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Math Help - Quadratic Equation

  1. #1
    ADY
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    Quadratic Equation

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

    With:

    a = 1, b = (-4) & c = (-3)

    What do we do when we need to square root the -4 ?

    using x=\frac{-(-4)\pm\sqrt{-4}}{2}
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  2. #2
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    Quote Originally Posted by ADY View Post
    using x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    With:

    a = 1, b = (-4) & c = (-3)

    What do we do when we need to square root the -4 ?

    using x=\frac{-(-4)\pm\sqrt{-4}}{2}
    b^2 - 4ac

    (-4)^2 - 4(1)(-3) = 16 + 12 = 28 , not -4
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  3. #3
    ADY
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    x=\frac{-(-4)\pm\{{5.921}}{2}

    ?
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  4. #4
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    Quote Originally Posted by ADY View Post
    x=\frac{-(-4)\pm\{{5.921}}{2}

    ?
    \frac{4 \pm \sqrt{4 \cdot 7}}{2}

    \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}
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  5. #5
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    First of all, if you evaluate x=b^2-4ac, with the numbers you provided, you end up with \sqrt{16-(-12)} = \sqrt{28} . So no problem here, i guess.


    In the other case of a square root of a negative number, you will have to work with Complex numbers. Let us suppose you have to solve x=\frac{-(-4)\pm\sqrt{-4}}{2}. The next step is to write \sqrt{-4} as \sqrt{4i^2}. If solve this root, you get 2i. Now we have x=\frac{-(-4)\pm2i}{2}
    . Let's simplify all that and we get x=2\pm i

    Greets,
    Leslon
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