Difference between revisions of "2005 AMC 12B Problems/Problem 12"
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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>\ | + | <math>\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math> |
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| − | |||
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| + | ==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{\ | + | 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>. |
| − | + | 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 2a and 2b and the roots of <math>x^2 + px + m = 0</math> are a and b, then using Vieta's | + | 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>2a + 2b = -m</cmath> | ||
<cmath>a + b = -p</cmath> | <cmath>a + b = -p</cmath> | ||
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and substituting the fourth equation into the third equation gives | and substituting the fourth equation into the third equation gives | ||
<cmath>n = 4(m)</cmath> | <cmath>n = 4(m)</cmath> | ||
| − | Therefore, <math>n = 8p</math>, so <math>\frac{n}{p} | + | 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 
has roots twice those of 
, and none of 
and 
is zero. What is the value of 
?
Solutions
Solution 1
Let 
have roots 
and 
. Then
![\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]](http://latex.artofproblemsolving.com/d/6/3/d6347a5ee25d5b941694ca5cd21c35997e6d31f8.png)
so 
and 
. Also, 
has roots 
and 
, so
![\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]](http://latex.artofproblemsolving.com/5/0/1/501f7835ffc0da1ceef39eaa3eebb8784db663d4.png)
and 
and 
. Thus 
.
To test that this actually works, consider the quadratics 
.
Solution 2
If the roots of are and and the roots of are and , then using Vieta's formulas,
![\[2a + 2b = -m\]](http://latex.artofproblemsolving.com/0/8/e/08ecb4147711eacc9e46c3025ecf06f134ac14d1.png)
![\[a + b = -p\]](http://latex.artofproblemsolving.com/0/2/b/02b64c3d1b6523bbb51defb309c6d0549b692b29.png)
![\[2a(2b) = n\]](http://latex.artofproblemsolving.com/8/6/c/86c586dee72aa6367bb10d8204969336903bd179.png)
![\[a(b) = m\]](http://latex.artofproblemsolving.com/d/1/4/d14005e53d35d168169a64841d4ceef3b3b42600.png)
Therefore, substituting the second equation into the first equation gives
![\[m = 2(p)\]](http://latex.artofproblemsolving.com/b/1/e/b1ee6cc0acfd106f4f1d26965b77854e3878ea4f.png)
and substituting the fourth equation into the third equation gives
![\[n = 4(m)\]](http://latex.artofproblemsolving.com/0/b/b/0bbe2b999bd85bf79835ac644e6dfb2a4587e4d9.png)
Therefore, 
, so 
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. 
