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Math Help - Solve 3 quadratics simultaneously given (x,y)

  1. #1
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    Solve 3 quadratics simultaneously given (x,y)

    I am trying to solve a system of quadratic equations.

    Given:
    y1 = a*x1^2 + b*x1 + c
    y2 = a*x2^2 + b*x2 + c
    y3 = a*x3^2 + b*x3 + c

    (x1, y1), (x2, y2), (x3, y3)

    I need to solve for a,b,c.

    Trying to use Gaussian Elimination does not seem to give any success.

    Does anyone have any idea where to find a solution to this? I have checked out the web on solving for the three-point quadratic, but nothing shows the algorithm.
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  2. #2
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    You can solve this with a computer easily by using Cramer's rule as the algorithm.

    Note: As line as (x1,y1),(x2,y2),(x3,y3) are not on the same line the determinant is non-zero.
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  3. #3
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    Hello, Hydra!

    I am trying to solve a system of quadratic equations.

    Given: . \begin{array}{c}<br />
y_1 \:= \:ax_1^2 + bx_1 + c \\<br />
y_2 \:= \:ax_2^2 + bx_2 + c \\<br />
y_3 \:= \:ax_3^2 + bx_3 + c\end{array}

    . . (x_1,\,y_1),\;(x_2,\,y_2),\;(x_3,\,y_3)

    I need to solve for a,\,b,\,c.

    First of all, this system of equations is not quadratic.

    We have a linear system: . \begin{vmatrix}x_1^2 & x_1 & 1 & | & y_1 \\ x_2^2 & x_2 & 1 & | & y_2 \\ x_3^2 & x_3 & 1 & | & y_3\end{vmatrix}


    As ThePerfectHacker suggested, Cramer's Rule would be the simplest approach.

    Curious . . . Do you REALLY want a general formula for a 3-by-3 system?
    . . It is truly ugly . . .

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post


    Curious . . . Do you REALLY want a general formula for a 3-by-3 system?
    . . It is truly ugly . . .

    I know two ways, I think, to find the Determinant of a 3-by-3 matrix. The general co-factor expansion method, and this other one that I don't even know the name for. Were you referring to something else?
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Were you referring to something else?
    He was talking about the solution for x,y,z to the above system. Any way you do it is will lead to a long equation answer.

    The only simple think to compute is the determinant which is the Virtumonde Determinant.
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  6. #6
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    Solving 3 Quadratics simultaneously

    After reading the various replies to this query, I do not think I stated my intentions clearly. I am trying to solve for a parabola that passes through three points.

    Given 3 points (x1, y1), (x2, y2), (x3, y3), where x1=x, x2=x+1, x3=x+2

    I want to solve the three equations simultaneously to get one generic equation for those three points. So obviously the three equations to solve simultaneously for would be

    y1 = a*x1^2 + b*x1 + c = a*x^2 + bx + c
    y2 = a*x2^2 + b*x2 + c = a*(x+1)^2 + bx + b + c = ax^2 + 2ax + a + bx + b + c
    y3 = a*x3^2 + b*x3 + c = a*(x+2)^2 + bx + 2b + c = ax^2 + 4ax + 4a + bx + 2b + c

    Sorry for any confusion caused.
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  7. #7
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    y1 = a*x1^2 + b*x1 + c = a*x^2 + bx + c
    y2 = a*x2^2 + b*x2 + c = a*(x+1)^2 + bx + b + c = ax^2 + 2ax + a + bx + b + c
    y3 = a*x3^2 + b*x3 + c = a*(x+2)^2 + bx + 2b + c = ax^2 + 4ax + 4a + bx + 2b + c


    These are 3 very friendly equations, because they all have + c on the end. By doing 2nd - 1st and 3rd - 2nd you get rid of the c at a stroke and have two eqns with two unknowns.
    These 2 equations both have + b on the end so you can take one away from the other and find "a".
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hydra3847 View Post
    After reading the various replies to this query, I do not think I stated my intentions clearly. I am trying to solve for a parabola that passes through three points.

    Given 3 points (x1, y1), (x2, y2), (x3, y3), where x1=x, x2=x+1, x3=x+2

    I want to solve the three equations simultaneously to get one generic equation for those three points. So obviously the three equations to solve simultaneously for would be

    y1 = a*x1^2 + b*x1 + c = a*x^2 + bx + c
    y2 = a*x2^2 + b*x2 + c = a*(x+1)^2 + bx + b + c = ax^2 + 2ax + a + bx + b + c
    y3 = a*x3^2 + b*x3 + c = a*(x+2)^2 + bx + 2b + c = ax^2 + 4ax + 4a + bx + 2b + c

    Sorry for any confusion caused.
    You need to solve the system:
    y_1 = ax^2 + bx + c
    y_2 = a(x + 1)^2 + b(x + 1) + c
    y_3 = a(x + 2)^2 + b(x + 2) + c

    Given the points (x, y_1), (x + 1, y_2), (x + 2, y_3)

    The solution method is the same as you've been told. Set up your equations in matrix form:
    \left ( \begin{matrix} x^2 & x & 1 \\ (x + 1)^2 & (x + 1) & 1 \\ (x + 2)^2 & (x + 2) & 1 \end{matrix} \right ) \cdot \left ( \begin{matrix} a \\ b\\ c \end{matrix} \right ) = \left ( \begin {matrix} y_1 \\ y_2 \\ y_3 \end{matrix} \right )

    Cramer's rule would probably be the simplest method to do this.

    -Dan
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