# 2005 AMC 12B Problems/Problem 12

The following problem is from both the 2005 AMC 12B #12 and 2005 AMC 10B #16, so both problems redirect to this page.

## Problem

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 $n/p$? $\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}$

## Solutions

### Solution 1

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 $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\mathrm{(D)}\ 8}$.

Indeed, consider the quadratics $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$.

### Solution 2

If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$, then using Vieta's formulas, $$2a + 2b = -m$$ $$a + b = -p$$ $$2a(2b) = n$$ $$a(b) = m$$ Therefore, substituting the second equation into the first equation gives $$m = 2(p)$$ and substituting the fourth equation into the third equation gives $$n = 4(m)$$ Therefore, $n = 8p$, so $\frac{n}{p} = 8 = \boxed{\textbf{D}}$

~ pi_is_3.14

## See also

 2005 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2005 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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