Sets of Polynomial roots For a real number $k,$ let $A$ be the set of roots of $$x^2 + (k - 1) x - 2(k + 1) = 0,$$and let $B$ be the set of roots of $$(k - 1) x^2 + kx + 1 = 0.$$(These roots may be complex.) Find the number of values of $k$ so that $|A \cup B| = 3.$
I don't know how to start this. Can someone help?

The roots of the two equations are
$$\frac{1}{2}\left(-(k-1) \pm \sqrt{(k-1)^2 + 8 (k+1)}\right) = \frac{1}{2}(-(k-1) \pm (k+3)) = \{2, -(k+1)\}$$
$$\frac{1}{2(k-1)} \left(-k \pm \sqrt{k^2 - 4(k-1)}\right) = \frac{1}{2(k-1)}(-k \pm (k-2)) = \{-\frac{1}{k-1}, -1\}$$
so you want
$$\{2, -(k+1), -\frac{1}{k-1}, -1\}$$
to have cardinality $3$.
