Question

La ecuación $2000x^6 + 100x^5 + 10x^3$ $+ x - 2 = 0$ tiene exactamente dos raíces reales, una de las cuales es $\frac{m + \sqrt{n}}{r}$ , donde $m$ , $n$ , y $r$ son enteros, $m$ y $r$ son primos entre sí, y $r \gt 0.$ Halla $m + n + r.$

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 $m + n + r.$

Respuesta

Solución:

Agrupa la ecuación como $\begin{aligned} &2(1000x^6 - 1) \\ &\quad {}+ x(100x^4 + 10x^2 + 1) = 0. \end{aligned}$ Como $1000x^6 - 1 = (10x^2)^3 - 1$ $= (10x^2 - 1)$ $(100x^4 + 10x^2 + 1)$ , el lado izquierdo se factoriza como $\begin{aligned} &(100x^4 + 10x^2 + 1) \\ &\quad \big(2(10x^2 - 1) + x\big) \\ &= (100x^4 + 10x^2 + 1) \\ &\quad (20x^2 + x - 2). \end{aligned}$

El factor cuártico siempre es positivo, así que las dos raíces reales son las raíces de $20x^2 + x - 2 = 0$ , a saber $x = \frac{-1 \pm \sqrt{161}}{40}$ . La raíz de la forma $\frac{m + \sqrt{n}}{r}$ es $\frac{-1 + \sqrt{161}}{40}$ , con $m = -1$ , $n = 161$ , $r = 40$ , y $\gcd(-1, 40) = 1$ . Así $m + n + r = -1 + 161 + 40$ $= 200$ .

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 Problema 13

Solución:

El Problema 13 en otros años