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Absolute Value problems Please help!

  1. Jan 2, 2005 #1
    (|2x-3| + x) / (x^2 - 3x + 2) < 1

    |(x^2 - 5x + 4) / ( x^2 - 4)| =< 1

    Can somebody help me with this quadratic inequalities, please... If you have time, please also give me an idea of how you solved them... Thank you!
  2. jcsd
  3. Jan 2, 2005 #2


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    I've never done that before but here's how I would proceed...

    [tex]|2x-3|+x < x^2-3x+2 \Leftrightarrow |2x-3| < x^2-4x+2 \Leftrightarrow -x^2+4-2<2x-3<x^2-4x+2 \Leftrightarrow -x^2+2x+1<0<x^2-6x+5[/tex]

    We want to find the x satisfying both these inequalities at the same time. Use the formula to find the roots of a quadratic poly on both inequalities and help yourself with the graph of the functions to find the wanted intervals. I find that the solution is the union of the intervals [itex] (1-\sqrt{2},1)U(1+\sqrt{2},5)[/itex]
    Last edited: Jan 2, 2005
  4. Jan 3, 2005 #3


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    A continuous function can change sign only where it is equal to 0 and functions like these are continuous everywhere they are defined- that is everywhere except where a denominator is 0.

    A very good way of solving such problems is: first determine where they are equal to 0. Next determine where they are not defined- where the denominators are equal to 0. Those numbers divide the number line into intervals. Finally, choose one number in each of the intervals and check to see whether the inequality is true or false at that point. If the inequality is true for that point, it is true for the entire interval.
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