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Math Help - inequality confucion

  1. #1
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    inequality confucion

    Solve this inequality .

    \frac{6}{|x|+1}<|x|

    squarring both sides , i got

    \frac{36}{x^2+2|x|+1}<x^2 , rearranging them

    \frac{36}{x^2+2|x|+1}-\frac{x^2(x^2+2|x|+1)}{x^2+2|x|+1}<0

    \frac{36-x^4-2x^2|x|-x^2}{x^2+2|x|+1}<0 , till here am i correct , and i don see a way to solve it from here .
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    Solve this inequality .

    \frac{6}{|x|+1}<|x|

    squarring both sides , i got

    \frac{36}{x^2+2|x|+1}<x^2 , rearranging them
    Squaring unnecessarily complicates this simple problem. Instead take the following route:

    \frac{6}{|x|+1}<|x|

    \frac{6}{|x|+1} - |x| < 0

    \frac{6 - |x|^2 - |x|}{|x|+1} < 0

    But |x| + 1 > 0, thus we can multiply it on both sides without affecting the inequality. Thus

    6 - |x|^2 - |x| < 0

    And so on....
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  3. #3
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    Re :

    Quote Originally Posted by Isomorphism View Post
    Squaring unnecessarily complicates this simple problem. Instead take the following route:

    \frac{6}{|x|+1}<|x|

    \frac{6}{|x|+1} - |x| < 0

    \frac{6 - |x|^2 - |x|}{|x|+1} < 0

    But |x| + 1 > 0, thus we can multiply it on both sides without affecting the inequality. Thus

    6 - |x|^2 - |x| < 0

    And so on....
    Yeah , i make things complicated for myself .
    From there , i got x>3 and x<-3 .... is it correct .
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    Yeah , i make things complicated for myself .
    From there , i got x>3 and x<-3 .... is it correct .
    Unfortunately no.

    From 6-|x|^2-|x|<0 ......you'll get:

    |x|^2+|x|-6>0~\implies~(|x|+3)(|x|-2)>0

    A product of 2 factors is positive if both factors are positive or both factors are negative. Since |x|+3>0 is true for all x, the second case is impossible. Therefore (keep in mind that |x|+3>0 is true for all x):

    |x|-2>0~\implies~|x|>2

    Now use the definition of the absolute value:

    |x|>2~\iff~x>2~\vee~-x>2~\iff~\boxed{x<-2~\vee~x>2}

    By the way:
    From there , i got x>3 and x<-3
    You'll never find a number which is greater than 3 and simultaneously is smaller than -3.
    Last edited by earboth; January 16th 2009 at 08:42 AM.
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