2019 AMC 8

Solution 1

We have that if and only if . If , then , giving 2 solutions. If , then , giving 2 more solutions. All four of these solutions work, so the answer is . Further, the equation is a quartic in , so by the Fundamental Theorem of Algebra, there can be at most four real solutions.

Solution 2

We can expand to get , so now our equation is . Subtracting from both sides gives us . Now, we can factor the left-hand side to get . If and/or equals , then the entire left side will equal . Since the solutions can be both positive and negative, we have solutions: (we can find these solutions by setting and equal to and solving for ). So, the answer is .