The discussion revolves around solving the equation $$||x|-2|+|x+1|=3$$. Participants explore different cases for the variable x to determine the correct solution set, focusing on the implications of absolute values in various intervals.
Participants do not reach a consensus on the initial solution, and multiple competing views regarding the intervals and cases remain unresolved.
The discussion includes various assumptions about the behavior of absolute values in different intervals, but these assumptions are not fully resolved or agreed upon.
Petrus said:Hello MHB,
solve $$||x|-2|+|x+1|=3$$
and we find that $$x=0,-2,2,-1$$
I got problem to find the 'case',
Regards,
$$|\rangle$$
I don't mean those are the answer, but those are the point we should check $$\geq$$ or $$\leq$$, I hope you did understand.I like Serena said:That doesn't look like the right solution...
Talking about cases, what if x<-2? Or -2<x<-1? And what about -1<x<0? Or perhaps 0<x<2? And x>2?
Petrus said:I don't mean those are the answer, but those are the point we should check $$\geq$$ or $$\leq$$, I hope you did understand.
Can I also check this one insted of -1<x<0
-2<x<0?
Thanks, got the correct answer now :)I like Serena said:Sure you can. It's just that |x+1| does something funny at x=-1.