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Quadratic dillema

  1. Jun 1, 2004 #1
    We were recently reviewing quadratic equations in my algebra 1 class. As my teacher simplified equation after equation on the board, I began to get this nagging feeling there was something incorrect.
    I have pin pointed where I believe an error was made.

    At this point in solving a quadratic equation, (6 +/- 2 root24)/2, my teacher simply cancels out the 6 and the 2 at once. I disagree here. It is a rule that you cannot cancel each component of an equation where a term is seperated by a + or - sign (of course, one can cancel the two since it is being multiplied with the "root24"). Instead, it I believe one must factor out a 2 from the numerator, then cancel out the 2 in the denominator.
    my way:

    1.)
    6 +/- 2 root24 2(3 +/- root24)
    -------------- = ---------------- = 3 +/- root24
    2 2

    2.)
    The way my teacher does it:

    6 +/- 2 root24 6/2 +/- 2/2 root24 = 3 +/- root24
    --------------=
    2


    I realize that essentially, when you factor (as I did) , you are dividing each term, seperately by 2. However on the second example, one is dividing each term by the exact same integer.


    is my analysis correct or incorrect?
     
  2. jcsd
  3. Jun 1, 2004 #2

    Integral

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    It is very difficult to figure out your equations. This site has very nice equation capabilities. I suggest that you read this thread
     
  4. Jun 1, 2004 #3
    Factoring the 2 first or distributing the 1/2 first doesn't change things. Multiplication is distributive.

    cookiemonster
     
    Last edited: Jun 1, 2004
  5. Jun 1, 2004 #4

    ahrkron

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    I wonder if using [ code ][ /code ] will help:

    Code (Text):

    1.)
    6 +/- 2 root24        2(3 +/- root24)
    -------------- =    ---------------- = 3 +/- root24
          2                           2
       
    2.)                                              
    The way my teacher does it:

    6 +/- 2 root24     6/2 +/- 2/2 root24 = 3 +/- root24
    --------------=
           2
     
     
  6. Jun 1, 2004 #5
    How about

    [tex]\frac{6 \pm 4\sqrt{6}}{2} = \frac{2(3 \pm 2\sqrt{6})}{2} = 3 \pm 2\sqrt{6}[/tex]

    and

    [tex]\frac{6 \pm 4\sqrt{6}}{2} = \frac{6}{2} \pm \frac{4\sqrt{6}}{2} = 3 \pm 2\sqrt{6}[/tex]

    cookiemonster
     
  7. Jun 1, 2004 #6
    thank you for your clarification, cookie. I totally understand now.
     
  8. Jun 2, 2004 #7
    You know I’ve never seen how the quadratic equation is derived, or a proof for it. Would some one post (or link) one please?
     
  9. Jun 2, 2004 #8

    Zurtex

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    It is just a generalisation of completing the square method: http://mathworld.wolfram.com/QuadraticEquation.html
     
  10. Jun 2, 2004 #9
    I feel silly for asking now…
     
  11. Jun 2, 2004 #10

    arildno

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    Really, why?
    It's not your fault that too many math teachers say "that's just the way it is" (usually to cover up their own ignorance/lack of understanding.)
     
  12. Jun 4, 2004 #11
    BRAVO!!!, right on the money there. lol
     
  13. Jun 4, 2004 #12
    My favorite: "The proof is left as an exercise to the reader." :)
     
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