Example 6 If the following equations: $x^{2}-a x+4=0, x^{2}+(a-2) x+4=0, x^{2}+2 a x+a^{2}+$ $1=0$ have at least one equation with real roots. Try to find the range of real number $a$.

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If $x^{2}-a x+4=0$ has real roots, it is equivalent to $(-a)^{2}-4 \times 4 \geqslant 0$, i.e., $a \geqslant 4$ or $a \leqslant-4$; if $x^{2}+(a-2) x+4=0$ has real roots, it is equivalent to $(a-2)^{2}-4 \times 4 \geqslant 0$, i.e., $a \geqslant 6$ or $a \leqslant-2$; if $x^{2}+2 a x+a^{2}+1=0$ has real roots, it is equivalent to $(2 a)^{2}-4\left(a^{2}+1\right) \geqslant 0$, i.e., $a \in \varnothing$. Therefore, $a \geqslant 4$ or $a \leqslant-2$ is the solution.

Solution Guide: If the sets of real numbers $a$ for which the three equations have real roots are denoted as $A$, $B$, and $C$ respectively, then the set of real numbers $a$ required is $A \cup B \cup C$. This example can also be considered from another perspective, i.e., finding the range of real numbers $a$ for which all three equations have no real roots, and then finding its complement, which is the "complement method".