |a|$ but this is not true for complex numbers. So putting the absolute value operation in would make the proof less general.
We could write the proof as
$$
\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}
$$
$$
\pm\left(x+{\frac {b}{2a}}\right)={\frac {\pm\sqrt{b^{2}-4ac}}{\pm2a}}
$$
$$
x+{\frac {b}{2a}}={\frac {\pm\sqrt{b^{2}-4ac}}{2a}}
$$ 
But generally it is considered sufficient to put in just a single $\pm$ from the start rather than putting in one for each square root and then removing the redundant ones.
A: One could take the square root as $2|a|$ instead, which would lead to:
$$
x+\frac {b}{2a} = \pm{\frac {\sqrt{b^{2}-4ac}}{2|a|}} \quad\iff\quad x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} \tag{1}
$$
However, given that $\,|a|\,$ is either $\,a\,$ or $\,-a\,$ it follows that $\,\pm|a|=\pm a\,$, so the formula simplifies to:
$$
x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} =  -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{\color{red}{2a}}} = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \tag{2}
$$
$(1)\,$ and $\,(2)\,$ are entirely equivalent, but $\,(2)\,$ is more convenient to use.
A: When taking the square root we put a $\pm$ on the right hand side to account for the two roots, so it is unnecessary to strip off the sign of $a$, as we will put it back anyways.
A: If you put
$x
=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
into
$ax^2+bx+c$,
since
$x^2
=\dfrac{b^2\mp2b\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2}
=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
$
you get
$\begin{array}\\
ax^2+bx+c
&a\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
+b\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}+c\\
&=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a}
+\dfrac{-2b^2\pm 2b\sqrt{b^2-4ac}}{4a}+c\\
&=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}-2b^2\pm 2b\sqrt{b^2-4ac}+4ac}{4a}\\
&=0\\
\end{array}
$
If you use $|a|$,
it won't work
since you can't
combine the terms.
A: my preference for remembering and using the quadratic formula (and electrical engineers seem to do that often) is to remember the root quadratic equations as:
$$ x^2 \ + \ b\,x \ + \ c \ = \ 0 $$
which has solution:
$$
 x \ = \ \begin{cases}
 -\tfrac{b}{2} \pm \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 > c \\
\\
  -\tfrac{b}{2}  \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 = c \\
\\
  -\tfrac{b}{2} \pm i \sqrt{c - \left(\tfrac{b}{2}\right)^2} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 < c \\
\end{cases}$$
normalizing out the "$a$" does not make the quadratic equation less general.  the only degrees of freedom are $b$ and $c$, so that means normally (except for a double root), there are two independent solutions.
A: Alternatively, noting $a\ne 0$:
$$\begin{align}\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}-4ac}{4a^{2}}} \iff \\
4a^2\left(x+{\frac {b}{2a}}\right)^{2}&=b^{2}-4ac \iff \\
\left(2ax+b\right)^{2}&=b^{2}-4ac \iff \\
2ax+b&=\pm \sqrt{b^2-4ac} \iff \\
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.\end{align}$$
