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Thread: A strange property of the Quadratic ecuation

  1. #1
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    A strange property of the Quadratic ecuation

    Ive found a strange and difficult to prove formula that derived from two quadratic equations.

    "If two quadratic equations:$\displaystyle Ax^2+Bx+C=0$ and $\displaystyle Dx^2+Ex+F=0$ share one solution.
    then: $\displaystyle (CD-FA)^2=(BF-EC)(AE-BD)$.......(1) is always true."

    If they are equivalent, It is known that their coefficients are proportional:$\displaystyle \frac{A}{D}=\frac{B}{E}=\frac{C}{F}$.
    since the two quadratic equations above share just one solution, I suppose they are partially equivalent.

    I've been trying to prove the formula in (1). I apologize for the lack of content about the problem.
    Any suggestion will be appreciated.
    Last edited by rochosh; May 13th 2012 at 04:28 PM.
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  2. #2
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    Re: A strange property of the Quadratic ecuation

    I came to a proof yesterday. I post my conclusion below(despite not being a difficult problem to be highlighted) so perhaps someone interested in the problem can see it.

    $\displaystyle Ax^2+Bx+C=0$....(1) $\displaystyle Dx^2+Ex+F=0$....(2)

    If the two equations above have just one solution in common,
    then. Let be: $\displaystyle {r_1 , r}$ the roots of (1), and $\displaystyle {r_2 , r}$ the roots of (2).

    because of the relation between roots and coefficients of a quadratic equation
    we have:

    $\displaystyle r_1 + r =-B/A $ and $\displaystyle r_2 + r =-E/D$

    solving for $\displaystyle r_1$ and $\displaystyle r_2 $repectively:

    $\displaystyle r_1 =\frac{-B-Ar}{A} $....(I) and $\displaystyle r_2 =\frac{-E-Dr}{D}$.....(II)

    dividing (I) and (II):

    $\displaystyle \frac{r_1}{ r_2}=\frac{D(B+Ar)}{A(E+Dr)}$....(3)

    subtracting (I) and (II):

    $\displaystyle r_1-r_2=\frac{AE-DB}{AD}$....(4)


    because of another relation between roots and coefficients
    we have:

    $\displaystyle r_1 =C/Ar $.....(III) and $\displaystyle r_2 =F/Dr$....(IV)

    dividing (III) and (IV):

    $\displaystyle \frac{r_1}{r_2}= \frac{CD}{FA}$....(5)

    subtracting (III) and (IV):

    $\displaystyle r_1-r_2=\frac{CD-FA}{ADr}$....(6)

    From (5) and (3) :

    $\displaystyle r=\frac{FB-CE}{CD-FA}$.....(7)

    From (4) and (6) :

    $\displaystyle r=\frac{CD-FA}{AE-DB}$......(8)


    Finally, from (7) and (8):

    $\displaystyle (CD-FA)^2=(AE-DB)(FB-CE)$

    If there are corrections or misunderstandings, feel free to reply, they are wellcome.
    Last edited by rochosh; May 17th 2012 at 06:21 PM.
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