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    Right, this is going to make me sound monumentally stupid, but here goes.

    Determine the centre and radius of:


    2x^2 - 16x + 2y^2 + 4y = 1

    (sorry for lack of latex)

    All I need to know, is how to complete the square when there is a co-efficient infront of the x/y^2

    Then im fine..

    So far for the x part I have:


    2(x-4)^2 then im stuck.


    Help Please.
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    The band is called Circle Takes the Square and is far more interesting than this maths problem will ever be.
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    I personally just divide through by the coefficient and carry on from there:

    x^2 - 8x + y^2 + 2y = 0.5
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    You don't complete the square in the same way as with quadratics. Treat the x and y parts seperately.

    e.g.  2x^2 - 16 x = 2(x-4)^2 -32
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    (Original post by EasyTiger)
    All I need to know, is how to complete the square when there is a co-efficient infront of the x/y^2
    Since it is the same coefficient, you could divide through both sides by it
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    Thanks for both of you. The way benswellday stated was the way I was trying to do, and I got 2(x-4)^2-32 aswell. But just didnt think it was correct.

    Thanks for the advice though.
 
 
 
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Updated: November 12, 2008

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