AMC12 2005 B

AMC12 2005 B · Q12

AMC12 2005 B · Q12. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$, and none of $m,n,$ and $p$ is zero. What is the value of $n/p$?

二次方程 $x^2+mx+n$ 的根是 $x^2+px+m$ 的根的两倍，且 $m,n,$ 和 $p$ 都不为零。$n/p$ 的值是多少？

(A) 1 1

(B) 2 2

(D) 8 8

(E) 16 16

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $x^2 + px + m = 0$ have roots $a$ and $b$. Then \[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\] so $p = -(a+b)$ and $m = ab$. Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$, so \[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\] and $m = -2(a+b)$ and $n = 4ab$. Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\textbf{(D) }8}$. To test that this actually works, consider the quadratics $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$.

设 $x^2 + px + m = 0$ 的根为 $a$ 和 $b$。则 \[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\] 所以 $p = -(a+b)$ 且 $m = ab$。另外，$x^2 + mx + n = 0$ 的根为 $2a$ 和 $2b$，因此 \[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\] 从而 $m = -2(a+b)$ 且 $n = 4ab$。于是 $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\textbf{(D) }8}$。为验证确实可行，可考虑二次式 $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$。

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