Why does $|x^2| = x^2$, only if $x$ is positive? (according to Wolfram Alpha) I don't understand why Wolfram says $|x^2|$ is equal to $x^2$, but only when $x$ is positive. Since we're squaring the value of $x$, the input to $abs$ will always be positive, and the result will be identical to the input. Here's my proof:
Lets take two possible values of $x$: $k$, and $-k$, and compare their equations: $|k^2|$ and $|(-k)^2|$
It's trivial that $(-k)^2$ = $k^2$, since $(-k)^2$ = $(-1)^2 \cdot k^2$ = $1 \cdot k^2$ = $k^2$.
So $|(-k)^2|$ = $|(-1)^2 \cdot k^2|$ = $|1 \cdot k^2|$ = $|k^2$|.
So it then follows that since $|k^2| = k^2$ when $k$ is positive,  $|k^2| = k^2$ for all values $k$.
Where am I going wrong?

Assuming that we're looking at the page linked by The Chaz 2.0:
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I read this as telling the truth but not the whole truth.  In other words, if $x$ is positive then $|x^2| = x^2$.  Wolfram Alpha is not saying that if $x^2 = |x^2|$ then $x$ is positive.  That would be false, as you point out.
I am not sure why Wolfram Alpha provides this incomplete information.  It does try to be general and not assume that variables have strictly real values.  Notice that $|i^2| \neq i^2$.  As for why it doesn't say “Alternate form assuming $x$ is real,” who knows.
