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)?

Actually, there're cyclic properties for higher order equations.
For cubic  $x^{3}+3px=2q$,
$$x_{k}=e^{\frac{2i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+
        e^{\frac{4i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$ for $k=0,1,2$.
We have same set of roots by writting $x_{k}$ as 
$$e^{\frac{4i\pi k}{3}} \sqrt[3]{q+\sqrt{p^{3}+q^{2}}}+
  e^{\frac{2i\pi k}{3}} \sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$$
For biquadratic $x^{4}-2ax^{2}+b=0$,
$$x=\pm \sqrt{a \pm \sqrt{a^{2}-b}}=
\pm \sqrt{\frac{a+\sqrt{b}}{2}} \pm \sqrt{\frac{a-\sqrt{b}}{2}}$$
Four combinations of $\pm$ signs giving four distinct roots.
Also de Moivre's quintic $x^{5}+5ax^{3}+5a^{2}x=2b$,
$$x_{k}=e^{\frac{2i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+
        e^{\frac{8i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$ for $k=0,1,2,3,4$.
It also gives same set of roots by writing $x_{k}$ as
$$x_{k}=e^{\frac{4i\pi k}{5}} \sqrt[5]{b+\sqrt{a^{5}+b^{2}}}+
        e^{\frac{6i\pi k}{5}} \sqrt[5]{b-\sqrt{a^{5}+b^{2}}}$$
Last but not least, there are $n$ roots for $n$ degree polynomial equation.  Redundancy of root (don't mess up with multiple roots) means you've done an extra cycle, but make sure you don't miss any.
