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Trigonometry Question (1 Viewer)

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Start date Dec 13, 2012

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#1

How do you do this question? I dont even know where to start apart from factorising.

$$Show that$\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$

Carrotsticks

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#2

$\sqrt{8 - 4 \sqrt{3}} \\\\ = \sqrt{6 - 2 \times 2 \times \sqrt{3} + 2} \\\\ = \sqrt{(\sqrt{6})^2 - 2 \times \sqrt{12} + ( \sqrt{2} )^2} \\\\ = \sqrt{(\sqrt{6})^2 - 2 \times \sqrt{6} \times \sqrt{2} + ( \sqrt{2})^2} \\\\ = \sqrt{ ( \sqrt{6} - \sqrt{2})^2 } = \pm (\sqrt{6} - \sqrt{2}) \\\\ $But $ \sqrt{8 - 4 \sqrt{3}} > 0 $ so we take the positive case, which is $ \sqrt{6} - \sqrt{2}$

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