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Homework Help: Solving an inquality

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data

    So this is the question


    2. Relevant equations


    3. The attempt at a solution

    I tried it, the solution seems right, but i don't know if my approach is correct.

  2. jcsd
  3. Feb 11, 2012 #2


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    Think about these absolute values like this:

    At x=1 or x=2, one of the absolute values is 0, so you can call these the "roots" (thought not technically roots, I can't think of a more appropriate name for them right now), so what you want to do is check all possibilities around those roots, and the roots themselves.


    For this range, both [itex]x-1[/itex] and [itex]x-2[/itex] will be negative, so the inequality you need to solve would be [tex]-(x-1)-(x-2)>1[/tex]

    Here you will have [itex]x-1>0[/itex] and [itex]x-2<0[/itex] so what you need to solve is [tex](x-1)-(x-2)>1[/tex]

    For this value, both are positive so it should be clear what you need to solve here.

    And then always check the "roots" themselves. Plug in the values of x=1 and x=2. By this point, you've checked all possible cases and should have your solution set.
  4. Feb 11, 2012 #3
    The way I learned it:

    To solve |ax+b|>k, solve ax+b>k and ax+b<-k. (This is basically what you did.)

    Then I would plug test values into the original equation to see if it makes a true or false statement. You would use the intervals (-[itex]\infty[/itex],1);(1,2);(2,[itex]\infty[/itex]).

    The values from those intervals that make true statements give you your solution set.
  5. Feb 11, 2012 #4


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    edit: I just realized that your question is a special case. What if it was instead [tex]|x-1|+|x-2|>2[/tex] or [tex]|x-1|+|x-2|>0[/tex] ? For the first what you need to do is check when [tex]|x-1|+|x-2|=2[/tex] and then check each interval around that.
    Last edited: Feb 11, 2012
  6. Feb 11, 2012 #5
    Thanks Mentallic and Adaptation!
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