Example 6 (6th "Hope Cup" Competition Question) Given that the equation $x^{2}-a x+a^{2}-4=0$ has two distinct real roots and one positive root, the range of values for $a$ is $\qquad$ .
Solve: Fill in $-2 \leqslant a \leqslant 2$. Reason: The necessary and sufficient condition for the equation $x^{2}-a x+a^{2}-4=0$ to have two distinct real roots and one positive root is $\left\{\begin{array}{l}\Delta=(-a)^{2}-4\left(a^{2}-4\right)>0, \\ a^{2}-4<0,\end{array}\right.$ Solving it yields $\left\{\begin{array}{l}-\sqrt{\frac{16}{3}}<a<\sqrt{\frac{16}{3}}, \text { hence } a \text { 's range is }-2 \leqslant a \leqslant 2 . \\ -2 \leqslant a \leqslant 2 .\end{array}\right.$