Difference between revisions of "2022 AMC 10B Problems/Problem 7"
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− | + | {{duplicate|[[2022 AMC 10B Problems/Problem 7|2022 AMC 10B #7]] and [[2022 AMC 12B Problems/Problem 4|2022 AMC 12B #4]]}} | |
− | <math> | + | ==Problem== |
+ | For how many values of the constant <math>k</math> will the polynomial <math>x^{2}+kx+36</math> have two distinct integer roots? | ||
− | <math> | + | <math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</math> |
− | + | ==Solution 1== | |
+ | Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By Vieta's Formula, we have <math>p+q=-k</math> and <math>pq=36.</math> | ||
− | This shows that p and q must be | + | 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> |
+ | 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 | |
− | + | ~MRENTHUSIASM | |
+ | |||
+ | == See Also == | ||
+ | {{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 17:10, 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.