2005 AMC 12B Problems/Problem 12
- The following problem is from both the 2005 AMC 12B #12 and 2005 AMC 10B #16, so both problems redirect to this page.
Contents
Problem
The quadratic equation 
has roots twice those of 
, and none of 
and 
is zero. What is the value of 
?
Solution
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
Realize that if first quadratic has roots twice those of the second, then the sum of its roots will be twice those of the second and the product of its roots will be four times those of the second
Using Vieta's formula, the sum of the roots of 
is 
and the product of the roots is 
The sum of the roots of 
is 
and the product of the roots is 
We now have two equations:
![\[-2p = -m\]](http://latex.artofproblemsolving.com/1/6/2/16215192d58e09d581a5d804c877c971eee5a8c7.png)
![\[4m = n\]](http://latex.artofproblemsolving.com/8/9/1/891f6ddfcd84a6125e2490715ccd8d32350309f7.png)
Solving the first equation for p, we have
Thus $\frac{n}{p} = \frac{4m}{\frac{m}{2}} = 8
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 | |
