Common root of quadratic equations with parameter The problem: For which values of parameter $a$ the functions $f(x)$ and  $g(x)$ have at least $1$ common root.
$f(x)=x^2+ax+1=0$
$g(x)=x^2+x+a=0$
What i did was look at the 2 functions separately and find the roots then equal them and i get the answer.
My question: I am looking for a better way because if the functions were more difficult this method won't be practical.
 A: Suppose $x$ is the common root:
$$
\begin{cases}
\begin{align}
x^2 + ax + 1 = 0 \\
x^2+x+a=0
\end{align}
\end{cases}
$$
Subtracting the equations gives:
$$\require{cancel}
\cancel{x^2} + ax + 1 - (\cancel{x^2}+x+a) = 0 \quad\iff\quad (x-1)(a-1)=0
$$
Then:


*

*either $\,a=1\,$ in which case the equations are in fact identical;

*or $\,x=1\,$ is the common root which, after substituting back in either equation, gives $a=-2\,$.
A: You didn't say how you obtained your roots but possibly you used the formula for solving quadratic equations. This may illustrate an alternative.
Let the roots of $f$ be $p_1$ and $q_1$. Let the roots of $g$ be $p_2$ and $q_2$. 
For $f$
$$p_1+q_1=-a\\p_1q_1=1$$
For $g$
$$p_2+q_2=-1\\p_2q_2=a$$
Case 1:
1 root is common, say $p_1=p_2=p$ and $q_1\neq q_2$
Then
$$p+q_1=-a\\pq_1=1$$
and 
$$p+q_2=-1\\pq_2=a$$
Eliminating $p$ and then $q_1$ leads to $q_2=a$, and also then $q_1=1$ and $p=1$. The above equations then give $a=-2$.
Case 2:
Both roots equal $p_1=p_2=p, q_1=q_2=q$
Then as above
$$
p+q=-a \\ p+q=-1
$$
Subtracting gives $a=1$
A: Well, if their roots are equal, then they're equal at $0$. So, $$x^2+ax+1=0=x^2+x+a\implies ax+1=x+a\implies a(x-1)=x-1\implies a=1$$
This is basically an inspection where we find when the functions are exactly equal (have $2$ imaginary roots in this case). The other answer ($a=-2$) I'm not sure how to easily find.
A: Plotting $y=-x^2-x$ and $y=-x-\dfrac{1}{x}$, they intersect at $(x,y)=(1,2)$.
That is $a=2$ gives common root $x=1$.
Note the trivial case $a=1$ gives no real roots.

