HMMT 十一月 2020 · 冲刺赛 · 第 15 题

HMMT November 2020 — Guts Round — Problem 15

[9] For a real number r , the quadratics x + ( r − 1) x + 6 and x + (2 r + 1) x + 22 have a common real a root. The sum of the possible values of r can be expressed as , where a, b are relatively prime positive b integers. Compute 100 a + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#

解析

[9] For a real number r , the quadratics x + ( r − 1) x + 6 and x + (2 r + 1) x + 22 have a common real a root. The sum of the possible values of r can be expressed as , where a, b are relatively prime positive b integers. Compute 100 a + b . Proposed by: Andrew Yao Answer: 405 Solution: Let the common root be s . Then, 2 2 s + ( r − 1) s + 6 = s + (2 r + 1) s + 22 , 16 2 and s = − . Substituting this into s + ( r − 1) s + 6 = 0 yields r +2 256 16( r − 1) − + 6 = 0 . 2 ( r + 2) r + 2 2 After multiplying both sides by ( r + 2) , the equation becomes 2 256 − 16( r − 1)( r + 2) + 6( r + 2) = 0 , which simplifies into 2 5 r − 4 r − 156 = 0 . 4 Thus, by Vieta’s Formulas, the sum of the possible values of r is . 5