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Hard Inequalities

  1. Oct 23, 2012 #1
    I am looking at questions like
    1/(x+4)>x-4
    or 1/(x+7)>x-7
    I have no idea how to solve them,
    I have simplified to (-x^2+50) /(7+x)
    however I don't think it is correct and I don't know what do do from there.
     
  2. jcsd
  3. Oct 23, 2012 #2

    tiny-tim

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    hi alewisGB! :smile:

    (try using the X2 button just above the Reply box :wink:)

    obviously, you need to get rid of the fraction!

    in an ordinary equation, you'd just multiply both sides by the denominator

    but that might be negative (which would change the > to <),

    so, instead, multiply both sides by the square of the denominator …

    what do you get? :smile:
     
  4. Oct 24, 2012 #3
    for:
    1/(x+2)>x-2
    Step 1: x+2>x3+2x2-4x-8
    Step 2: 0>x3+2x2-5x-10
    Step 3: 0>(x+2)(x2-5)
    Step 4: if 0=(x+2)(x2-5)
    x = -2 or +√5 or -√5
    Step 5: Unsure, do input -2.01 and -1.99 and see which one is true? (same for +or-√5) or do I do something else
     
  5. Oct 24, 2012 #4

    tiny-tim

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    hi alewisGB! :smile:

    this way is easier :wink:

    1/(x+2)>x-2
    Step 1: x+2>(x+2)(x2-4)
    Step 3: 0>(x+2)(x2-5)
    Step 4: 0>(x+√5)(x+2)(x-√5)

    … and you now have this in the form 0 > (x-a)(x-b)(x-c) with a<b<c,

    sooo … ? :smile:
     
  6. Oct 24, 2012 #5
    I have got the answer of -√5 > x and 5 > x > -2
    I have not done further maths and although inequalities were covered they were not nearly this complex
    I have not seen the rule "0 > (x-a)(x-b)(x-c) with a<b<c"
    So I am not sure what that means
    Thank you for your help :)
     
  7. Oct 24, 2012 #6

    tiny-tim

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    oh, that's easy

    the RHS is the product of three numbers,

    and it can only be negative if one or all three of them are negative,

    which means either x < a or b < x < c :wink:
     
  8. Oct 24, 2012 #7
    Thank you ever so much, you have no idea how grateful I am. This will help me in now but also in the future :)
     
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