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Why (ax^2 + bx + c) = a(x-α)(x-β)?

  1. Jun 14, 2013 #1
    Say, 3x2 - 2x - 1 = 0
    Then: x1 = -1/3, x2 = 1
    Therefore: k * (x + 1/3) (x - 1) = 3x2 - 2x - 1
    Why is k = 3?

    Why, [itex]Ax^{2} + Bx + C = A (x-\alpha) (x-\beta)[/itex]?

    Thanks for help.
     
    Last edited by a moderator: Jun 14, 2013
  2. jcsd
  3. Jun 14, 2013 #2

    pwsnafu

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    Well the polynomial ##3x^2 - 2x -1 = (3x+1)(x-1) = 3 (x+1/3)(x-1)##.

    Also solving the equation ##3x^2-2x-1 = 0## is equivalent to ##x^2 - \frac{2}{3}x - \frac{1}{3} = 0##.

    And even further notice that when you take ##A(x-\alpha)(x-\beta)## and expand the coefficient of ##x^2## is equal to A.
     
  4. Jun 14, 2013 #3
    To answer the question:

    Well, you have in general
    [tex]x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

    and

    [tex]x_1=\frac{-b+ \sqrt{b^2-4ac}}{2a}[/tex]

    [tex]x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}[/tex]


    We also have

    [tex]a(x_1 + x_2)=a(\frac{-b+ \sqrt{b^2-4ac}}{2a} + \frac{-b- \sqrt{b^2-4ac}}{2a}) = a(\frac{-2b}{2a}) = a(\frac{-b}{a}) = -b[/tex]

    and if [itex]a(x_1+x_2) = -b[/itex],

    [tex]a(-x_1-x_2) = b[/tex]

    We also have

    [tex](x_1)\cdot(x_2)=(\frac{-b+ \sqrt{b^2-4ac}}{2a})\cdot(\frac{-b- \sqrt{b^2-4ac}}{2a})[/tex]

    which after a little work gets you [itex](x_1) \cdot (x_2)= \frac{c}{a}[/itex]. Equivalently we can have [tex]a(x_1)\cdot(x_2) = c[/tex]

    -----------------

    If we substitute our b and c into [itex]ax^2+bx+c[/itex] we get

    [tex]ax^2+bx+c = ax^2+ a(-x_1-x_2)x + a(x_1)\cdot (x_2)[/tex]
    [tex]= a(x^2 -x\cdot x_1 -x\cdot x_2 + (x_1)\cdot(x_2))[/tex]
    [tex]= a(x-x_2)(x-x_1)[/tex]

    In the above step you factorize, it should be clear how you get there if you try to expand[itex](x-x_2)(x-x_1)[/itex].

    So that's why [itex]ax^2+bx+c = a(x-x_2)(x-x_1)[/itex]
     
    Last edited: Jun 14, 2013
  5. Jun 14, 2013 #4
    Thank you for the answers.
     
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