19. Let $x$ be a real number such that $x^{2}-15 x+1=0$. Find the value of $x^{4}+\frac{1}{x^{4}}$.
19. Answer: 49727
Since $x^{2}-15 x+1=0, x+\frac{1}{x}=15$. Therefore
$$
\begin{aligned}
x^{4}+\frac{1}{x^{4}} & =\left(x+\frac{1}{x}\right)^{4}-4\left(x^{2}+\frac{1}{x^{2}}\right)-6 \\
& =\left(x+\frac{1}{x}\right)^{4}-4\left(x+\frac{1}{x}\right)^{2}+8-6 \\
& =15^{4}-4 \times 15^{2}+2=49727
\end{aligned}
$$