<s>[INST] Principal roots in the derivation of the quadratic formula The following is an abbreviation of a common derivation of the quadratic formula:
$$
ax^2+bx+c=0\\
\vdots\\
\sqrt{(x+\frac{b}{2a})^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\
x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
$$
Following the square root extraction, the equation should have three plus-minus signs:
$$
\pm(x+\frac{b}{2a})=\frac{\pm\sqrt{b^2-4ac}}{\pm2a}\\
$$
But instead there's only one. Is it because the signs cancel each other out, or are moved from one expression to the other (through multiplication in $-1$ or $\frac{-1}{-1}$) in six of the eight permutations that the three plus-minus signs allow; or is it a matter of choosing a principal root for some arcane reason beyond the scope of elementary algebra (eg. branch points)?
 [/INST] Cancellation and "moving" signs (as you put it) are one explanation for why only one $\pm$ sign is required. We know that
\begin{gather}
\frac{-p}{-q} = \frac{p}{q}, \\[.5ex]
-\left(\frac{p}{q}\right) = \frac{-p}{q} = \frac{p}{-q}, \\[.5ex]
p = -q \quad\text{if and only if}\quad -p = q.
\end{gather}
Using these facts we can show that the next four equations
are all equivalent, and hence by stating any one of them we also imply
that each of the others is true:
\begin{align}
x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{2a}, &
x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{-2a}, \\
-\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{2a}, &
-\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{-2a}.
\end{align}
The following four equations also are all equivalent to each other:
\begin{align}
x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{2a}, &
x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{-2a}, \\
-\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{2a}, &
-\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{-2a}.
\end{align}
By asserting the equation
$$\pm\left(x+\frac{b}{2a}\right)=\frac{\pm\sqrt{b^2-4ac}}{\pm2a},$$
then, assuming we can choose each of the $+$ or $-$ signs independently,
we assert that at least one of the previous eight equations is true.
So either one of the first four of those equations is true, in which case
$x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a}$
(as implied by any of those four equations),
or one of the other four of those equations is true, in which case
$x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a}$.
That is, the equation with three $\pm$ signs is simply asserting that
$$
x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a} \quad\text{or}\quad
x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a},
$$
which is just what we mean when we write
$$x+\frac{b}{2a} = \frac{\pm\sqrt{b^2-4ac}}{2a}.$$
It is simpler, however, if we apply the rules of sign cancellation, etc., earlier in the procedure, as detailed in the previous comments and answer, so that we never end up with three $\pm$ signs in the first place.
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