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Thread: absolute value

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
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    Re: absolute value

    $a=2$ has three solutions for $x$ ...
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  5. #5
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    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
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  6. #6
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    Re: absolute value

    I knew there was a reason to learn sketching absolute value functions ...
    Attached Thumbnails Attached Thumbnails absolute value-absvalue_sketch.jpg  
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  7. #7
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    Re: absolute value

    Quote Originally Posted by HallsofIvy View Post
    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.
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  8. #8
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    Re: absolute value

    Maybe you meant -a-1= a-5
    a=2
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  9. #9
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    Re: absolute value

    Is there anyone there?
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  10. #10
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    Re: absolute value

    Yes, you found a small typo from HallsofIvy. Did you really need confirmation of that? You already found the answer.
    Last edited by SlipEternal; May 18th 2017 at 03:30 AM.
    Thanks from Plato
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  11. #11
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    Re: absolute value

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