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Math Help - Square roots and imaginary numbers

  1. #1
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    Square roots and imaginary numbers

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  2. #2
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    Quote Originally Posted by tmac11522 View Post
     \bigg(\sqrt{3} - \sqrt{-4}\bigg)\times\bigg(\sqrt{6}-\sqrt{-8}\bigg)

     \bigg(\sqrt{3} - i\sqrt{4}\bigg)\times\bigg(\sqrt{6}-i\sqrt{8}\bigg)

    \sqrt{3} \bigg(\sqrt{6}-i\sqrt{8}\bigg) - i\sqrt{4}\bigg(\sqrt{6}-i\sqrt{8}\bigg)

    \sqrt{3}\sqrt{6}-i\sqrt{3}\sqrt{8} - i\sqrt{4}\sqrt{6}+i^2\sqrt{4}\sqrt{8}

    \sqrt{3\times6}-i\sqrt{3\times8} - i\sqrt{4\times6}-\sqrt{4\times8}

    \sqrt{18}-i\sqrt{24} - i\sqrt{24}-\sqrt{36}

    \sqrt{18}-\sqrt{36} - i\bigg(\sqrt{24} + \sqrt{24}\bigg)

    3\sqrt{2}-2\sqrt{9} - i\bigg(2\sqrt{24}\bigg)

    3\sqrt{2}-6 - i\bigg(2\times 2\sqrt{6}\bigg)

    3\sqrt{2}-6 - i\bigg(4\sqrt{6}\bigg)
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  3. #3
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    Hello, tmac11522!

    Simplify: . \left(\sqrt{3} - \sqrt{-4}\right)\left(\sqrt{6} - \sqrt{-8}\right)

    We have: . \left(\sqrt{3} - 2i\right)\left(\sqrt{6}-2\sqrt{2}\,i\right)

    . . . = \;\left(\sqrt{3}\right)\left(\sqrt{6}\right) + <br />
\left(\sqrt{3}\right) \left(-2\sqrt{2}\ i\right) + \left(-2i\right)\left(\sqrt{6}\right) +  \left(-2i\right)\left(-2\sqrt{2}\,i\right)

    . . . = \;\sqrt{18} - 2\sqrt{6}\,i - 2\sqrt{6}\,i + 4\sqrt{2}\,i^2

    . . . = \;3\sqrt{2} - 4\sqrt{6}\,i - 4\sqrt{2}

    . . . = \;-\sqrt{2} - 4\sqrt{6}\,i

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