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HMMT 二月 2013 · 代数 · 第 1 题

HMMT February 2013 — Algebra — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

Let xandybe real numbers with x > y such that x2y2+x2+y2+ 2xy= 40 and xy+x+y= 8. Find the value of x.

解析

英文解析

  1. Let x and y be real numbers with x > y such that x y + x + y + 2 xy = 40 and xy + x + y = 8. Findthe value of x .
    √
    2 2
    Answer: 3 + 7 We have ( xy ) + ( x + y ) = 40 and xy + ( x + y ) = 8. Squaring the secondequation and subtracting the first gives xy ( x + y ) = 12 so xy, x + y are the roots of the quadratica − 8 a + 12 = 0. It follows that { xy, x + y } = { 2 , 6 } . If x + y = 2 and xy = 6, then x, y are the roots of 2
    the quadratic b − 2 b + 6 = 0, which are non-real, so in fact x + y = 6 and xy = 2, and x, y are the roots 2
    √
    √
    2 6+ 28
    of the quadratic b − 6 b + 2 = 0. Because x > y , we take the larger root, which is = 3 + 7. 2