# Number of solutions of an equality with absolute value operator

Consider:

$$\left|\left|\left|x-1\right|-2\right|-4\right|=4$$

What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality were to hold, then $$| x - 1 |$$ should be 9. Because $$| x -1 |$$ will always be positive, and once we remove $$| x- 1 |$$, the only positive number that can lead to a possible solution is 9. So solving for $$| x - 1 |$$ gives me $$10$$ or $$-8$$, and I'm done. Number of solutions is 2.

But I was curious on what I'd do if I had different numbers. Say:

$$\left|\left|\left|x-1\right|-2\right|-5\right|=4$$

Now, Desmos says there are five possible solutions. How should one go about finding the number of solutions (or finding the solutions themselves) for something like this by hand? A general method would be appreciated.

• Each absolute value will double the maximum number of solutions. (Maximum number of solutions because zero may be an intermediate solution or two branches may give the same solution.) – Ertxiem - reinstate Monica Mar 24 '19 at 12:35

You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|\in\{8,0\},$$ for the $$0$$ case, it follows that $$|x-1|=2\iff x\in\{3,-1\}.$$

For the $$8$$ case, it follows that $$|x-1|-2=\pm 8\implies |x-1|\in\{10,-6\},$$

which is only possible for $$|x-1|=10\implies x-1=\pm 10,$$

which leads to $$x\in\{11,-9\}.$$

So, there are $$4$$ solutions in total, namely

$$x\in\{-9,-1,3,11\}.$$

Open one-by-one (remember, absolute value is always nonnegative): $$\left|\left|\left|x-1\right|-2\right|-4\right|=4 \iff \\ \left|\left|x-1\right|-2\right|-4=\pm 4 \iff \left|\left|x-1\right|-2\right|=0;8 \iff \\ |x-1|-2=\pm(0;8) \iff |x-1|=2;10 \iff \\ x-1=\pm(2;10) \iff x=-9;-1;3;11.$$