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Quadratic Equations in Vertex Form

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the roots and the vertex for 3x^2 + 5x - 2


    2. Relevant equations
    ax^2 + bx + c
    a(x - h) + k

    3. The attempt at a solution
    OK, this is what I attempted.

    3x^2 + 5x - 2 = 0
    3x^2 + 6x - x - 2 = 0
    3x(x+2) - 1(x+2) = 0
    x = 1/3 and x = -2

    OK, so I found the roots, however, I am having difficulty understanding finding the vertex of THIS problem. The book says the vertex should be (-5/6, -4 1/12), so here's what I did.

    3x^2 + 5x - 2 = 0
    x^2 + 5/3x - 2/3 = 0
    x^2 + 5/3x - (25/36) - (2/3 - 25/35) = 0
    (x + 5/6)^2 - 49/36

    therefore the vertex is -5/6 and -49/36; obviously I didn't get -4 1/12.

    Can somebody help me and explain to me what's going on?
     
  2. jcsd
  3. Jul 21, 2009 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    Yup, it's pretty simple. You went from

    x2+5/3x-25/36 (emphasis mine EDIT: Actually, a bold minus sign looks no different than a regular one...)

    to

    (x+5/6)2

    if yo factor this out, you'll see you get a +25/36. So your constant term is off because you put the negative 25/36 into the complete the square portion instead of the positive part
     
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