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Math Help - Nonlinear system, manually computing imaginary solutions

  1. #1
    Junior Member Greymalkin's Avatar
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    Nonlinear system, manually computing imaginary solutions

    x^2+y^2=4
    16x^2+9y^2=144

    The answer is not as important as how you arrived there, wolfram alpha can compute these. I would like to know how to solve manually.

    I keep getting the equation 25x^2-36x-108 from substitution(solved for formula 1, subbed into 2),
    elimination yields nothing because there are no real roots, the graph is a circle within an ellipse.
    Using the quadratic formula with the above equation gives me much different answers than wolfram.
    I get \frac{36\pm \sqrt{12096}}{50} Which does not make much sense.
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  2. #2
    Junior Member Greymalkin's Avatar
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    Re: Nonlinear system, manually computing imaginary solutions

    Another system I cannot solve manually is:
    5y^2-x^2=1
    xy=2

    Due to the same reasons, I think I am forcing myself into extraneous solutions somehow :S
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  3. #3
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    Re: Nonlinear system, manually computing imaginary solutions

    Hello, Greymalkin!

    \begin{array}{cccc}x^2+y^2&=&4 & [1] \\ 16x^2+9y^2 &=& 144 & [2] \end{array}

    \begin{array}{ccccccc}\text{Multiply -9}\times[1]: & \text{-}9x^2 - 9y^2 &=& \text{-}36 \\ \text{Add [2]:} & 16x^2 + 9y^2 &=& 144 \end{array}

    We have: . 7x^2 \:=\:108 \quad\Rightarrow\quad x^2 \:=\:\tfrac{108}{7} \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{6}{7}\sqrt{21}}

    Substitute into [1]: . \tfrac{108}{7} + y^2 \:=\:4 \quad\Rightarrow\quad y^2 \:=\:\text{-}\tfrac{80}{7}

    . . . . . . . . . . . . . . . y \:=\:\pm\sqrt{\text{-}\tfrac{80}{7}} \quad\Rightarrow\quad \boxed{y\:=\:\pm\tfrac{4}{7}i\sqrt{35}}
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    Re: Nonlinear system, manually computing imaginary solutions

    Hello agaibn, Greymalkin!

    \begin{array}{cc}5y^2-x^2\:=\:1 & [1] \\ xy\:=\:2 & [2] \end{array}

    From [2]: . y \,=\,\tfrac{2}{x}

    Substitute into [1]: . 5\left(\tfrac{2}{x}\right)^2 - x^2 \:=\:1  \quad\Rightarrow\quad \frac{20}{x^2} - x^2 \:=\:1

    . . 20 - x^4 \:=\:x^2 \quad\Rightarrow\quad x^4 + x^2 - 20 \:=\:0 \quad\Rightarrow\quad (x^2-4)(x^2+5) \:=\:0

    . . \begin{Bmatrix}x^2-4 \:=\:0 & \Rightarrow & x^2 \:=\:4 & \Rightarrow & x \:=\:\pm2 \\ x^2+5 \:=\:0 & \Rightarrow & x^2 \:=\:\text{-}5 & \Rightarrow & x \:=\:\pm i\sqrt{5} \end{Bmatrix}


    If x = \pm2, [2] becomes: . (\pm2)y \:=\:2 \quad\Rightarrow\quad y \:=\:\pm1

    If x = \pm i\sqrt{5, [2] becomes: . (\pm i\sqrt{5})}y \:=\:2 \quad\Rightarrow\quad y \:=\:\mp\tfrac{2i}{5}\sqrt{5}


    We have four solutions: . (\pm2,\,\pm1),\;\left(\pm i\sqrt{5},\:\mp\tfrac{2i}{5}\sqrt{5}\right)
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  5. #5
    Junior Member Greymalkin's Avatar
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    Re: Nonlinear system, manually computing imaginary solutions

    Thanks again Soroban!
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  6. #6
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Nonlinear system, manually computing imaginary solutions

    First system has a circle & ellipse that do not interset:


    Second system has two real and two complex roots:
    Attached Thumbnails Attached Thumbnails Nonlinear system, manually computing imaginary solutions-nonlinear-system.png   Nonlinear system, manually computing imaginary solutions-nonlinear-system.jpg  
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