PUMaC 2023 · 代数（B 组） · 第 6 题

PUMaC 2023 — Algebra (Division B) — Problem 6

The set of real values of a such that the equation x − 3 ax + (2 a + 4 a ) x − 5 a x + 3 a has exactly two nonreal solutions is the set of real numbers between x and y, where x < y. If x + y m can be written as for relatively prime positive integers m, n, find m + n. n $ % 10 10 X 2 πk

解析

The set of real values of a such that the equation x − 3 ax + (2 a + 4 a ) x − 5 a x + 3 a has exactly two nonreal solutions is the set of real numbers between x and y, where x < y. If x + y m can be written as for relatively prime positive integers m, n, find m + n . n Proposed by Frank Lu Answer: 8 First, we consider trying to factor this into quadratics. Notice that this equals 4 3 2 2 2 2 2 x − 3 tx + (2 t − 2 t ) x + t x − 3 t = ( x − tx + t )( x − 2 tx + 3 t ) . Therefore, to have two nonreal solutions, one of the discriminants of the quadratics needs to 2 be negative, and the other is nonnegative. In particular, it follows that we need t − 4 t < 2 2 2 0 and 4 t − 12 t ≥ 0 or t − 4 t ≥ 0 and 4 t − 12 t < 0 . For the former to hold, notice that we 2 need 0 < t < 4 , but t > 3 . The latter cannot hold, however: t − 4 t ≥ 0 implies that t ≥ 4 2 or t ≤ 0 , but 4 t − 12 t < 0 implies that 0 < t < 3 . Therefore, we see that a = 3 , b = 4 , and a + b = 7 = 7 / 1 . Our answer is thus 7 + 1 = 8 . $ % 10 10 X 2 πk