Difference between revisions of "2022 AMC 10B Problems/Problem 7"
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Each unordered pair gives a unique value of <math>k.</math> Therefore, there are <math>\boxed{\textbf{(B) }8}</math> values of <math>k,</math> namely <math>\pm37,\pm20,\pm15,\pm13.</math> | Each unordered pair gives a unique value of <math>k.</math> Therefore, there are <math>\boxed{\textbf{(B) }8}</math> values of <math>k,</math> namely <math>\pm37,\pm20,\pm15,\pm13.</math> | ||
| − | ~Stevens0209 | + | ~Stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math> |
| − | |||
| − | ~MRENTHUSIASM | ||
== See Also == | == See Also == | ||
Revision as of 17:14, 17 November 2022
- The following problem is from both the 2022 AMC 10B #7 and 2022 AMC 12B #4, so both problems redirect to this page.
Problem
For how many values of the constant 
will the polynomial 
have two distinct integer roots?
Solution 1
Let 
and 
be the roots of 
By Vieta's Formula, we have 
and 
This shows that and must be distinct factors of 
The possibilities of 
are ![\[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\]](http://latex.artofproblemsolving.com/9/a/9/9a995e75c8baf10fff9deffb17fa9f4cdd457393.png)
Each unordered pair gives a unique value of 
Therefore, there are 
values of 
namely 
~Stevens0209 ~MRENTHUSIASM ~
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
