1. ## absolute value

||x+3|-2|=a.There are 3 real numbers that supplies the equation.Then
what must be the value of a?

My work:

|x+3|=a+2 or |x+3|=-a+2

x+3=a+2 or x+3=-a-2 or x+3=-a+2 or x+3=a-2

x = a-1, -a-5, -a-1, a-5
But I couldn't find the answer.

2. ## Re: absolute value

There is no such a. If a<0, there are no solutions. If a=0, there are exactly 2 solutions, namely x=-5 and x=-1. If a>0 there are exactly 4 solutions.

3. ## Re: absolute value

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.

4. ## Re: absolute value

$a=2$ has three solutions for $x$ ...

5. ## Re: absolute value

Skeeter's exactly right. I'm going to take some time off and try and quit smoking that stuff.

If a<0, there are no solutions.
If a=0, there are exactly two solutions, x=-5 and x=-1
If 0<a<2, there are 4 solutions
If a=2, there are 3 solutions, x=-7, x=-3 and x=1
If a>2 there are 2 solutions

6. ## Re: absolute value

I knew there was a reason to learn sketching absolute value functions ...

7. ## Re: absolute value

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.
How did you get "a-2= -a- 5"?
Right side of the equation is ok. But I couldn't find the left side.

8. ## Re: absolute value

Maybe you meant -a-1= a-5
a=2

9. ## Re: absolute value

Is there anyone there?

10. ## Re: absolute value

Yes, you found a small typo from HallsofIvy. Did you really need confirmation of that? You already found the answer.

11. ## Re: absolute value

Many Thanks for your help. I asked that because I though I did something wrong. I am sorry.