\[x^{4} - 17x^{2} + 16 = 0\]
\[Пусть\ \ \ t = x^{2} \geq 0:\]
\[t^{2} - 17t + 16 = 0\]
\[t_{1} + t_{2} = 17;\ \ t_{1} \cdot t_{2} = 16\]
\[t_{1} = 16,\ \ t_{2} = 1.\]