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Math Help - Completing the square

  1. #1
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    Completing the square

    Hello,

    Can someone please show me how to complete this square?

    2x^2+y^2=2y+1

    Thanks!
    -Mark
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  2. #2
    Super Member Bacterius's Avatar
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    You are sure it isn't 2xy in the right-hand side ? If yes, then it is easy, if no, then there is a little more thinking to do. Have you tried using square identities to factor out both unknowns together ?
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    You are sure it isn't 2xy in the right-hand side ? If yes, then it is easy, if no, then there is a little more thinking to do. Have you tried using square identities to factor out both unknowns together ?
    Yes, the right side is what shows in the book.

    I was thinking, is this the right way to start it?

    2x^2 - 1 = 2y - y^2
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  4. #4
    Super Member Bacterius's Avatar
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    I think there should be a way to factorize y using a square identity, leaving alone a constant based on x (since you are completing the square) :

    2x^2 + y^2 = 2y + 1

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

    I know what follows is not very mathematic but perhaps it could be a good idea to try with some numbers substituting x and y and investigate what happens, if a pattern doesn't show up ... but you have to find working numbers ...
    Here is what I would do :

    I would say x = 1, therefore :

    2 - 2y + y^2 = 1

    y^2 - 2y + 2 = 1

    y^2 - 2y + 1 = 0

    Woah ! A square identity !

    (y - 1)^2 = 0

    And I successfully completed the square.

    Try with x = 2 :

    8 - 2y + y^2 = 1

    y^2 - 2y + 8 = 1

    Erm ... we have to do some working there :

    y^2 - 2y + 1 = -6 (substract 7 from both sides)

    And then the magic operates :

    (y - 1)^2 = -6

    Thus, it seems that regardless of what value x takes, we always have (y - 1)^2 = c, where c is dependent on x.

    Basically, we know that c = 1 - (2x^2 - 1) (since we always take off the left-hand side just the needed amount to leave 1 on the left-hand side)

    Which simplifies to : c = 2 - 2x^2, and finally, c = 2(1 - x^2).

    Now let's put it all together, and detail the steps more mathematically (what was shown up there was brainwork) :

    2x^2 + y^2 = 2y + 1

    y^2 - 2y = 1 - 2x^2

    But we need a little 1 on the left-hand side to successfully complete the square :

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

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

    Finally :

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

    I think that is what the book wants
    Last edited by Bacterius; November 23rd 2009 at 08:22 PM.
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  5. #5
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    Great stuff, Thank you!
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