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Thread: Complex Conjugate Roots and the Quadratic Formula

  1. #1
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    Complex Conjugate Roots and the Quadratic Formula

    Hello, there.

    I have the quadratic equation $\displaystyle x^2+2x+7=0$

    Using $\displaystyle \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$, I get $\displaystyle \begin{array}{*{20}c} {x = \frac{{ - 2 \pm \sqrt {-24} }}{{2}}} \\ \end{array}$.

    Now, I know the answer to this question is two complex conjugate roots, $\displaystyle x = - 1 \pm i \sqrt{6}$, my question is how? The square root of 24 isn't 6. I'm confused.

    I guess I'm just asking for the steps to get there.

    Thank you in advance.
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  2. #2
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    Re: Complex Conjugate Roots and the Quadratic Formula

    sqrt(-1)=i
    and
    sqrt(24)=2sqrt(6)

    so sqrt(-24)=i2sqrt(6)

    now you try.
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  3. #3
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    Re: Complex Conjugate Roots and the Quadratic Formula

    I'm still not sure I see where the $\displaystyle \sqrt{6}$ comes from..

    This is on the tip of my tongue from when I took Algebra, but I'm not sure I remember how to factor radicals.

    Edit: Nevermind I get it. $\displaystyle \sqrt{ab} = \sqrt{a} \sqrt{b}$ so $\displaystyle \sqrt{24} = \sqrt{4} \sqrt{6} = 2 \sqrt{6}$
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