I have stumbled upon the following two exercises:
For $a,b \in \mathbb{Z}$, consider $A=\{x\in\mathbb{R} | x^2+2ax+b=0\}$ and $B=\{x\in\mathbb{R} | x^2+2bx+a = 0\}$. Let $k$ be the number of elements of $M=\{(a,b)\in\mathbb{Z}^2 | A \cap B\text{ has exactly one element}\}$. Find $k$.
Find parameter $m\in\mathbb{R}$ such that equations $x^2+mx+1=0$ and $x^2+x+m=0$ have exactly one common real solution.
I was not able to solve any of these, but I recognize they're probably very similar in approach. My question is: what do I need to check to find when two parabolas have precisely one solution? For instance, in the second (2) problem, I initially looked for $$\begin{cases}\Delta_1 \geq 0 \\ \Delta_2 \geq 0\\x^2+mx+1=x^2+x+m \iff (m-1)x=m-1\iff x=1\end{cases} $$ However, doing this only got me $m \leq -2$, while the problem says only $m=-2$ is correct. Another question I have regarding the difference of the two equations, if that is indeed something I need to do, is how do I make sure by that, subtracting one from the other, I find common roots and not simply intersection points?
How about you start with this: two parabolas $p,q$ have exactly one common solution $a$ if there is $b\neq c$ s.t. \begin{align*} p(x) = (x-a)(x-b) && q(x) = (x-a)(x-c) \end{align*} which resolves into \begin{align*} p(x) = x^2-(a+b)x+ab && q(x) = x^2-(a+c)x+ac \end{align*} And now compare to what you are given. Hope that helps :)