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Math Help - Entering Quadratic

  1. #1
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    Entering Quadratic

     \frac{90}{s+5} + \frac{90}{s-5} = 2\frac{1}{8}

    If I want to enter this into quadratic formula, do I need to change it someway? The coefficients are in fractions and that's what confuses me.
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  2. #2
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    Hi reallylongnicknam.
    put the equation into this form a^2 +bx +C = 0



    bjh
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  3. #3
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    Firstly, multiple both sides by (s+5)(s-5). What do you get?
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  4. #4
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     s^2 -25 I'm thinking there is cross multipling with those 90s.
    Last edited by reallylongnickname; March 30th 2011 at 06:28 PM.
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  5. #5
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    I mean \displaystyle  \frac{90}{s+5} \times (s+5)(s-5)+ \frac{90}{s-5}\times (s+5)(s-5) = 2\frac{1}{8}\times (s+5)(s-5)

    What do you get?
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  6. #6
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     90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5)
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  7. #7
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    Good, now expand each term and group where they are like.

    What do you get?
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  8. #8
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    Quote Originally Posted by reallylongnickname View Post
     90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5)
    OK so far...continue...easier if you change 2 1/8 to 17/8
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  9. #9
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     -\frac{17}{8 }s^2 + 180s + 25 = 0

    Would I divide all coefficients by 8 to remove the fraction before I can plug this into the quadratic formula?
    Last edited by reallylongnickname; March 31st 2011 at 03:36 AM.
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  10. #10
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    You don't have to: a, b, and c in " ax^2+ bx+ c= 0" don't have to be integers but you will probably find the arithmetic easier.
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  11. #11
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    Quote Originally Posted by reallylongnickname View Post
     -\frac{17}{8 }s^2 + 180s + 25 = 0
    That is not correct.
    90(S - 5) + 90(S + 5) = 180S

    17/8(S - 5)(S + 5) = 17/8(S^2 - 25)

    So: (17/8)S^2 - (17/8)25 - 180S = 0
    17S^2 / 8 - 180S - 425 / 8 = 0
    Multiply by 8:
    17S^2 - 1440S - 425 = 0

    OK?
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