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Math Help - completing the square

  1. #1
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    completing the square

    I need some help please I'm trying to rewrite the below equation into the form of a hyperbola

    x^2-4x-y^2+6y-6=0

    on competing the square for x^2-4x i get

    x^2-4x = (x^2-4x+4)-4 = (x-2)^2-4 thats ok

    But I'm getting in a terrible mess when completing the square for -y^2+6y

    the final answer in the exercise book is

    (x-2)^2-(y-3)^2-4+9-6=0

    That is (x-2)^2-(y-3)^2=1

    So it would seem that the completed square form of
    -y^2+6y is (y-3)^2+9

    But I just can't seem to get this I end up with a negative 9 if someone could take me through this process step by step I would be eternally grateful
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  2. #2
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    Quote Originally Posted by macca101
    I need some help please I'm trying to rewrite the below equation into the form of a hyperbola

    x^2-4x-y^2+6y-6=0

    ...
    Hello,

    you've got this:
    x^2-4x\ \ -y^2+6y-6=0. Rearrange to:

    (x^2-4x+4) \ -(y^2-6y+9)=6+4-9

    Notice that there is a negative sign before the 2nd paranthese. Therefore the 9 which is added in the paranthese to get a complete square is actually subtracted. So you added 4 and subtracted 9 on the LHS of the equation and therefore you have to do the same on the RHS too.
    You'll get:

    (x-2)^2 \ - (y-3)^2=1

    Greetings

    EB
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  3. #3
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    eternally grateful, i am

    Thanks
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