The equation $$||x|-2|+|x+1|=3$$ has solutions at $$x=0, -2, 2, -1$$. The discussion emphasizes the importance of analyzing different cases based on the values of x, specifically the intervals $$x<-2$$, $$-2
Students studying algebra, educators teaching mathematical concepts, and anyone looking to improve their problem-solving skills in absolute value equations.
Petrus said:Hello MHB,
solve $$||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.