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Question

If $$a = 2 + \sqrt {3}$$, then find the value of $$a - \dfrac {1}{a}$$.


Solution

We have $$a = 2 + \sqrt {3}$$
$$\therefore \dfrac {1}{a} = \dfrac {1}{2 + \sqrt {3}} = \dfrac {1}{2 + \sqrt {3}} \times \dfrac {2 - \sqrt {3}}{2 - \sqrt {3}} = \dfrac {2 - \sqrt {3}}{(2)^{2} - (\sqrt {3})^{2}}$$

$$= \dfrac {2 - \sqrt {3}}{4 - 3} = \dfrac {2 - \sqrt {3}}{1} = 2 - \sqrt {3}$$

$$\therefore a - \dfrac {1}{a} = 2 + \sqrt {3} - 2 + \sqrt {3} = 2\sqrt {3}$$.

Mathematics

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