Question

find the sum of all possible integral values of p for which the equation |x+1/x-3| = p-3 has exactly 2 distinct solutions

Answer 1

21

Answer 2

1. Split $\left|x+\frac{1}{x}-3\right|=p-3$ into $x+\frac{1}{x}=p$ and $x+\frac{1}{x}=6-p$.

2. For $p\ge3$, the first equation always gives 2 solutions.

3. For the second equation to have no solutions, we need $|6-p|<2$ which is true for $p\in(4,8)$.

4. Also, $p=3$ makes both branches identical (giving 2 solutions).

5. Thus, valid integral $p$ are $3, 5, 6, 7$ whose sum equals 21.