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Thread: Solving Inequality

  1. #1
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    Solving Inequality

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

    im not sure about my result.

    as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

    the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

    which leads me to the first set of solutions: x > 2

    am i right here ?

    the 2nd case would give me the same just with flipped inequality sign

    so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

    somehow all my math programs give me the solution: -2/3 < x < 2
    i cant see why that is
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  2. #2
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    Quote Originally Posted by coobe View Post
    $\displaystyle \frac {-4x^2} {2-x} < 2-4x$

    im not sure about my result.

    as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

    the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

    which leads me to the first set of solutions: x > 2

    am i right here ?

    the 2nd case would give me the same just with flipped inequality sign

    so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

    somehow all my math programs give me the solution: -2/3 < x < 2
    i cant see why that is
    HI

    Obviously , $\displaystyle x\neq 2$

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

    $\displaystyle 0<2-4x+\frac{4x^2}{2-x}$

    $\displaystyle \frac{8x^2-10x+4}{2-x}>0 $

    Since $\displaystyle 8x^2-10x+4$ is complex , then $\displaystyle 2-x>0\Rightarrow x<2 $
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  3. #3
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    Quote Originally Posted by coobe View Post
    $\displaystyle \frac {-4x^2} {2-x} < 2-4x$

    im not sure about my result.

    as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

    the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

    which leads me to the first set of solutions: x > 2

    am i right here ?

    the 2nd case would give me the same just with flipped inequality sign

    so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

    somehow all my math programs give me the solution: -2/3 < x < 2
    i cant see why that is
    Clearly $\displaystyle x \neq 2$.


    Case 1: $\displaystyle 2 - x < 0$


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

    $\displaystyle -4x^2 > (2 - 4x)(2 - x)$

    $\displaystyle -4x^2 > 4 - 10x + 4x^2$

    $\displaystyle 8x^2 - 10x + 4 < 0$

    $\displaystyle x^2 - \frac{5}{4}x + \frac{1}{2} < 0$

    $\displaystyle x^2 - \frac{5}{4}x + \left(-\frac{5}{8}\right)^2 - \left(-\frac{5}{8}\right)^2 + \frac{1}{2} < 0$

    $\displaystyle \left(x - \frac{5}{8}\right)^2 + \frac{7}{64} < 0$


    This is clearly a contradiction.



    Case 2: $\displaystyle 2 - x > 0$.


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

    $\displaystyle -4x^2 < (2 - 4x)(2 - x)$

    $\displaystyle -4x^2 < 4 - 10x + 4x^2$

    $\displaystyle 8x^2 - 10x + 4 > 0$

    $\displaystyle \left(x - \frac{5}{8}\right)^2 + \frac{7}{64} > 0$


    This is true for all $\displaystyle x$.


    Thus, the solution to the inequality is $\displaystyle 2 - x > 0$, or $\displaystyle x < 2$.
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