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A ridiculous question

  1. Dec 23, 2004 #1

    Clausius2

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    I think all people here are going to laugh at me, but I'm going to be strong enough to formulate this question which should be made when I was a schoolboy. :blushing: :blushing:

    What is the demonstration of the 2nd order equation solution?

    [tex] ax^2+bx+c=0[/tex]

    then all teachers said:

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

    Where does it come from? How could I demonstrate that formula? We memorized it so well that we haven't ask ourselves how is proved.
     
  2. jcsd
  3. Dec 23, 2004 #2

    arildno

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    It's all about completion of squares:
    1) Multiply your equation with "4a":
    [tex](2ax)^{2}+2*(2ax)b+4ac=0[/tex]
    2) I'll leave to you the next steps..
    Hint: adding 0 in the form of [tex]b^{2}-b^{2}[/tex] might work nicely..:wink:
     
  4. Dec 23, 2004 #3

    Clausius2

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    Well, now I realise I will never win the Fields Medal I think I'm going to delete this thread. :rolleyes: :yuck: :cry:

    Or doing it better, I will leave this here to show the people what kind of guy is an engineer guru here. :rofl: :rofl:

    EDIT: MERRY CHRISTMAS!!
     
    Last edited: Dec 23, 2004
  5. Dec 23, 2004 #4

    arildno

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    BUENO NATIVITAD?
    Does that mean "Merry Christmas" in Spanish?
    That's what I intended, at least..:smile:
     
  6. Dec 23, 2004 #5

    Clausius2

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    My question was horrible, but your is spanish is worst. :biggrin:

    Merry Christmas!!=FELIZ NAVIDAD!! in spanish

    Sorry, but I haven't found any norwegian dictionary over here... :uhh:
     
  7. Dec 23, 2004 #6
    It can be also done like that:
    By dividing both sides by 'a' we get:
    [tex]x^2+\frac{b}{a}x+\frac{c}{a}=0 [/tex]
    Taking the constant term to the right side of the equation and then completing the square. This will give us the following equation:
    [tex]x^2+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2[/tex]
    This is equal to:
    [tex](x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2)}[/tex]
    By making x the subject of the formula we get:

    [tex] x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
     
    Last edited: Dec 23, 2004
  8. Dec 23, 2004 #7

    Clausius2

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    A bit more elegant, Deathknight. :wink:
     
  9. Dec 23, 2004 #8

    arildno

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    GOD JUL!!!!!!!!!!!!!!1
     
  10. Dec 23, 2004 #9
    find the vertex by the use of the derivative and apply the symmetric shape to find the zeros.
     
  11. Dec 23, 2004 #10
    This makes me kinda glad my high school teacher showed us derivations of everything.
     
  12. Dec 23, 2004 #11

    kreil

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    You can work backwards and show it as well:

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

    [tex](x+\frac{b+\sqrt{b^2-4ac}}{2a})(x+\frac{b-\sqrt{b^2-4ac}}{2a})=0[/tex]

    [tex]x^2+{\frac{bx-x\sqrt{b^2-4ac}}{2a}}+{\frac{bx+x\sqrt{b^2-4ac}}{2a}}+\frac{b^2-b^2+4ac}{4a^2}=0[/tex]

    [tex]x^2+\frac{2bx}{2a}+\frac{4ac}{4a^2}=0[/tex]

    [tex]x^2+\frac{bx}{a}+\frac{c}{a}=0[/tex]

    [tex]ax^2+bx+c=0[/tex]
     
  13. Dec 23, 2004 #12
    Yeah, I like that way of doing it a lot better than completing the squares.
     
  14. Dec 23, 2004 #13
    Can someone show the derivations for the solutions of cubic and quartic equations as well?
     
  15. Dec 24, 2004 #14
    But, um, Deathknight DID complete the square?
     
  16. Dec 24, 2004 #15

    dextercioby

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    Check this out.
    cubic

    depressed quartic

    full quartic

    Daniel.
     
  17. Dec 24, 2004 #16
    I leaning towards his version of it other than the way we're all taught to do it.
     
  18. Dec 24, 2004 #17

    Hurkyl

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    The problem is that it isn't useful in other situations where you want to complete the square... such as when doing this integral:

    [tex]
    \int \frac{1}{3x^2 + 6x - 7} \, dx
    [/tex]
     
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