Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as | + | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as <math>(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)</math>. |
For any polynomial <math>f(x)</math>, you can create a new polynomial <math>f(x+2)</math>, which will have roots that instead have the value subtracted. | For any polynomial <math>f(x)</math>, you can create a new polynomial <math>f(x+2)</math>, which will have roots that instead have the value subtracted. | ||
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-ev2028 | -ev2028 | ||
+ | ~Latex by eevee9406 | ||
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+ | ==Solution 2== | ||
+ | Let <math>f(x)=x^3 + 2x^2 - x + 3</math>. Then | ||
+ | <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)=(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)=f(2i)f(-2i)</math>. | ||
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+ | We find that <math>f(2i)=-8i-8-2i+3=-10i-5</math> and <math>f(-2i)=8i-8+2i+3=10i-5</math>, so <math>f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}</math>. | ||
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+ | ~eevee9406 | ||
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==Solution 3== | ==Solution 3== | ||
First, denote that | First, denote that | ||
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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 18:10, 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
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.