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Math Help - matching the terms in partial fractions

  1. #1
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    matching the terms in partial fractions

    does anybody know where this "1" comes from?
    Attached Thumbnails Attached Thumbnails matching the terms in partial fractions-img_3571.jpg  
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  2. #2
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    Re: matching the terms in partial fractions

    expanding the RHS yields the quadratic terms 2Au^2 + Cu^2 = (2A+C)u^2

    note the coefficient of u^2 on the LHS is 1 , hence by matching coefficients 2A+C = 1
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  3. #3
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    Re: matching the terms in partial fractions

    u2 = (1)(u2), that is, in the polynomial u2, the coefficient of u2 is 1 (this is usually suppressed when we write polynomials).
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    Re: matching the terms in partial fractions

    Thank you very much.
    Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

    matching the terms in partial fractions-img_3572.jpg
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  5. #5
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    Re: matching the terms in partial fractions

    Quote Originally Posted by cxz7410123 View Post
    Thank you very much.
    Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

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    LHS is {\color{red}0}x^2 + 0x + 3

    you're matching coefficients, remember?
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