There are exactly three real numbers $x$ for which $\left(x-\frac{5}{x}\right)$ is the reciprocal of $(x-4)$. What is the sum of these three real numbers?
Two numbers are reciprocals if their product is 1 . Therefore, we are looking for real numbers $x$ for which $\left(x-\frac{5}{x}\right)(x-4)=1$.

Multiplying through by $x$ gives $\left(x^{2}-5\right)(x-4)=x$. Expanding gives $x^{3}-4 x^{2}-5 x+20=x$ which can be rearranged to get $x^{3}-4 x^{2}-6 x+20=0$.

Given that the problem states that there are exactly three real numbers with the given condition, these three real numbers must be the solutions to the cubic equation above. A cubic equation has at most three real roots, which means this cubic must have exactly three real roots. The sum of the roots of a cubic is the negative of the coefficient of $x^{2}$, which means the answer is 4 . However, in the interest of presenting a solution that does not assume the question "works", we will factor this cubic to find the three real numbers.

A bit of guessing and checking reveals that $2^{3}-4\left(2^{2}\right)-6(2)+20=0$, which means $x-2$ is a factor of the cubic. We can factor to get

$$
(x-2)\left(x^{2}-2 x-10\right)=0
$$

By the quadratic formula, the other two roots of the cubic are

$$
\frac{2 \pm \sqrt{2^{2}-4(-10)}}{2}=1 \pm \sqrt{11}
$$

The three real numbers are $2,1-\sqrt{11}$, and $1+\sqrt{11}$ which have a sum of

$$
2+(1-\sqrt{11})+(1+\sqrt{11})=4
$$

ANSWER: 4