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Completing the squarehelp

  1. Nov 11, 2012 #1
    Hi, how do you complete this square x2+y2=2x

    To get this result (x-1)2+y2=1
     
  2. jcsd
  3. Nov 11, 2012 #2

    LCKurtz

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    Write it as ##(x^2 - 2x\quad\quad)+y^2 = 0## Then figure out what number you can add to both sides to fill in the blank and make the quantity in parentheses a perfect square.
     
  4. Nov 11, 2012 #3

    symbolipoint

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    Using number properties to have each variable on either one side or the other, you obtain:

    x2-2x=y2

    The expression on the left side is not a square; you WANT a square. The expression can be factored,

    x(x-2) = y2.

    Notice you can represent the left hand side as a rectangular area with a picture (if you wanted). One side is x long and the other side is x-2 long. Nevermind the negative or subtraction; but if you could split the "-2" in half and reposition one of those halves along the other part of the length, x, (this really needs a picture for you to see), then you would see a missing square piece on a corner. The area of this missing square piece is (2/2) by (2/2), or 1 by 1 square units. This is 1 square unit, to "complete the square".

    Continuing then, add 1 to both sides of the equation:

    x2-2x+1=y2+1

    As I said, a picture will help to show this completion of the square. Can you finish the problem from here?
    EDIT: I posted this too quickly and I believe I made a sign mistake.
     
  5. Nov 11, 2012 #4
    Ahh Yess If you add one one both sides it works :)
    thank you!!
     
  6. Nov 11, 2012 #5

    symbolipoint

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    That's it. When you find the quantity, be sure to add it to both sides of the equation.
    EDIT: Good, you found it already.
     
  7. Nov 11, 2012 #6
    By the way, i always feel like i get competent understandable explanations in this forum. Are some of you teachers? Or just very devoted in helping others understand?
     
  8. Nov 12, 2012 #7
    Although this already has been answered, I would just like to give something I believe is missing:
    As said: x^2-2x = -y^2.
    As said also, x^2-2x = x(x-2), now were looking for a perfect square, you can do any number, but easy ones will be (x-2)^2. Now we calculate (x-2)^2 = x^2-2x+4. So we need to add 4 to both sides, that is x^2-2x+4 = 4-y^2. Taking the square root (this is what we planned everything for, we made sure the left side will be a nice root):
    (x-2)^2 = (2+y)(2-y), taking the square root yields:
    x-2 = sqrt((2+y)(2-y)
    so x = 2+ sqrt((2+y)(2-y)

    Bonaparte
     
  9. Nov 12, 2012 #8

    LCKurtz

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    You have some of everything: Other students, High School teachers, graduate students, and both active and retired university professors as well as working professionals.
     
  10. Nov 12, 2012 #9

    Mentallic

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    [tex](x-2)^2=x^2-4x+4[/tex]

    These steps aren't needed because the problem was to complete the square, not to solve for x.
     
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