6.129. For what value of $a$ do the equations $x^{2}+a x+8=0$ and $x^{2}+x+a=0$ have a common root?
## Solution.

Let $x_{1}$ be the common root, then

$$
\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+8=0, \\
x_{1}^{2}+x_{1}+a=0
\end{array} \Rightarrow a x_{1}-x_{1}+8-a=0, x_{1}=\frac{a-8}{a-1}\right.
$$

From the second equation of the system, we have

$$
\begin{aligned}
& \left(\frac{a-8}{a-1}\right)^{2}+\left(\frac{a-8}{a-1}\right)+a=0, \frac{a^{3}-24 a+72}{(a-1)^{2}}=0 \Leftrightarrow\left\{\begin{array}{l}
a^{3}-24 a+72=0, \\
a \neq 1,
\end{array}\right. \\
& a^{3}+216-216-24 a+72=0,\left(a^{3}+216\right)-24 a-144=0, \\
& \left(a^{3}+6^{3}\right)-24(a+6)=0 \\
& (a+6)\left(a^{2}-6 a+36\right)-24(a+6)=0 \\
& (a+6)\left(a^{2}-6 a+12\right)=0
\end{aligned}
$$

from which $a=-6$. For the quadratic equation, $D<0, \varnothing$.

Answer: $a=-6$.