1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

An inequality with absolute values

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve Ix+3I>2
    *I is used for absolute value notation

    3. The attempt at a solution
    Considering both
    a) Ix+3I > 0 then Ix+3I= x+3
    b) Ix+3I < 0 then Ix+3I= -(x+3)

    when solved this would yield to;

    a) x>-3 and x>-1
    b) x<-5 and x<-3

    from my general reasoning i think the answer should be x>-1 and x<-5. Why are the solutions x>-3 and x<-3 omitted? Is it because the other two include a broader range?

    thanks for your help.
     
  2. jcsd
  3. Sep 6, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Hi theself, welcome to PF.

    You can find an absolute value key (vertical line) on the keyboard.

    So the problem is: Solve |x+3|>2

    The absolute value is never negative. The two cases are x+3≥0 (x>-3) and x+3<0 (x<-3).

    You wrote correctly that |x+3|=x+3 in the first case and |x+3|=-(x+3) in the second case.

    In the first case, you got x>-1 if x>-3 is true. x>-1 is more strict requirement than x>-3. So x>-1 is solution.

    In the second case, you got x<-5 if x<-3. x<-5 is more strict condition than x<-3 so x<-5 is solution.

    "Why are the solutions x>-3 and x<-3 omitted?"

    They are not solutions. Substitute x=-2 for example: (it is greater then -3). |x+3|=1 less then 2. Choose x=-4: (it is less then -3) |-4+3|=|-1|=1.




    ehild
     
  4. Sep 6, 2012 #3

    Mark44

    Staff: Mentor

    Why? Use the | character instead.
    How did you get what you have just above?
    |x + 3| > 2
    <==> x + 3 > 2 OR x + 3 < -2
    Can you take it from here?
     
  5. Sep 6, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The simplest way to solve a general inequality is to first solve the corresponding equation. If |x+ 3|= 2 then either x+ 3= 2 or x+ 3= -2. In the first case x= -1 and in the second x= -5.

    The key point here is that, since absolute value is continuous, we can go from "<" to ">", and vice-versa, only at "=". That is, the two points, -5 and -1, divide the real numbers into three intervals and on each interval only one of "<" or ">" can apply. And we need only check one point in each interval to see which.

    For example, -6< -5 and |-6+3|= |-3|= 3 which is larger than 2, not less. The original inequality is false at x= -6 and so false for all x less than -5. -4 lies between -5 and -1 and |-4+3|= |-1|= 1 which is less than 2. The original inequality is true at x= -4 and so true for all x between -5 and -1. Finally, 0 is larger than -1 and |0+ 3|= |3|= 3 which is larger than 2, not less. The original inequality is false at x= 0 and so false for all x larger than -1.

    By the way, every keyboard I have ever seen has a "|" just above the return key. Are you using a keyboard that doesn't?
     
  6. Sep 6, 2012 #5
    Is the if condition there because we define it like that at the beginning ?
     
  7. Sep 6, 2012 #6
    Well, I kind of did it in a long way;
    If |x+3|> 0 then |x+3|= x+3, and substituted the term (x+3) to where |x+3| is found
    this gives
    if x> -3, x > -1
    and did the same for the other condition
     
  8. Sep 6, 2012 #7
    Wouldn't using the equation create a problem as you can not do everything you do to an equation to an inequality?
    For eg: dividing both sides by - 1

    * Also I apologize for my clumsiness for not finding the key for "|"
     
  9. Sep 6, 2012 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I think the answer is "yes"
    The result x>-1 came from the condition that x>-3. x>-3 and x>-1 are not two solutions. Those x values for which -3<x<-1 do not satisfy the original inequality. In similar way, when we supposed that x<-3 we got the result x<-5. Again, the original inequality is not true for -5<x<-3.

    ehild
     
    Last edited: Sep 6, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: An inequality with absolute values
Loading...