Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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<cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath> | <cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath> | ||
<cmath>=\fbox{(D) 125}</cmath> | <cmath>=\fbox{(D) 125}</cmath> | ||
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==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:46, 8 November 2024
Contents
[hide]Problem
The roots of are and What is the value of
Solution 1
You can factor as .
For any polynomial , you can create a new polynomial , which will have roots that instead have the value subtracted.
Substituting and into for the first polynomial, gives you and as for both equations. Multiplying and together gives you .
-ev2028
~Latex by eevee9406
Solution 2
Let . Then .
We find that and , so .
~eevee9406
Solution 3
First, denote that Then we expand the expression ~lptoggled
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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.