Q. If $ x = 7-4\sqrt{3}$, then the value of $ \left ( x+\frac{1}{x} \right )$ is:
Answer: $ 14$
Notes: $ x = 7 – 4\sqrt{3}$ $ \therefore \frac{1}{x} = \frac{1}{7 – 4\sqrt{3}}$ $ = \frac{\left ( 7+4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}$ $ = \frac{\left ( 7+4\sqrt{3} \right )}{49-48} = \left ( 7+4\sqrt{3} \right )$ $ \therefore x + \frac{1}{x}$ $ = 7-4\sqrt{3} + 7+4\sqrt{3} = 14$ Hence option [C] is the right answer.