What do you mean by $\operatorname{root}(x)$ ?
What's the value of $\operatorname{root}(252+ \operatorname{root}(8+ root(64$))) ?
Is it $+16$ or $+ 16$ and $-16$ ?
I have seen Wikipedia.
It is given that : $\operatorname{root}(x^2) = \operatorname{mod}(x)$ where $x$ can be $-ve$ or $+ve$.
Here , $\operatorname{root}(256) = \operatorname{mod}(16) = \operatorname{mod}(-16) = 16$
Am I right??
In general, the $n$th root of a number $a$ refers to the solutions of the equation $$\tag{1} x^n = a $$ In particular, the square root, which is the second root, is a solution to the equation $$ x^2 = a $$ Without any additional modifiers, the word root of a number refers to this square root. For example, the root of $256$ can be thought of as the solution to the equation $$ x^2 = 256 $$ It's fairly easy to see that this has the solutions $16$ and $-16$. But if you're using a calculator (be it a pocket calculator or a computer), normally a calculator doesn't know how to list all the solutions. In calculators, the root is calculated by \begin{equation}\tag{2} x = \sqrt[n]{a} \end{equation} And the square root is again same as $n=2$. The difference between Equations (1) and (2) is that the result of the calculation in the second equation is either always positive (if $n$ is even) or takes the sign of $a$ (if $n$ is odd). Therefore, if you ask a calculator, the root of $256$ is $\sqrt{256} = 16$.
By definition for $a\ge0$ we have $\sqrt a =b\ge 0$ such that $b^2=a$.
Then when we consider the case $\sqrt {x^2}$ we have two cases
$x\ge 0\implies \sqrt x^2=x$
$x< 0\implies \sqrt x^2=-x$
that is
