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Math Help - Simple algebra

  1. #1
    Senior Member Mukilab's Avatar
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    Simple algebra

    x^2-8x+23=(x-p)^2+q for all values of x

    Find p and q, does this mean I can just use x as anything such as 0?

    if so using simultaneous equations I got -8=p^2-p^2

    What now?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    x^2-8x+23=(x-p)^2+q for all values of x

    Find p and q, does this mean I can just use x as anything such as 0?

    if so using simultaneous equations I got -8=p^2-p^2

    What now?
    No, you should complete the square on the left hand side to get it into the form on the right, then find p and q by comparing coefficients

    To get you started x^2+bx = \left(x+\frac{b}{2}\right)^2 - \frac{b^2}{4}<br />
    Last edited by e^(i*pi); May 20th 2010 at 12:29 PM. Reason: error in expansion
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  3. #3
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    No, you should complete the square on the left hand side to get it into the form on the right, then find p and q by comparing coefficients

    To get you started x^2+bx = \left(x+\frac{b}{2}\right)^2 - \frac{b^2}{4}<br />
    Do not know what you mean by
    " complete the square on the left "
    or
    "comparing coefficients"
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    Do not know what you mean by
    " complete the square on the left "
    or
    "comparing coefficients"
    Completing the square is covered in post 3 of this topic: http://www.mathhelpforum.com/math-he...tic-curve.html

    Comparing coefficients says that two equations are equal if and only if the coefficients on each side match.

    For example given that ax^2+bx+c = x^2 - 12x+1 comparing coefficients tells you that a = 1, b=-12 and c = 1
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  5. #5
    Senior Member Mukilab's Avatar
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    I have several problems with 'completing the square', please read that thread.

    Thanks for telling me about comparing the coefficient
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