Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
| − | You can factor (p^2 + 4)(q^2 + 4)(r^2 + 4) as ( | + | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as (p−2i)(p+2i)(q−2i)(q+2i)(r−2i)(r+2i). |
| − | For any polynomial f(x), you can create a new polynomial f(x+2), 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. |
| − | Substituting x-2 and x+2 into x for the first polynomial, gives you 10i-5 and -10i-5 as c for both equations. Multiplying 10i-5 and -10i-5 together gives you 125 | + | Substituting <math>x-2</math> and <math>x+2</math> into <math>x</math> for the first polynomial, gives you <math>10i-5</math> and <math>-10i-5</math> as <math>c</math> for both equations. Multiplying <math>10i-5</math> and <math>-10i-5</math> together gives you <math>\boxed{\textbf{(D) }125}</math>. |
-ev2028 | -ev2028 | ||
Revision as of 18:00, 8 November 2024
Contents
Problem
The roots of 
are 
and 
What is the value of ![\[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]](http://latex.artofproblemsolving.com/9/3/8/938ecf6869b053c8335864493848733a3f242e2e.png)
Solution 1
You can factor 
as (p−2i)(p+2i)(q−2i)(q+2i)(r−2i)(r+2i).
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
Solution 2
Let 
. Then
.
We find that 
and 
, so 
.
~eevee9406
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. 
