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)
 
(22 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #12]] and [[2005 AMC 10B Problems|2005 AMC 10B #16]]}}
 
== Problem ==
 
== Problem ==
 
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===
 
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  
  
 
<cmath>x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,</cmath>
 
<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>
 
<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}} = \boxed{\textbf{(D) }8}</math>.  
  
−
Indeed, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>. See [[Vieta's formulas]].  
+
To test that this actually works, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
 +
 
 +
===Solution 2===
 +
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 ==
 +
{{AMC10 box|year=2005|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2005|num-b=11|num-a=13|ab=B}}
 
{{AMC12 box|year=2005|num-b=11|num-a=13|ab=B}}
−
 
+
* [[Vieta's Formulas]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

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"