Difference between revisions of "2005 AMC 12B Problems/Problem 12"

Revision as of 18:46, 27 August 2008

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

Indeed, consider the quadratics $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$ . See Vieta's formulas.

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

@@ Line 9: / Line 9: @@
 <cmath>x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,</cmath>
-so <math>p = -(a+x)</math> and <math>m = ab</math>. Also, <math>x^2 + mx + n = 0</math> has roots <math>2a</math> and <math>2b</math>, so
+so <math>p = -(a+b)</math> and <math>m = ab</math>. Also, <math>x^2 + mx + n = 0</math> has roots <math>2a</math> and <math>2b</math>, so
 <cmath>x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,</cmath>
-and <math>m = -2(a+b)</math> and <math>n = 4ab</math>. Thus <math>\frac{n}{p} = \frac{4ab}{-(a+x)} = \frac{4m}{\frac{m}{2}} = 8 \Longrightarrow \textbf{(D)}</math>.
+and <math>m = -2(a+b)</math> and <math>n = 4ab</math>. Thus <math>\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = 8 \Longrightarrow \textbf{(D)}</math>.
 Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>. See [[Vieta's formulas]].
 == See also ==