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Inequality problem

  1. Nov 21, 2009 #1
    http://img403.imageshack.us/img403/6198/48289623.png [Broken]

    Find possible values of x.

    Can somone explain me step by step how to do that?

    I would be grateful.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 21, 2009 #2

    Mentallic

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    You'll have to solve two sets of inequalities.

    First solve [tex]\frac{2}{x-2}\leq \frac{x+2}{x-2}[/tex]

    and then either solve [tex]\frac{2}{x-2}\leq 1[/tex] OR [tex]\frac{x+2}{x-2}\leq 1[/tex]

    And of course the set of solutions need to be valid in both cases, so some cases will be scrapped.

    e.g. if you have [itex]a\leq b\leq c[/itex] (a,b,c any function of x)
    and you solve [itex]a\leq b[/itex] and end up with the result [itex]x\leq 5[/itex]
    and then you solve [itex]b\leq c[/itex] and end up with [itex]x> 2[/itex]

    then you can conclude that the answer is [itex]2<x\leq 5[/itex]
     
  4. Nov 21, 2009 #3
    Thank you for replying Mentallic.

    Sorry for ignorance but I have some more doubts.

    how do I solve [tex]
    \frac{x+2}{x-2}\leq 1
    [/tex]?

    Should I do x+2 <= 1, find the answer and make intersection with answer of x-2 <= 1?
    I don't remember how do solve a quotient-inequality...

    Thank you
     
  5. Nov 21, 2009 #4

    Mentallic

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    Ahh sorry, I just assumed you were comfortable with solving one set of inequalities and the problem here was actually what to do when it's in the format of two inequalities.

    There are also a variety of ways to solve these, so I guess you could take your pick of your favourite :smile: But it's always nice to learn the other methods too.


    1) Like you were kind of thinking, you can take cases in the denominator separately. There are a few cases in this, so I'll give you the idea with the first one:

    Assume the denominator is positive i.e. [itex]x>2[/itex]

    Now since the denominator is positive by our assumption, we can multiply both sides of the equation and it won't affect the inequality sign.

    [tex]x+2\leq x-2[/tex]

    Now we solve this normally since there is no threat of dividing by a negative. So we get [itex]2\leq -2[/itex] but this is obviously wrong, so this means our first assumption that [itex]x>2[/itex] is wrong.

    Now you need to try for the case that the denominator is negative (you don't need to do it for 0, since the denominator cant be 0).


    2) The problem with inequalities is that the sign must flip around if you multiply/divide by a negative number. And since we are doing this for x, there will be values that are both positive and negative. But any number squared is always positive, so this is what we can do.

    [tex]\frac{x+2}{x-2}\leq 1[/tex]

    multiply both sides of the fraction by the same thing: [tex]\frac{(x+2)(x-2)}{(x-2)^2}\leq 1[/tex]

    and now you can see the denominator will always be positive, so we multiply through by normal. Then you'll need to re-arrange and make the quadratic equal zero. Do you know how to solve quadratic inequalities?
     
  6. Nov 21, 2009 #5
    Thank you for replying again Mentallic, your explanation helped me a lot.
     
  7. Nov 21, 2009 #6

    Mentallic

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    You're welcome :smile:
     
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