Difference between revisions of "2005 AMC 12B Problems/Problem 12"
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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 == | == Video Solution == |
Latest revision as of 12:30, 24 August 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
so and . Also, has roots and , so
and and . Thus .
Indeed, consider the quadratics .
Solution 2
If the roots of are and and the roots of are and , then using Vieta's formulas, Therefore, substituting the second equation into the first equation gives and substituting the fourth equation into the third equation gives 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.