Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
 
(13 intermediate revisions by 6 users not shown)
Line 3: Line 3:
 
The [[quadratic equation]] <math>x^2+mx+n</math> has roots twice those of <math>x^2+px+m</math>, and none of <math>m,n,</math> and <math>p</math> is zero. What is the value of <math>n/p</math>?
 
The [[quadratic equation]] <math>x^2+mx+n</math> has roots twice those of <math>x^2+px+m</math>, and none of <math>m,n,</math> and <math>p</math> is zero. What is the value of <math>n/p</math>?
  
−
<math>\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}</math>
+
<math>\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math>
−
 
 
−
== Solution ==
 
−
 
 
  
 +
==Solutions==
 
===Solution 1===
 
===Solution 1===
 
Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then  
 
Let <math>x^2 + px + m = 0</math> have roots <math>a</math> and <math>b</math>. Then  
Line 17: Line 15:
 
<cmath>x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,</cmath>
 
<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+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\mathrm{(D)}\ 8}</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}} = \boxed{\textbf{(D) }8}</math>.  
  
−
Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
+
To test that this actually works, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
  
 
===Solution 2===
 
===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 <math>2a</math> and <math>2b</math> and the roots of <math>x^2 + px + m = 0</math> are <math>a</math> and <math>b</math>, then using Vieta's formulas,
 +
<cmath>2a + 2b = -m</cmath>
 +
<cmath>a + b = -p</cmath>
 +
<cmath>2a(2b) = n</cmath>
 +
<cmath>a(b) = m</cmath>
 +
Therefore, substituting the second equation into the first equation gives
 +
<cmath>m = 2(p)</cmath>
 +
and substituting the fourth equation into the third equation gives
 +
<cmath>n = 4(m)</cmath>
 +
Therefore, <math>n = 8p</math>, so <math>\frac{n}{p}= \boxed{\textbf{(D) }8}</math>
 +
 
 +
== Video Solution ==
 +
https://youtu.be/3dfbWzOfJAI?t=1023
 +
 
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 01:57, 11 November 2024

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 $n/p$?

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

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

To test that this actually works, 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 $\frac{n}{p}= \boxed{\textbf{(D) }8}$

Video Solution

https://youtu.be/3dfbWzOfJAI?t=1023

~ pi_is_3.14

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_12&oldid=233216"