# Math Help - A strange property of the Quadratic ecuation

1. ## A strange property of the Quadratic ecuation

I´ve found a strange and difficult to prove formula that derived from two quadratic equations.

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

If they are equivalent, It is known that their coefficients are proportional: $\frac{A}{D}=\frac{B}{E}=\frac{C}{F}$.
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.

2. ## 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.

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

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

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

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

solving for $r_1$ and $r_2$repectively:

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

dividing (I) and (II):

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

subtracting (I) and (II):

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

because of another relation between roots and coefficients
we have:

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

dividing (III) and (IV):

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

subtracting (III) and (IV):

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

From (5) and (3) :

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

From (4) and (6) :

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

Finally, from (7) and (8):

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

If there are corrections or misunderstandings, feel free to reply, they are wellcome.