Is the sign of the square root dependent on the argument inside it?

[jk22]

Could it be said that since $a=A(f(x))\sqrt{f(x)}$, with $A(x)\in\{1,-1\}$ then $a^2=f(x)$,, that $a$ is the square root of $f(x)$ ?

In other words could the sign of the root depend on the argument inside it ?

Else it would have to be chosen by human free will and to be blocked for the rest of the calculation ?

 

[jedishrfu]

No, the principal square root of a number is positive real or positive imaginary based on whether the argument inside is a positive or negative number.

However, we realize that there are two roots the principal root and -1 times the principal root.

As an example, both 2 and -2 are roots of the square root of 4. 2 is the principal root and -2 is another root since (-2)*(-2) = 4 just as 2 * 2 = 4

In the case of a negative -4 then the roots are 2i and -2i

More fun stuff here:

https://en.wikipedia.org/wiki/Square_root