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Need help Proving quadriatic equation

  1. Jun 19, 2004 #1
    I need help proving the quadriatic equation... this is all i got up to:

    ax(squared)+bx+(b/2)quantity squared= -c+(b/2)quantity squared

    :frown:

    sorry.. i kind of dont know how to use the other codes!
    :confused:
     
  2. jcsd
  3. Jun 19, 2004 #2

    AKG

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    Work your way backwards, then reverse the steps so you know the way forwards.

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

    Multiply both sides by 2a

    [tex]2ax = -b \pm \sqrt{b^2 - 4ac}[/tex]

    Add b to both sides

    [tex]2ax + b = \pm \sqrt{b^2 - 4ac}[/tex]

    Square both sides

    [tex]4a^2x^2 + 4abx + b^2 = b^2 - 4ac[/tex]

    Subtract [itex]b^2[/itex] from both sides

    [tex]4a^2x^2 + 4abx = - 4ac[/tex]

    Add 4ac to both sides

    [tex]4a^2x^2 + 4abx + 4ac = 0[/tex]

    Divide both sides by 4a

    [tex]ax^2 + bx + c = 0 \ \dots \ (a \neq 0)[/tex]

    So, to go forwards, do the opposite of those actions in reverse order:

    Multiply both sides by 4a
    Subtract 4ac from both sides
    Add [itex]b^2[/itex] to both sides
    Square root both sides
    Subtract b from both sides
    Divide both sides by 2a
     
  4. Jun 19, 2004 #3

    matt grime

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    Or better is to complete the square on the original eqaution.
     
  5. Jun 20, 2004 #4
    Exactly what I was going to say. Just complete the square with variables.
     
  6. Jun 20, 2004 #5

    HallsofIvy

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    Actually, the original post was trying to complete the square. Unfortunately, she was doing it wrong:

    After writing ax2+ bx= -c, divide both sides by a: x2+ (b/a)x= -c/a.

    NOW complete the square: the coefficient of x is (b/a) so we square half of that and add (b/2a)2 to both sides:
    x2+ (b/a)x+ (b/2a)2= (b/2a)2- c/a

    princesscharming26, do you understand WHY you add that square?

    It's because x2+ (b/a)x+ (b/2a)2= (x+ b/2a)2.

    Now you have (x+ b/2a)2= b2/4a- c/a= (b2- 4ac)/(2a).
     
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