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Thread: 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!!!

    $\displaystyle x^2 + y^2 = 9$
    $\displaystyle y=2-x^2$
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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!!!

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

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

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

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

    then for the other one,
    $\displaystyle -2.193=2-x^2$
    but $\displaystyle x^2=-4.193$ 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 $\displaystyle 0=y^2-y-7$

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

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

    then for the other one,
    $\displaystyle -2.193=2-x^2$
    but $\displaystyle x^2=-4.193$ 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
    $\displaystyle 3.193=2-x^2$
    $\displaystyle x^2=-1.193$=> no solution
    $\displaystyle -2.193=2-x^2$
    but $\displaystyle x^2=4.193$ => 2 solutions : sqrt(4.193) and -sqrt(4.193)
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