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Math Help - Quadratic-Quadratic Systems Algebraically Question

  1. #1
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    Quadratic-Quadratic Systems Algebraically Question

    Solve Each System:

    x2 + y2 = 20
    y = x2

    This might be a dumb question question, but 20 isn't a square number, so how do i find the radius to graph a circle? Please help!
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  2. #2
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    Re: Quadratic-Quadratic Systems Algebraically Question

    Quote Originally Posted by Vince604 View Post
    Solve Each System:

    x2 + y2 = 20
    y = x2

    This might be a dumb question question, but 20 isn't a square number, so how do i find the radius to graph a circle? Please help!
    It has r= \sqrt{20}
    Last edited by Plato; December 21st 2012 at 01:01 PM.
    Thanks from Vince604
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  3. #3
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    Re: Quadratic-Quadratic Systems Algebraically Question

    No, it doesn't. It has radius sqrt{20}= sqrt{4(5)}= 2 sqrt{5} which is about 4.472.

    (Been hitting the eggnog a little early?)
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  4. #4
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    Re: Quadratic-Quadratic Systems Algebraically Question

    Hello, Vince604!

    Solve:
    . . x2 + y2 = 20 . [1]
    . . . . y = x2 . . . [2]

    Substitute [2] into [1]: .y + y2 = 20 . . y2 + y - 20 = 0

    . . . (y - 4)(y + 5) .= .0 . . y = 4, -5


    Substitute into [2]:

    . . 4 = x2 . . x = +2

    . .-5 = x2 . No real roots


    Therefore: .(x, y) .= .(+2, 4)
    Thanks from Vince604
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  5. #5
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    Re: Quadratic-Quadratic Systems Algebraically Question

    Quote Originally Posted by Soroban View Post
    Hello, Vince604!


    Substitute [2] into [1]: .y + y2 = 20 . . y2 + y - 20 = 0

    . . . (y - 4)(y + 5) .= .0 . . y = 4, -5


    Substitute into [2]:

    . . 4 = x2 . . x = +2

    . .-5 = x2 . No real roots


    Therefore: .(x, y) .= .(+2, 4)
    Thanks, it makes a lot of sense now.
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