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Solving Rational Equations

  1. Oct 16, 2011 #1
  2. jcsd
  3. Oct 16, 2011 #2


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    Without actually giving you all the details of the process for #i,

    Subtract x from both sides;
    Multiply both sides by (x+2);

    The left-hand member can be expressed as x2+x-6.
    You need to consider two cases. (x+2) positive, and (x+2) negative. Also, as you might have noticed, x cannot be -2.
  4. Oct 16, 2011 #3


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    Need parentheses! This looks like
    [tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]
    Furthermore, looking at your work, it looks like the the numerator was originally 4x + 4. So which is it? 4x - 4 or 4x + 4? Please double check the problem.
  5. Oct 16, 2011 #4
    Sorry, the question is (4x+4)/(x-1) < x^2 + 3x + 2 and by the way how did you write the problem as:

    [tex]4x - \frac{4}{x} - 1 < x^2 + 3x + 2[/tex]
  6. Oct 16, 2011 #5


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    There's a button in the editing window (the last one on the 2nd row, the one that looks like the summation symbol) that allows you to type in LaTeX. You should learn it -- makes the expressions/equations/inequalities much more readable.

    Anyway, I think your problem here is that you set up the sign chart too early. You have this:
    [tex]\frac{(x+1)(x+2)(x-1)\text{ }(-4x-4)}{x-1} < 0[/tex]
    ... and the problem is that it looks like the last factor (-4x-4) is multiplied by the other 3 factors, when in fact it is added (or subtracted). And so you didn't get the correct zeros for the numerator when looking for the critical points.

    What you'll need to do is take the numerator (which should be written like this):
    [tex](x+1)(x+2)(x-1) - 4x - 4[/tex]
    multiply out the 3 factors, combine like terms with the -4x - 4, and THEN set equal to zero. Since this will be a cubic, you'll have to try the Rational Roots test. (The 3 roots are all integers, which makes things a little easier.)
  7. Oct 16, 2011 #6
    OMG!! Thank you so much! This question was driving me crazy and thanks to you I finally solved it!! :)
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