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Math Help - quadratic equation

  1. #1
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    quadratic equation

    three points : A=3,2 B=1,2 C=0,k with k being a constant

    AC=5BC

    FIND THE VALUES OF K

    my teacher said to use a quadratic please help
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  2. #2
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    Re: quadratic equation

    Quote Originally Posted by kitobeirens View Post
    three points : A=3,2 B=1,2 C=0,k with k being a constant
    AC=5BC
    FIND THE VALUES OF K
    my teacher said to use a quadratic please help
    Solve (0-3)^2+(k-2)^2=25[(0-1)^2+(k-2)^2]
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  3. #3
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    Re: quadratic equation

    Quote Originally Posted by Plato View Post
    ...... =25[(0-1)^2+(k-2)^2]
    The 25 should be 5.
    You'll get 2 valid solutions.
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  4. #4
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    Re: quadratic equation

    Quote Originally Posted by Wilmer View Post
    The 25 should be 5.
    You'll get 2 valid solutions.
    Think again: {\left[{5\sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {k - 2} \right)}^2}} } \right]^2} = 25\left[ {{{\left( {0 - 1} \right)}^2} + {{\left( {k - 2} \right)}^2}} \right]
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  5. #5
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    Re: quadratic equation

    My bad. Jumped on "5" and got nice integer solutions: k=1 or k=3;
    both "looked" good on my quickly sketched diagram!

    BUT: if the correct "25" is used, then : k = [6 +- sqrt(-6)] / 3;
    so no real solution.
    Perhaps the OP made a typo....
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  6. #6
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    Re: quadratic equation

    Quote Originally Posted by Wilmer View Post
    My bad. Jumped on "5" and got nice integer solutions: k=1 or k=3;
    both "looked" good on my quickly sketched diagram
    BUT: if the correct "25" is used, then : k = [6 +- sqrt(-6)] / 3;
    so no real solution.
    Perhaps the OP made a typo....
    I suspect that whoever wrote the question made the mistake.
    The 5 does give a very nice solution. So it should have been AC=\sqrt5 BC.
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