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Completing the square

  1. Aug 24, 2011 #1
    Problem: -x^2 + 4x

    - ( x^2 - 4x + (4/2)^2 ) - 4
    = - (x - 2) - 4

    Vertex is at (2, -4)

    but the vertex is obviously at (2,4)...
    what did I do wrong.
     
  2. jcsd
  3. Aug 24, 2011 #2

    Mark44

    Staff: Mentor

    The error is here. You have actually added -4 inside the parentheses to complete the square, so to keep the expression equal, you need to add + 4.
     
  4. Aug 24, 2011 #3
    how did I add -4... then the expression would be x^2 - 4x - 4 which is not a perfect square...? Does it have something to do with x^2 being negative?
     
  5. Aug 24, 2011 #4

    Mark44

    Staff: Mentor

    Let's go through the steps.

    -x2 + 4x
    = -(x2 - 4x)
    = -(x2 - 4x + 4) + 4
    I put in +4 inside the parentheses, but due to the minus sign out front, I have actually added -4, so to balance, I have to add + 4.

    = -(x - 2)2 + 4
     
  6. Aug 25, 2011 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Another way of looking at it is to do both addition and subtraction inside the parentheses:
    [tex]-(x^2- 4x)= -(x^2- 4x+ 4- 4)= -(x^2- 4x+ 4)-(-4)= -(x- 2)^2+ 4[/tex]
     
  7. Aug 25, 2011 #6

    PeterO

    User Avatar
    Homework Helper

    or if you like

    -x2 + 4x
    = -[x2 - 4x]
    = -[x2 - 4x + 4 - 4]
    = -[(x - 2)2 - 4]
    = -(x - 2)2 + 4
     
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