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The Quadratic Equation Process, is this wrong?

  1. Feb 9, 2005 #1
    Hellooooo everybody. i have a very distinct feeling that there's something very wrong here...

    OK, what i'm trying to do is to get to the quadratic equation through the standard form of a polynomial equation in the second degree. :smile:

    watch this, and go to this website. it lists the process very briefly http://mathworld.wolfram.com/QuadraticEquation.html...in this picture http://mathworld.wolfram.com/qimg558.gif isn't it wrong to put the x+b/2a inside the parenthesis with the WHOLE thing raised to the second power?? i mean doesn't that mean that the WHOLE THING INCLUDING the 'x' gets squared?? that's wrong!! aha!! that's wrong!! i knew it!! i knew it!! :biggrin: :biggrin: Nah I'm just kidding there but i really do want to know if that's a typo or not. I think that when completing the square, you should only raise the "b" term in (the standard form) to the second power. like this:

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

    [tex](\frac{b}{2})^2[/tex]

    subtract the 'c' and bring it over to the other side, then complete the square:
    [tex]ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2[/tex]

    is this right, as opposed to the ones shown on that website? :uhh: :uhh:
     
    Last edited: Feb 9, 2005
  2. jcsd
  3. Feb 9, 2005 #2

    StatusX

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    The site is right, but I'm not sure what you're trying to do. What is your next step?
     
  4. Feb 9, 2005 #3

    dextercioby

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    I'm sure what's on that site is not wrong.How about u finish your calculations...???

    Daniel.
     
  5. Feb 9, 2005 #4
    Hey, how's it going? :smile: i'm gonna go do that, i'll get back at you all. :biggrin:
     
  6. Feb 9, 2005 #5
    okay guys im in a pot-hole. :confused: :confused:

    here's my work:

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

    [tex]ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2[/tex]

    [tex]ax^2+bx+(\frac{b^2}{4})=-c+(\frac{b^2}{4})[/tex]

    [tex]\left(x-\frac{b}{2a}\right)^2=\left(\frac{-c+\frac{b^2}{4}}{a}\right)[/tex]

    Then i took the square root of both sides and got this:

    (x-(b/2a))=\sqrt(-c+(b^2/4)/ \sqrt(a)

    then i did that thing where you can't have a square root on the bottom (forgot what its called), and to do that, i multiplied the top and bottom of the right side of the equation with the square root of a over the square root of a (1 in essence).


    then it wasn't looking like it was going in the right direction...help. :uhh: :uhh: :frown:
     
  7. Feb 9, 2005 #6

    StatusX

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    After you implicity divided by a between the last two (latex) steps, you no longer had a perfect square on the left. You need that last term to be b^2/4a^2. That's why they divided by a first. Once you correct that, your method becomes identical to theirs.
     
  8. Feb 9, 2005 #7

    dextercioby

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    Do you agree that the RHS of your last equation (written in tex) can be written
    [tex] \frac{b^{2}-4ac}{4a} [/tex]

    Now extract square root...

    Daniel.

    P.S.Yes,an "a" is missing in the numerator...
     
  9. Feb 9, 2005 #8
    i looked and looked and looked at it, but it didn't click.

    how on earth did they go from here:
    http://mathworld.wolfram.com/qimg557.gif

    to here:
    http://mathworld.wolfram.com/qimg558.gif

    what happened to the x in the left in the second step?? i mean, it was, (b/a)x :confused: :confused: :confused:
     
  10. Feb 9, 2005 #9

    StatusX

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    expand the squared term and you'll see
     
  11. Feb 9, 2005 #10
  12. Feb 9, 2005 #11
    hahah yeah, thanks man, i just saw that. :wink: :rolleyes:
     
  13. Feb 9, 2005 #12

    dextercioby

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    As an advice for future "events",ALWAYS TRUST MATHWORLD from WOLFRAM.COM.

    I haven't seen one single mistake on their site.

    Daniel.
     
  14. Feb 9, 2005 #13
    HOLY MOLY!!

    its like it all just revealed itself right before my eyes!! IT WORKS!! BY GOD ALMIGHTY IT WORKS!! thanks for ur help! :) :smile: :smile: :biggrin:

    thanks for the tip Daniel. i'll keep that in mind. mathworld rules. :cool: :cool: :wink:
     
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