Is $\lvert\sqrt{x}\rvert^2$ equal $\lvert x\rvert$?

Question

I know we have $\lvert x\rvert =\sqrt{x^2}$, but what is $$ \lvert\sqrt{x}\rvert^2 \quad ? $$ Is it $\lvert\sqrt{x}\rvert^2=\sqrt{\lvert x \rvert}^2=\lvert x\rvert$? Or maybe $\lvert\sqrt{x}\rvert^2 =x$?

Answer 1

Note that $f(x) = |x|$ and $g(x)=|\sqrt x|^2$ are equal functions if and only if $x\in R^+\cup \{0\}$. $$$$ Also, if $x\in R$, $\sqrt x$ is not defined for $x<0$. $$$$ If any $2$ functions $f(x)$ and $g(x)$ are equal functions, then their domain of definition must be the same (say $D$), and the values of $f(x)$ and $g(x)$ must be the same for all $x\in D$.

Answer 2

No, $\sqrt x$ is not defined for the negatives. Anyway, in the positives,

$$(\sqrt x)^2=x.$$

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By the way, the absolute value is doubly useless, because the square root is always positive and the square "discards" the sign.

Answer 3

$\sqrt{x}$ is defined for $x \ge 0$ and $\sqrt{x} \ge 0$, hence $|\sqrt{x}|= \sqrt{x}$ , therefore $\lvert\sqrt{x}\rvert^2 =\sqrt{x}^2=x=|x|.$