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 $n/p$?

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

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

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

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 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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