16. Let $a, b, c$ be real numbers with $c \neq 1$. It is known that the two equations $x^{2}+a x+1=0$ and $x^{2}+b x+c=0$ have a common real root, and so do the two equations $x^{2}+x+a=0$ and $x^{2}+c x+b=0$. Find the value of $a+b+c$.
設 $a 、 b 、 c$ 為實數, 其中 $c \neq 1$ 。已知 $x^{2}+a x+1=0$ 和 $x^{2}+b x+c=0$ 這兩條方程有一個公共實根, 而 $x^{2}+x+a=0$ 和 $x^{2}+c x+b=0$ 這兩條方程亦有一個公共實根。求 $a+b+c$ 。
16. -3
16. Let $m$ be the common root of $x^{2}+a x+1=0$ and $x^{2}+b x+c=0$. Clearly $m \neq 0$. Since $c \neq 1$, we must have $a \neq b$ as $m^{2}+a m+1=m^{2}+b m+c=0$. This also gives $m=\frac{c-1}{a-b}$. Similarly, if $n$ is the common root of $x^{2}+x+a=0$ and $x^{2}+c x+b=0$, then $n=\frac{a-b}{c-1}$. It follows that $m n=1$. As $m$ is a root of the equation $x^{2}+a x+1=0$, which has product of roots $1, n$ is also a root of this equation.
Now $n$ is a common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$, so we have $n^{2}+a n+1=n^{2}+n+a=0$, which simplifies to $(a-1)(n-1)=0$. Clearly $a \neq 1$, for otherwise the equation $x^{2}+a x+1=0$ will have no real root. Thus we must have $n=1$ and $n^{2}+n+a=0$ gives $a=-2$. Since $m n=1$, we have $m=1$ and $m^{2}+b m+c=0$ gives $b+c=-1$. Thus $a+b+c=-3$.