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Thread: Solve this inequality

  1. #1
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    Solve this inequality

     |3x+1|>\sqrt{x+1}

    My working :

    There will be 2 cases here .

    Case 1 : 3x+1>\sqrt{x+1}

     9x^2+6x+1>x+1

     x(9x+5)>0

    Case 2 : 3x+1<-\sqrt{x+1}

    (3x+1)^2>x+1

    x(9x+5)>0 , Isn't this the same as case 1 ..

    I think i made a mistake at case 2 . Can someone check for me .. Thanks .
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  2. #2
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    second case is irrelevant since both sides are positive, so you are free to square them and solve the quadratic inequality, provided of course that we require x\ge-1.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    second case is irrelevant since both sides are positive, so you are free to square them and solve the quadratic inequality, provided of course that we require x\ge-1.
    Thanks Krizalid .

    So i will only need to solve x(9x+5)>0 to get my solution right ?

    If so , my answer is (-\frac{5}{9},\infty)

    But the answer given is  {x:x>0 or -1<x<-\frac{5}{9}} .. Puzzled again
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  4. #4
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    that's because you need to restrict the square root.

    solution set for the quadratic inequality is \left] -\infty ,-\frac{5}{9} \right[\cup \left] 0,\infty  \right[. but this is not the full solution, 'cause on the other hand we require x\ge-1, thus the solution set it's actually \left] -\infty ,-\frac{5}{9} \right[\cup \left] 0,\infty  \right[\cap \left[ -1,\infty  \right[=\left[ -1,-\frac{5}{9} \right[\cup \left] 0,\infty  \right[, which is our full solution set.
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  5. #5
    Super Member dhiab's Avatar
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    Another resolution
    Attached Files Attached Files
    Last edited by dhiab; Jun 12th 2009 at 02:27 PM.
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  6. #6
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    you did the same i did.
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