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Math Help - Impossible equation?

  1. #1
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    Impossible equation?

    I've been working on it for a few days, and I can't seem to find a possible answer. My maths teacher told me it was not impossible (and she's not the joking kind).

    x^2 + y^2 = 4
    y= -x+3

    So let me show you what I tried...

    x^2 + y^2 = 4
    x^2 + (-x+3)^2 = 4
    x^2 + (-x+3)(-x+3) = 4
    x^2 + (x^2-3x-3x+9) = 4
    2x^2 -6x + 5 = 0

    Then...

    \frac {-b \pm \sqrt {b^2-4ac}}{2a} so \frac {6 \pm \sqrt {36-40}}{4} so \frac {6 \pm \sqrt {-4}}{4}

    And \sqrt {-4} is obviously the problem here.

    Any ideas? Thanks.
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  2. #2
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    Look at the graph. Do the two intersect?
    Attached Thumbnails Attached Thumbnails Impossible equation?-cir_ln.gif  
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  3. #3
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    Quote Originally Posted by Plato View Post
    Look at the graph. Do the two intersect?
    Ohhh, thanks! Tracing it on a graph is always easier, now I get it =p Well, time to bash my teacher now!
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  4. #4
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    we are currently looking at uses of the discriminant in school. according to the text book there are no real roots but 2 complex roots when its < 0.

    can these roots be shown. haven't got to this part yet at school.
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  5. #5
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    so so
     x_{1,2} =\frac{6 \pm \sqrt{-4}}{4} = \frac {6 \pm 2i}{4}

     x_1 = \frac {3}{2} + \frac {i}{2}

    x_2 = \frac{3}{2} - \frac {i}{2}

    i = \sqrt{-1}
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  6. #6
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    Quote Originally Posted by sammy28 View Post
    we are currently looking at uses of the discriminant in school. according to the text book there are no real roots but 2 complex roots when its < 0.

    can these roots be shown. haven't got to this part yet at school.
    Yes, two answers are found when it's < 0. Here is an example:

    a=3
    b=5
    c=2

    \frac {-b \pm \sqrt {b^2-4ac}}{2a}

    \frac {-5 \pm \sqrt {5^2-4(3)(2)}}{2(3)}

    \frac {-5 \pm \sqrt {25-24}}{6}

    \frac {-5 \pm \sqrt {1}}{6}

    \frac {-5 \pm {1}}{6}

    So here are both of the answers:

    \frac {-5 - {1}}{6} and \frac {-5 +{1}}{6}

    Simplified, you get:

    -1 and \frac {-2}{3}
    Last edited by Sodapop; April 12th 2009 at 02:37 PM.
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