Question

If $x=\left(7+4\sqrt{3}\right),\mathrm{then}\left(x+\frac{1}{x}\right)$=?
(a) $8\sqrt{3}$
(b) 14
(c) 49
(d) 48

Answer 1

(b) 14
x = $7+4\sqrt{3}$
Thus, we have:
$$\begin{gathered} \frac{1}{x}=\frac{1}{7+4\sqrt{3}} \\ =\frac{1}{7+4\sqrt{3}}\times \frac{1}{7+4\sqrt{3}} \\ =\frac{7-4\sqrt{3}}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \\ =7-4\sqrt{3} \end{gathered}$$
∴ $\left(x+\frac{1}{x}\right)=\left(7+4\sqrt{3}\right)+\left(7-4\sqrt{3}\right)=7+7=14$