$|2-|1-x||=|x^2-1|$ equation, incorrect solution When I solved this equation I got $x_1 = -1$, $x_2 = 0$ and $x_3 = \frac{-1+\sqrt{17}}{2}$
I solved it like this: I first sketched a graph and I could see that the first two solutions were trivial, I got the third one by solving $x^2-1 = -(x-3)$ for $x &gt; 1$, since I could see from my graph that there was an intersection.
Wolfram alpha only displays the first two solutions and doesn't mention the third one, is my solution to the whole problem incorrect?
I also have a question, how do you view if someone solves a problem by sketching a graph and solves it from there and doesn't do all the required calculation work to show that the expression inside every absolute value can be negative and what happens if it is positive and does everything from there.
Thank you for your answers.
Alpha does all computations in the complex and it turns out that the solution includes a whole locus with a terribly complicated expression (because the equation is equivalent to a quartic one). But that locus certainly includes the third real root.
Now the following Alpha request will give you the three real roots: https://www.wolframalpha.com/input/?i=real+roots+of+abs%282-abs%28x-1%29%29-abs%28x%5E2-1%29
Your method is very good but it is not considered a rigorous proof. You can use the graph to turn this into a rigorous proof. We see on the graph that there are only three cases to study: $x\lt -1$ and  $-1\le x \le 1$ and $1 \le x$. Each of these three cases is very easy to study.