Question

方程 $2000x^6 + 100x^5 + 10x^3$ $+ x - 2 = 0$ 恰有两个实根，其中一个是 $\frac{m + \sqrt{n}}{r}$ ，其中 $m$ 、 $n$ 、 $r$ 是整数， $m$ 与 $r$ 互质，且 $r \gt 0$ 求 $m + n + r$ 。

The equation $2000x^6 + 100x^5 + 10x^3$ $+ x - 2 = 0$ has exactly two real roots, one of which is $\frac{m + \sqrt{n}}{r},$ where $m,$ $n,$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r \gt 0.$ Find $m + n + r.$

答案

答案：200

知识点：多项式立方和与立方差因式分解二次方程

难度评级：2920

解答：

把方程左边分组为因为 $1000x^6 - 1 = (10x^2)^3 - 1$ $= (10x^2 - 1)$ $(100x^4 + 10x^2 + 1)$ ，左边可分解为 $\begin{aligned} &2(1000x^6 - 1) \\ &\quad {}+ x(100x^4 + 10x^2 + 1) = 0. \end{aligned}$ $\begin{aligned} &(100x^4 + 10x^2 + 1) \\ &\quad \big(2(10x^2 - 1) + x\big) \\ &= (100x^4 + 10x^2 + 1) \\ &\quad (20x^2 + x - 2). \end{aligned}$

四次因子恒为正，所以两个实根是 $20x^2 + x - 2 = 0$ 的根，即 $x = \frac{-1 \pm \sqrt{161}}{40}$ 。符合 $\frac{m + \sqrt{n}}{r}$ 形式的根为 $\frac{-1 + \sqrt{161}}{40}$ ，其中 $m = -1$ 、 $n = 161$ 、 $r = 40$ ，且 $\gcd(-1, 40) = 1$ 。因此 $m + n + r = -1 + 161 + 40$ $= 200$ 。

Group the equation as $\begin{aligned} &2(1000x^6 - 1) \\ &\quad {}+ x(100x^4 + 10x^2 + 1) = 0. \end{aligned}$ Since $1000x^6 - 1 = (10x^2)^3 - 1$ $= (10x^2 - 1)$ $(100x^4 + 10x^2 + 1),$ the left side factors as $\begin{aligned} &(100x^4 + 10x^2 + 1) \\ &\quad \big(2(10x^2 - 1) + x\big) \\ &= (100x^4 + 10x^2 + 1) \\ &\quad (20x^2 + x - 2). \end{aligned}$

The quartic factor is always positive, so the two real roots are the roots of $20x^2 + x - 2 = 0,$ namely $x = \frac{-1 \pm \sqrt{161}}{40}.$ The root of the form $\frac{m + \sqrt{n}}{r}$ is $\frac{-1 + \sqrt{161}}{40},$ with $m = -1,$ $n = 161,$ $r = 40,$ and $\gcd(-1, 40) = 1.$ Thus $m + n + r = -1 + 161 + 40$ $= 200.$

2000 AIME II 第 13 题

答案：200

难度评级：2920

解答：

其他年份的第 13 题