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Math Help - Rearranging a fixed point equation

  1. #1
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    Rearranging a fixed point equation

    Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x

    But then rearranges to give x^2+3x-28=0

    I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

    Any help/hints greatly appreciated! Thanks.
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  2. #2
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    \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x

    \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=\frac{8}{8}x

    \frac{-1}{8}x^2+\frac{-3}{8}x+\frac{7}{2}=0

    x^2+3x+\frac{-56}{2}=0

    x^2+3x-28=0
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  3. #3
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    Quote Originally Posted by NitrousUK View Post
    Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x

    But then rearranges to give x^2+3x-28=0

    I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

    Any help/hints greatly appreciated! Thanks.
    If we multiply through by 8 (the LCD of 2 and 8)

    -x^2 + 5x + 28 = 8x

    The 3x comes from 8x-5x when simplifying

    x^2+3x-28 = 0
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  4. #4
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    Quote Originally Posted by NitrousUK View Post
    Hi, sorry if this is a silly/simple question! But i've been banging my head against it for a while. There's an example of a fixed point equation in my course book that gives \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x

    But then rearranges to give x^2+3x-28=0

    I can see they're dividing through by the 'a' coefficient for the quadratic, but I don't understand the operation involved in turning the 'b' coefficient from 5/8 divide by 1/8, which gives me 5, into the 3 when the RHS (which is 'x' over 1/8) is transposed.

    Any help/hints greatly appreciated! Thanks.

    \frac{-1}{8}x^2+\frac{5}{8}x+\frac{7}{2}=x
    multiply by 8
    -x^2+5x+28=8x
    multiply by -1

    x^2-5x-28=-8x
    x^2-5x-28+8x=0
    x^2-3x-28=0
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  5. #5
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    Multiply by the denominator to remove the fractions! Of course..

    Thank you all for the very speedy and helpful reply!!
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