Square roots and imaginary numbers

**Mush** · January 25th 2009, 05:02 PM

Originally Posted by tmac11522

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

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

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

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

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

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

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

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

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

$3\sqrt{2}-6 - i\bigg(4\sqrt{6}\bigg)$

**Soroban** · January 25th 2009, 08:17 PM

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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