Difference between revisions of "2022 AMC 10B Problems/Problem 7"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By Vieta's | + | Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By [[Vieta's formulas]], we have <math>p+q=-k</math> and <math>pq=36.</math> |
This shows that <math>p</math> and <math>q</math> must be distinct factors of <math>36.</math> The possibilities of <math>\{p,q\}</math> are <cmath>\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.</cmath> | This shows that <math>p</math> and <math>q</math> must be distinct factors of <math>36.</math> The possibilities of <math>\{p,q\}</math> are <cmath>\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.</cmath> |
Revision as of 17:17, 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 formulas, we have and
This shows that and must be distinct factors of The possibilities of are 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.