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Solving quadratic equation

  1. Nov 16, 2013 #1
    I know for ##ax^2+bx+c=0##,
    [tex]x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}[/tex]

    Using ##x^2-3x-4=0##, we know it is equal to ##(x+1)(x-4)=0##. So ##x=-1## or ##x=4##.

    but using the formula:

    [tex]x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}=\frac{3^+_-\sqrt{9+4}}{2}=\frac{3^+_-\sqrt{13}}{2}[/tex]

    I cannot get -1 and 4!!! What happened?
     
  2. jcsd
  3. Nov 16, 2013 #2

    Mentallic

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    How are you calculating the discriminant [itex]\Delta = b^2-4ac[/itex]? Because you shouldn't be getting 13.
    c = -4, not -1 which it seems that you've confused it for.
     
  4. Nov 16, 2013 #3
    b=-3, a=1 and c=-4, so ##b^2-4ac##= 9+16=25.

    Yes, I am missing the moon.

    Thanks
     
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