Question:
If $x=(7+4 \sqrt{3})$ then $\left(x+\frac{1}{x}\right)=?$
(a) $8 \sqrt{3}$
(b) 14
(c) 49
(d) 48
Solution:
Given: $x=(7+4 \sqrt{3})$
$\frac{1}{x}=\frac{1}{7+4 \sqrt{3}}=\frac{1}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}=\frac{7-4 \sqrt{3}}{49-48}=7-4 \sqrt{3}$
$\left(x+\frac{1}{x}\right)=7+4 \sqrt{3}+(7-4 \sqrt{3})=14$
Hence, the correct answer is option (b).