If \( x=7+4 \sqrt{3} \) then find the value of \( x+\frac{1}{x} \)

To find the value of \( x + \frac{1}{x} \) where \( x = 7 + 4\sqrt{3} \), we first calculate \( \frac{1}{x} \). Start by rationalizing the denominator: \[ x = 7 + 4\sqrt{3} \] \[ \frac{1}{x} = \frac{1}{7 + 4\sqrt{3}} \cdot \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \] Now, calculate the denominator using the difference of squares: \[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \] Thus, we have: \[ \frac{1}{x} = 7 - 4\sqrt{3} \] Now we can find \( x + \frac{1}{x} \): \[ x + \frac{1}{x} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14 \] Therefore, the value of \( x + \frac{1}{x} \) is: \[ \boxed{14} \]