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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. J. Bałaziński
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  • Move right side to the left and now you have absolute values only on one side :D – Markoff Chainz Jul 12 '19 at 11:35
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    Squaring both sides of $|f(x)|=|g(x)|$ gives us $[f(x)]^2=[g(x)]^2$. This may be useful. – Hussain-Alqatari Jul 12 '19 at 11:36
  • Ok, thanks for answering, guys, but I still don't know how to deal with the problem in the 1st part of the question :( – A. J. Bałaziński Jul 12 '19 at 11:38
  • Here, for example equation of the form |f(x)|=g(x) was solved in multiple cases. https://math.stackexchange.com/questions/1955061/equation-with-double-absolute-value – A. J. Bałaziński Jul 12 '19 at 11:56
  • @AleksyBałaziński: I understand your confusion, we do need to add condition $g(x) \ge 0$ for the first case. It will help you to avoid fake solutions. It's a good practice to consider domain of the equation from the start. – Vasili Jul 12 '19 at 12:18

3 Answers3

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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.

A. J. Bałaziński
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To your last question: $$|f(x)|=|g(x)|$$ squaring this equation and using that $$a^2-b^2=(a+b)(a-b)$$ we get $$(f(x)-g(x))(f(x)+g(x))=0$$

Dr. Sonnhard Graubner
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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 $$f(x)=\pm g(x).$$

Bernard
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  • Ok, that's what I've been told about. However f(x)=+g(x) only when f(x) is bigger or equal zero. So if, somehow, we get solution x=a, and f(a) is less than zero we should say that a is not the solution for the equation. In your method (and simultaneously my teacher's) we don't check anything... Is there a mistake in my reasoning? – A. J. Bałaziński Jul 12 '19 at 11:51
  • Are you mentioning the first or the second equation? – Bernard Jul 12 '19 at 12:00
  • The first one, absolute value only on f(x) – A. J. Bałaziński Jul 12 '19 at 12:02
  • Well if you have a solution $a$ of the system, as $(a)\ge 0$, $f(a)$ can't be negative. That's why there's noting to check. – Bernard Jul 12 '19 at 12:04
  • If a=2 and f(x)=-x, then f(a)=-2 which is obviously less than zero. I found a question where someone had problem analogous to mine which was solved using multiple cases. https://math.stackexchange.com/questions/1955061/equation-with-double-absolute-value – A. J. Bałaziński Jul 12 '19 at 12:07
  • Of course, but $a=2$ does not satisfy the 2nd condition: $g(a)\ge 0$, so it has to be deleted. – Bernard Jul 12 '19 at 12:27