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Math Help - Inequalities

  1. #1
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    Inequalities

    Can anybody help me to solve this?

    1. \frac{6}{\mid x \mid + 1} < \mid x \mid

    2. \frac{6}{\mid x \mid} < \mid 2x-1 \mid

    3. \mid x \mid < \mid x^2-7x+6 \mid

    Answer:
    1. x<-2, x>2

    2. \{x:x<-\frac{3}{2}\} \cup \{x:x>2\}

    3. x<4-\sqrt{10}, 3-\sqrt{3}<x<3+\sqrt{3}, x>4+\sqrt{10}
    Last edited by cloud5; July 2nd 2009 at 03:46 AM.
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    Can anybody help me to solve this?

    1. \frac{6}{\mid x \mid + 1} < \mid x \mid

    2. \frac{6}{\mid x \mid} < \mid 2x-1 \mid

    Answer:
    1. x<-2, x>2

    2. \{x:x<-\frac{3}{2}\} \cup \{x:x>2\}
    1. \frac{6}{\mid x \mid + 1} < \mid x \mid or 0 < \mid x \mid^2 + \mid x \mid - \,6\;\;\;\Rightarrow\;\;\; 0 < \left( \mid x \mid + 3\right)\left( \mid x \mid -2\right) which only happens if \mid x \mid < -3 (never) or \mid x \mid > 2, your answer.

    2. \frac{6}{\mid x \mid} < \mid 2x-1 \mid or \mid 2x-1 \mid \mid x \mid > 6 or \mid 2x^2 - x \mid > 6. Solutions of this are 2x^2-x > 6 or 2x^2-x <-6 (never). For the first 2x^2-x > 6 or (2x+3)(x-2) > 0 which gives x < - \frac{3}{2} and x > 2.
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