Example 1 Given $a, b \in \mathbf{R}$, the equation about $x$
$$
x^{4}+a x^{3}+2 x^{2}+b x+1=0
$$

has one real root. Find the minimum value of $a^{2}+b^{2}$.
【Analysis】This is a quartic equation, and parameters $a, b$ are not easy to handle. We might as well regard $a, b$ as the main variables and $x$ as a parameter. Since $a^{2}+b^{2}$ represents the square of the distance from the moving point $P(a, b)$ to the origin, and $P(a, b)$ lies on the line
$$
x^{3} a + x b + x^{4} + 2 x^{2} + 1 = 0,
$$

then
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}} \geqslant \frac{x^{4}+2 x^{2}+1}{\sqrt{x^{6}+x^{2}}} = \frac{x^{4}+1+2 x^{2}}{|x| \sqrt{x^{4}+1}} \\
= \frac{\sqrt{x^{4}+1}}{|x|} + \frac{2|x|}{\sqrt{x^{4}+1}} \geqslant 2 \sqrt{2}.
\end{array}
$$

The equality holds if and only if $x = \pm 1$.
Therefore, $\left(a^{2}+b^{2}\right)_{\text {min }} = 8$.