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Math Help - Solving a system graphically and numerically

  1. #1
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    Post Solving a system graphically and numerically

    Okay so I'm having trouble with a system involving a circle and a parabola. My book tells me to solve the system graphically and support my answer numerically. I don't know how to do this and really appreciate the help!!!

    x^2 + y^2 = 9
    y=2-x^2
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  2. #2
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    Quote Originally Posted by LoveBeachForever View Post
    Okay so I'm having trouble with a system involving a circle and a parabola. My book tells me to solve the system graphically and support my answer numerically. I don't know how to do this and really appreciate the help!!!

    x^2 + y^2 = 9
    y=2-x^2
    The system leads to
    x^2 = 9 - y^2
    y=2-(9 - y^2)=-7 + y^2

    Second equation gives y^2 - y - 7 = 0 which is a second degree equation
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  3. #3
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    Okay so if 0=y^2-y-7

    then I can use the quadratic and get y=\frac{1 \pm \sqrt 29}{2}

    which is y=3.193 or -2.193
    using that and plugging it back in i get:
    3.193=2-x^2
    x^2=1.193
    x=1.092 so the solution is at (3.193,1.092) correct?

    then for the other one,
    -2.193=2-x^2
    but x^2=-4.193 which leaves me with an imaginary number. So I'm assuming that it means that just the other one is the solution. Am I right?
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  4. #4
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    Quote Originally Posted by LoveBeachForever View Post
    Okay so if 0=y^2-y-7

    then I can use the quadratic and get y=\frac{1 \pm \sqrt 29}{2}

    which is y=3.193 or -2.193
    using that and plugging it back in i get:
    3.193=2-x^2
    x^2=1.193
    x=1.092 so the solution is at (3.193,1.092) correct?

    then for the other one,
    -2.193=2-x^2
    but x^2=-4.193 which leaves me with an imaginary number. So I'm assuming that it means that just the other one is the solution. Am I right?
    No
    3.193=2-x^2
    x^2=-1.193=> no solution
    -2.193=2-x^2
    but x^2=4.193 => 2 solutions : sqrt(4.193) and -sqrt(4.193)
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