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Math Help - Another problem child

  1. #1
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    Another problem child

    The problem child I scanned...
    (Note: The scanner cut off part of the problem, thats a 4 on the left side there.)

    While I work on correcting some problems, this was another one from earlier I was having problems with. Anyone know the right way to solve it?

    This should also be the last problem for now I ask for steps on how to do it. (Until I possibly stumble upon another one later...)


    I also forgot to post this earlier, its 2 other problems I wanted to check on if I did it right (if I didn't please point out where). Although i'm now noticing I didn't quite finish #2 yet, but am I at least on the right track?

    More children with problems...
    Last edited by SailorCardKnight; November 24th 2009 at 09:08 AM. Reason: corrected 2nd link
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  2. #2
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    for the 1st problem, start by factoring the numerators and denominators of both fractions ... then do the division by multiplying by the reciprocal of the divisor.
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  3. #3
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    = \frac{4x^2 - 9}{3x^2 + 2x} x \frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}

    = \frac{(2x)^2 - 3^2}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{(2x- 3)(2x + 3)}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{2x- 3}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{x + 4}

    could be done further than that.....
    3x^2 - 7x + 6 cannot be factorized since its discriminant b^2 - 4ac < 0
    .....so may be something here is written wrong....
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  4. #4
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    Quote Originally Posted by ukorov View Post
    = \frac{4x^2 - 9}{3x^2 + 2x} x \frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}

    = \frac{(2x)^2 - 3^2}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{(2x- 3)(2x + 3)}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{2x- 3}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{x + 4}

    could be done further than that.....
    3x^2 - 7x + 6 cannot be factorized since its discriminant b^2 - 4ac < 0
    .....so may be something here is written wrong....
    *looks* 3x^2-7x+6 is actually supposed to be 3x^2-7x-6, thats the only part I saw written wrong, could that be it?

    Any word about those problems I did?
    Last edited by SailorCardKnight; November 24th 2009 at 11:33 AM.
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  5. #5
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    Quote Originally Posted by ukorov View Post
    = \frac{4x^2 - 9}{3x^2 + 2x} x \frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}

    = \frac{(2x)^2 - 3^2}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{(2x- 3)(2x + 3)}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}

    = \frac{2x- 3}{x(3x + 2)} x \frac{3x^2 - 7x + 6}{x + 4}

    could be done further than that.....
    3x^2 - 7x + 6 cannot be factorized since its discriminant b^2 - 4ac < 0
    .....so may be something here is written wrong....
    Yes,it's 3x^2-7x-6 which can be factorized to (x-3)(3x+2)
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