Solve for the values of $x$ in $|x+k|=|x|+k$ where $k$ is a positive real number The question asks me for which values of the real number $x$ is $|x+k|=|x|+k$ where $k$ is a positive real number.
How do I go about this? Can I square both sides to get rid of the absolute value signs? When I do it this way, I get a single $x$ sandwiched in the middle of a quadratic equation which I don't know what to do with, after I square and expand $|x|+k$. Is there another way of solving this problem?

with conditioning on $x$ like a switch-case command we'll get:
$case1): \; $ if $x>k$ 
$$  \Rightarrow x+k=x+k $$
so all real numbers greater that $k$ are in solution interval. now assume:
$case2): \; 0 <x <k \;$    then we get:
$$x+k=x+k$$ so this interval is in solution too. then
$case3): \; -k<x<0 \;$    and we have:
$$x+k=-x+k$$ thus there is no number in this interval other that $x=0$ in solution.then assume:
$case 4): \; x <-k \;$ and we get:
$$-x-k=-x+k$$ so this interval is in solution if $k=0$ otherwise this interval is not in solution.
thus the solution interval is $[0, \infty) $ for $ k > 0$ 
if $k=0$ the solution is The Whole Real Line because the problem reduces to trivial problem $|x|=|x|$ which is always true.
