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

Revision as of 20:49, 22 February 2019

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

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

@@ Line 22: / Line 22: @@
 ===Solution 2===
-If the roots of <math>x^2 + mx + n = 0</math> are twice the roots of <math>x^2 + px + m = 0</math>, then the sum of the roots of the first equation is also twice the sum of the roots of the second equation, and the product of the roots of the first equation is <math>2(2) = 4</math> times the product of the roots of the second equation. Using Vieta's equations, <math>-m = 2(-p)</math>, and <math>n = 4(m)</math>. Substituting, <math>n = 8p</math>,  so <math>\frac{n}{p} = 8 = \boxed{\textbf{D}}</math>
+If the roots of <math>x^2 + mx + n = 0</math> are 2a and 2b and the roots of <math>x^2 + px + m = 0</math> are a and b, then using Vieta's equations,
+<math>2a + 2b = -m</math>,
+<math>a + b = -p</math>,
+<math>2a(2b) = n</math>
+<math>a(b) = m</math>.
+Therefore, substituting the second equation into the first equation gives
+<math>m = 2(p)</math>,
+and substituting the fourth equation into the third equation gives
+<math>n = 4(m)</math>.
+Substituting, <math>n = 8p</math>,  so <math>\frac{n}{p} = 8 = \boxed{\textbf{D}}</math>
 == See also ==