Algebra Precalculus – Why Can’t |x| = x Be Solved Conventionally?

Question

I warn the reader beforehand that, perhaps, this question is laughably simple; yet I need assistance answering it. It is taught in schools how one can solve equations containing absolute values. The method is simple: take for example $|5x| = 10$. This is true in 2 ways only: when $5x$ equals $10$, and when $5x$ equals $-10$. The logic of the reasoning is simple and straightforward, and it should seem that such simple logic must apply to exactly all other cases in which absolute values are to be resolved. Yet what about $|x| = x$? The equation is obviously true for whenever $x$ is positive (this can also be seen by the definition for absolute values). Then why exactly doesn't the method shown in school provide the sought-for answer? On the contrary, it provides an incorrect solution. For $|x| = x$, according to the method, is true when $x = x$ or when $x = –x$. This is obviously erroneous. I only want to know why exactly the error occurs? And if there is a general case to when equations containing absolute values shouldn't be resolved using such a method. Thank you in advance.

Answer 1

The problem is you are not using the "conventional" method correctly.

To solve $|f(x)|=g(x)$, the "conventional method" is:

1. Solve $f(x)\geq 0$ and $f(x)=g(x)$; and
2. Solve $f(x)\leq 0$ and $-f(x)=g(x)$.

So your original problem should really be done by doing:

1. Find all solutions to $x\geq 0$ and $5x=10$;
2. Then find all solutions to $x\lt 0$ and $-5x=10$.

The first case leads to $x=2$, the second to $x=-2$.

For $|x|=x$, you want to:

1. Find all solutions to $x\geq 0$ and $x=x$;
2. Then find all solution sto $x\lt 0$ and $-x=x$.

The first consists of all $x\geq 0$; the second is empty. So the solution set is precisely $[0,\infty)$.