Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Then the value of $\alpha^{4}+3 \beta$ is $\qquad$
(2003, National Junior High School Mathematics Competition, Tianjin Preliminary Contest)
Explanation: $\alpha^{4}+3 \beta$ is not a symmetric expression of the two roots of the equation, and it is obviously impossible to directly substitute it using Vieta's formulas. We can construct the dual expression $\beta^{4}+3 \alpha$ of $\alpha^{4}+3 \beta$, and calculate the sum and difference of the two expressions, then solve for the value through a system of equations. Because
$$
\begin{array}{l}
\alpha^{4}+3 \beta+\beta^{4}+3 \alpha \\
=\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-2(\alpha \beta)^{2}+3(\alpha+\beta) \\
=10, \\
\left(\alpha^{4}+3 \beta\right)-\left(\beta^{4}+3 \alpha\right) \\
=(\alpha-\beta)\left\{(\alpha+\beta)\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]-3\right\} \\
=0,
\end{array}
$$

Therefore, $\alpha^{4}+3 \beta=\frac{10+0}{2}=5$.