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Math Help - 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 x^2+2x+7=0

    Using \begin{array}{*{20}c}    {x = \frac{{ - b \pm \sqrt {b^2  - 4ac} }}{{2a}}}  \\ \end{array}, I get \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, 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 \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. \sqrt{ab} = \sqrt{a} \sqrt{b} so \sqrt{24} = \sqrt{4} \sqrt{6} = 2 \sqrt{6}
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