Originally Posted by
HallsofIvy From ||x+ 3|- 2|= a, we have either
1) |x+ 3|- 2= a so |x+ 3|= a+ 2 or
2) |x+ 3|- 2= -a so |x+ 3|=-a+ 2.
From (1) |x+ 3|= a+ 2 we have either
1a) x+ 3= a+ 2 so x= a- 1 or
1b) x+ 3= -(a+ 2)= -a- 2 so x= -a- 5.
From (2) |x+ 3|= -a+ 2 we have either
2a) x+ 3= -a+ 2 so x= -a- 1 or
2b) x+ 3= a- 2 so x= a- 5.
Okay, those are the values you have. Now, the problem says "There are 3 real numbers that supplies the equation" (it really should be "satisfies" the equation) so two of those expressions must be the same.
Could a-2= -a- 5? That is the same as 2a= -3 so that a= -3/2. In that case the solutions are x= -3/2- 2= 3/2- 5= -7/2, x= 3/2- 1= 1/2, and x= -3/2- 5= -13/2. We have 3 distinct solutions so this is a valid solution.
Could a- 1= -a- 1? That is the same as 2a= 0 so that a= 0. In that case the solutions are x= 0- 1= -0- 1= -1, x= -0- 5= -5, and a- 5= -5. So that is NOT a solution- there are only two distinct solutions, not 3.
Of course, we could not have a- 1= a- 5 or -a- 5= -a- 1. And -a- 5= a- 5 gives the previous a= 0 answer.