Example 8 (1986 National High School League Test Question) Let real numbers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0, \\
b^{2}+c^{2}+b c-6 a+6=0 .
\end{array}\right.
$$

Then the range of values for $a$ is ( ).
A. $(-\infty,+\infty)$
B. $(-\infty, 1] \cup[9,+\infty)$
C. $(0,7)$
D. $[1,9] \quad$
Solution: Choose D. Reason: From the given, we have
\[
\left\{\begin{array}{l}
b c=a^{2}-8 a+7, \\
b+c= \pm(a-1) \text{. }
\end{array}\right.
\]
Thus, we can construct the equation
\[
x^{2} \pm(a-1) x+a^{2}-8 a+7=0,
\]

Therefore, \(\Delta=[ \pm(a-1)]^{2}-4\left(a^{2}-8 a+7\right) \geqslant 0\),
which simplifies to \(a^{2}-10 a+9 \leqslant 0\),
hence \(1 \leqslant a \leqslant 9\).