Difference between revisions of "2005 AMC 12B Problems/Problem 12"
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===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 equations, | 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, | ||
| − | + | <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 | 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 | 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} = 8 = \boxed{\textbf{D}}</math> |
| − | |||
| − | |||
== See also == | == See also == | ||
Revision as of 13:02, 23 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.
Problem
The quadratic equation 
has roots twice those of 
, and none of 
and 
is zero. What is the value of 
?
Solution
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 
.
Indeed, consider the quadratics 
.
Solution 2
If the roots of are 2a and 2b and the roots of are a and b, then using Vieta's equations,
![\[2a + 2b = -m\]](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 
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
