How to solve equations with absolute value on both sides? Assume we are given an equation $|f(x)|$=g(x). From the definition of the absolute value we know that:$$
|f(x)|= \begin{cases} +f(x), & \mbox{if } f(x) \geq 0 \\ -f(x), & \mbox{if } f(x)<0 \end{cases}
$$
This implies that in order to solve it we need to consider two cases, namely when $f(x) \geq0$ and $f(x) < 0$. In the first one, for instance, we have $f(x)=g(x)$. We obtain some solutions, but shall we keep only these $x$'s that make $f(x) \geq 0$? It seems logical. In my class however, we solved problems like that one simply by making an alternative, i.e. $$
f(x)=g(x) \vee f(x)=-g(x)
$$
without checking if $x$ from the first "case" makes $f(x)$ bigger or equal zero, and if $x$ form the second one makes $f(x)$ smaller than zero. That really baffles me...
Also, how do you solve $|f(x)|=|g(x)|$?
Thanks in advance!
 A: To your last question:
$$|f(x)|=|g(x)|$$ squaring this equation and using that $$a^2-b^2=(a+b)(a-b)$$ we get
$$(f(x)-g(x))(f(x)+g(x))=0$$
A: The first equation is simply equivalent to
$$f(x)=\pm g(x)  \;\textbf{ and } \;g(x)\ge 0.$$
The second is, even more simply, equivalent to $$f(x)=\pm g(x).$$
A: First of all, thank you all for answering my question. That's my summary. 
Equation $|f(x)|=g(x)$ is equivalent to: $$
f(x)=g(x) \mbox{ if } f(x) \geq 0
$$ and $$
f(x)=-g(x) \mbox{ if } f(x)<0.
$$
Because $g(x)$ can take any value (it has't been specified) we have to check if the solutions obtained in above cases make function $f$ greater or equal to zero or less than zero, respectively. If, however, function $g$ is defined in such way that $$
\forall g(x) \geq 0
$$
there is no need to check whether the obtained solutions are in appropriate intervals - that is because the only place when two functions ($|f| \mbox{ and } g$) can meet lays above the $x$-axis. Thus if $\forall g(x) \geq 0$ then the equation $|f(x)|=g(x)$ simplifies into an alternative: $$
f(x)=g(x) \vee f(x)=-g(x). 
$$
The same, of course, holds if $g(x)=c$, where $c$ is some constant and $c \geq 0$. Then we have $$
f(x)=c \vee f(x)=-c.
$$
If $c<0$ no solutions exist.
