10. Let $x_{1}, x_{2}, x_{3}$, and $x_{4}$ be the solutions of the equation $x^{4}-2 x^{3}-7 x^{2}-2 x+1=0$. What is the value of $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}$?
(A) 1
(B) $\frac{1}{2}$
(C) 2
(D) 4
(E) 7.
10. The answer is (C). It is possible to explicitly determine the values of the solutions of the equation. Indeed, it is a reciprocal equation of the first kind (equations in which the coefficient of $x^{k}$ is equal to that of $x^{n-k}$, where $n$ is the degree of the equation and $k=0,1, \ldots, n$.) and they have the characteristic that if they are satisfied by a certain value $x_{0}$, they are also satisfied by $1 / x_{0}$, hence the name "reciprocal". The solution of equations of this type can be reduced to that of three quadratic equations: the equation can be written in the form $x^{2}\left[\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)-7\right]=0$. By setting $y=x+\frac{1}{x}$, we arrive at $\left(y^{2}-2\right)-2 y-7=0$ from which two values of $y$ and four of $x$ are determined. This method, however, leads to rather laborious calculations: the values of the solutions $x_{1}, x_{2}, x_{3}$, and $x_{4}$ (and therefore also of their reciprocals) are in fact

$$
x_{1,2}=\frac{1+\sqrt{10}}{2} \pm \frac{\sqrt{5}+\sqrt{2}}{2} \quad \text { and } \quad x_{3,4}=\frac{1-\sqrt{10}}{2} \pm \frac{\sqrt{5}-\sqrt{2}}{2} .
$$

pairwise reciprocals, so that

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=x_{2}+x_{1}+x_{4}+x_{3}=1+\sqrt{10}+1-\sqrt{10}=2 .
$$

One arrives much more quickly at the result by remembering that if $x_{1}, x_{2}, x_{3}$, and $x_{4}$ are solutions of the equation, then $x^{4}-2 x^{3}-7 x^{2}-2 x+1=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$. By performing the calculations on the right-hand side, one notices that the coefficient of the third degree term of the product is $-\left(x_{1}+x_{2}+x_{3}+x_{4}\right)$, which, by comparison with the left-hand side, is -2. Since the equation is reciprocal, we can immediately conclude that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=2$.

Finally, note that it is also possible to solve the exercise without considering that the equation is reciprocal: by always calculating the product $\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$, one observes that

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=\frac{x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}}{x_{1} x_{2} x_{3} x_{4}}=\frac{- \text { coefficient of the first degree term }}{\text { constant term }}=\frac{2}{1} .
$$