jmc

We rearrange our equation: $x^2-7x+1=0$. Then, we use the quadratic equation to solve for $x$: $$x=\frac{7\pm \sqrt{(-7{)}^2-(4)(1)(1)}}{2}=\frac{7\pm 3\sqrt{5}}{2}.$$ The two possible values of $x$ are reciprocals of each other. Here's why: $$\begin{array}{rl}\frac{1}{(7+3\sqrt{5})/2}& =\frac{2}{7+3\sqrt{5}}\\ & =\frac{2}{7+3\sqrt{5}}\cdot \frac{7-3\sqrt{5}}{7-3\sqrt{5}}\\ & =\frac{2(7-3\sqrt{5})}{7^2-(3\sqrt{5}{)}^2}=\frac{2(7-3\sqrt{5})}{4}=\frac{7-3\sqrt{5}}{2}.\end{array}$$

Thus, our answer is $$\frac{7+3\sqrt{5}}{2}+\frac{7-3\sqrt{5}}{2}=\boxed{7}.$$

- OR -

We want the sum of $x$ and its reciprocal. This is $$x+\frac{1}{x}=\frac{x^2+1}{x}.$$ We are given that $x^2+1=7x$. Therefore, our answer is $$\frac{x^2+1}{x}=\frac{7x}{x}=\boxed{7}.$$