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Math Help - partial fractions

  1. #1
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    partial fractions


    Could someone please explain how to solve this?
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  2. #2
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    Quote Originally Posted by Haris View Post

    Could someone please explain how to solve this?
    Take out a factor of a quarter and you get:

     \frac{1}{4}.\frac{1}{x^2-\frac{9}{4}}

    This is a difference of two squares:

     \frac{1}{4}.\frac{1}{x^2 -(\frac{3}{2})^2} = \frac{1}<br />
{4}.\frac{1}{(x-\frac{3}{2})(x+\frac{3}{2})}

    Is that better?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Haris View Post

    Could someone please explain how to solve this?
    note that \frac 1{4x^2 - 9} = \frac 1{(2x + 3)(2x - 3)} = \frac A{2x + 3} + \frac B{2x - 3}

    Now multiply through by (2x + 3)(2x - 3), you get

    1 = A(2x - 3) + B(2x + 3) ................(1)

    now plug in x = \frac 32 and solve for B, then set x = - \frac 32 and solve for A.
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  4. #4
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    (2x+3)-(2x-3)=6.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    (2x+3)-(2x-3)=6.
    haha, ok, fine Kriz

    without partial fractions: \frac 1{(2x + 3)(2x - 3)} = \frac 16 \cdot \frac 6{(2x + 3)(2x - 3)} = \frac 16 \cdot \frac {2x + 3 - (2x - 3)}{(2x + 3)(2x - 3)} = \frac 16 \left( \frac 1{2x - 3} - \frac 1{2x + 3} \right)

    happy now?
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  6. #6
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    Yeah.
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