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Thread: 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 $\displaystyle 2xy$ 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 $\displaystyle 2xy$ 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?

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

    $\displaystyle 2x^2 + y^2 = 2y + 1$

    $\displaystyle 2x^2 - 2y + y^2 = 1$

    I know what follows is not very mathematic but perhaps it could be a good idea to try with some numbers substituting $\displaystyle x$ and $\displaystyle y$ 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 $\displaystyle x = 1$, therefore :

    $\displaystyle 2 - 2y + y^2 = 1$

    $\displaystyle y^2 - 2y + 2 = 1$

    $\displaystyle y^2 - 2y + 1 = 0$

    Woah ! A square identity !

    $\displaystyle (y - 1)^2 = 0$

    And I successfully completed the square.

    Try with $\displaystyle x = 2$ :

    $\displaystyle 8 - 2y + y^2 = 1$

    $\displaystyle y^2 - 2y + 8 = 1$

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

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

    And then the magic operates :

    $\displaystyle (y - 1)^2 = -6$

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

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

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

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

    $\displaystyle 2x^2 + y^2 = 2y + 1$

    $\displaystyle y^2 - 2y = 1 - 2x^2$

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

    $\displaystyle y^2 - 2y + 1 = 1 - 2x^2 + 1$

    $\displaystyle y^2 - 2y + 1 = 2 - 2x^2$

    Finally :

    $\displaystyle (y - 1)^2 = 2 - 2x^2$

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