Find the range of $a$ for $a^2 - bc - 8a + 7 = 0$ and $b^2 + c^2 + bc - 6 a + b = 0$. Let $a, b, c$ be real numbers such that
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + b = 0\\
\end{cases}
Find the range of $a$.
By adding the two equations I have $$(a - 7)^2 + (b + \frac 12)^2 + c^2 = \frac {169}4,$$
thus $a \in [\frac 12, \frac {27}2]$. Then I found that $a$ cannot obtain the boundaries so $a \in (\frac 12, \frac {27}2)$. But this is only a rough restriction. Exactly how do I solve this?
Edit: When I got this question from my friend, it was a $b$, but after I saw the official solution, it should be a $6$ instead. So the official answer I posted is actually not the answer to the question here but the answer to the original question. I'm really sorry for this mistake.

This is the "official" solution which I found later:
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + \mathbf6= 0\\
\end{cases}
is equivalent to
\begin{cases}
bc=a^2 - 8a + 7\\
b+c=\pm(a-1)\\
\end{cases}
So $b,c$ are the roots of the equation
$$x^2\pm(a-1)x+(a^2-8a+7)=0$$
A necessary and sufficient condition that $b,c\in\mathbb{R}$ is
$$\Delta=(a-1)^2-4(a^2-8a+7)\ge0$$ Solving this gives the answer $a\in[1,9]$.
