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 $n/p$ ?

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

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 $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$ .

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 equations,

$2a + 2b = -m$ ,

$a + b = -p$ ,

$2a(2b) = n$

$a(b) = m$ .

Therefore, substituting the second equation into the first equation gives

$m = 2(p)$ ,

and substituting the fourth equation into the third equation gives

$n = 4(m)$ .

Substituting, $n = 8p$ , so $\frac{n}{p} = 8 = \boxed{\textbf{D}}$

2005 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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