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Thread: Another inequality problem

  1. #1
    Junior Member
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    Another inequality problem

    Solve the below inequality:

    $\displaystyle
    ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3
    $

    I tried solving this as below :

    squaring both side and after solving

    $\displaystyle
    (5x^2-9x+18) < 0
    $

    and the factors of the above equation comes to be
    x= (9 +(81 - 360)^1/2)/10,(9 -(81 - 360)^1/2)/10

    which turns out to be complex number

    Please let me know how to proceed further.

    Any help would be appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by a69356 View Post
    Solve the below inequality:
    $\displaystyle
    ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3
    $

    I tried solving this as below :

    squaring both side and after solving

    $\displaystyle
    \color{red}(5x^2-9x+18) < 0$ here is a mistake.
    It should be $\displaystyle 5x^2-9x-18 <0$.
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  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
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    Hi buddy,
    Quote Originally Posted by a69356 View Post
    Solve the below inequality:

    $\displaystyle
    ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3
    $

    I tried solving this as below :

    squaring both side and after solving

    $\displaystyle
    (5x^2-9x+18) < 0
    $
    This is already false.

    $\displaystyle
    ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3
    $

    squaring leads to

    $\displaystyle 1 - \frac{x+2}{x^2} < \frac 4 9$

    Multiply by x^2

    $\displaystyle x^2 - (x+2) < \frac 4 9 x^2$

    $\displaystyle x^2 - \frac 4 9 x^2 - (x+2)<0$

    $\displaystyle \frac 5 9 x^2 -x-2 < 0$

    solve this and you are nearly done. You still have to consider the domain of f(x)

    Beaten to Plato



    Yours Rapha
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