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Thread: Quadratic Equation

  1. #1
    ADY
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    Quadratic Equation

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

    With:

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

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

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

    With:

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

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

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

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

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

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

    $\displaystyle \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$
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  5. #5
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    First of all, if you evaluate $\displaystyle x=b^2-4ac$, with the numbers you provided, you end up with $\displaystyle \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 $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$. The next step is to write $\displaystyle \sqrt{-4}$ as $\displaystyle \sqrt{4i^2}$. If solve this root, you get $\displaystyle 2i$. Now we have $\displaystyle x=\frac{-(-4)\pm2i}{2}$
    . Let's simplify all that and we get $\displaystyle x=2\pm i$

    Greets,
    Leslon
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