Is the square of a square root always non-negative? ($\sqrt{x}^2 = |x|?$)

Question

I know that $\sqrt{x^2} = |x|.$

But does $\sqrt{x}^2 = |x|?$

It seems like it should, since $$\sqrt{x}^2 = \sqrt{x} \cdot \sqrt{x}$$ and $\sqrt{x}$ is non-negative, so their product should be non-negative.

But I haven't seen a rule like this, so I'm guessing it isn't the case, and am just wondering why.

Answer 1

I'm assuming we're dealing with real numbers only here (no complex numbers). Then the statement $\sqrt{x}^2=|x|$ is true… but we don't need a "rule" like this because we can do even better. The expression $\sqrt{x}$ is well-defined over the real numbers only if $x\ge0$, and therefore $\sqrt{x}^2=x$ — we don't need the absolute value here because $x$ is already non-negative.

EDIT: In other words, the property $\sqrt{x}^2=x$ is true whenever both sides (especially the left-hand side) are well-defined. But it's not true otherwise; in particular, it does not apply when $x$ is negative. And using the absolute value on the right-hand side wouldn't help with that.

Answer 2

$f(x)=(\sqrt x)^2$ is not the same function that $g(x)=|x|$.

Both have the same rank $[0,+\infty[$ but its domains are distinct: $f$ has domain the non-negative numbers while $g$ has all the real as domain.