<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] The  solutions of $Y^2=(b^2-4 a c)/4 a^2\;$ are $\;Y=+\sqrt {b^2-4 a c}/2 a\;$ and $\;Y=-\sqrt {b^2-4 a c}/2 a.$ 
If $(x+b/ 2 a )^2=Y^2$ then $x+b/2 a$ must be equal to one of these solutions or to the other. 
So $x+b/2 a=\pm \sqrt {b^2-4 a c}/2 a$. 
</s>