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Math Help - equations with unknowns in the denominator

  1. #1
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    equations with unknowns in the denominator

    Solve the inequality \frac{3}{x(x-2)}<1
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    MHF Contributor red_dog's Avatar
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    \frac{3}{x(x-2)}<1\Leftrightarrow\frac{3}{x(x-2)}-1<0\Leftrightarrow\frac{-x^2+2x+3}{x(x-2)}<0

    \Leftrightarrow\frac{-(x-3)(x+1)}{x(x-2)}<0\Leftrightarrow\frac{(x-3)(x+1)}{x(x-2)}>0

    Now make the sign of every factor in a table, then use the rule of signs and choose the intervals.
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  3. #3
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    Sorry, but its the first time I've seen it solved that way. I don't get what I'm supposed to do next? What is the rule of signs?
    Last edited by requal; September 5th 2009 at 01:15 AM. Reason: adding stuff
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  4. #4
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    Quote Originally Posted by requal View Post
    Solve the inequality \frac{3}{x(x-2)}<1
    \Rightarrow \frac{x(x - 2)}{3} > 1 \Rightarrow x(x - 2) > 3 \Rightarrow x^2 - 2x - 3 > 0.

    Solve this.
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  5. #5
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    My favorite method of solving "complicated" inequalities is to solve the equation first. \frac{3}{x(x-2)}= 1 gives 3= x(x-2) or x^2- 2x- 3= (x- 3)(x+ 1)= 0 so x= 3 or x= -1.

    The point of that is that a rational function (fraction) can "swap" an inequality on where it is "= " or where the denominator is 0. Here you know that happens at x= -1, x= 0, x= 2, and x= 3. That separates the real numbers into intervals: (-\infty, -1), (-1, 0), (0, 2), (2, 3), and (2, \infty).

    Now you just need to check one value in each interval to see whether it is ">" or "<" for every number in that interval.

    For example, x= -2 is in (-\infty, -1) and \frac{3}{-2(-2-2)}= \frac{3}{8}< 1 so \frac{3}{x(x-2)}< 1 for all x in (-\infty, -1).

    -1/2 is in (-1, 0) and \frac{3}{-\frac{1}{2}(-\frac{1}{2}-2}= \frac{3}{\frac{3}{4}}= 4> 1. \frac{2}{x(x-2)}> 1 for all x in (-1, 0).

    1 is in (0, 2) and \frac{1}{1(1-2)}= -1< 1 so \frac{3}{x(x-2)}< 1 for all x in (0, 2).

    Now try x= \frac{5}{2} (in (2, 3)) and x= 4 (in (3, \infty)).

    It's not hard to see that "> 1" and "< 1" alternate on the intervals. The "rule of signs" red dog referred to simply says that a product and/or quotient involving an even number of negatives is positive an a product involving an odd number of negatives is negative. After you write this as a single fraction as red dog did, you need to check whether each factor is positive or negative. Here, every time we "cross" on of those four numbers, exactly one term changes sign so the sign of the fraction changes.
    Last edited by mr fantastic; September 5th 2009 at 02:46 PM. Reason: Fixed some latex
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