2022 AMC 10B Problems/Problem 7
- The following problem is from both the 2022 AMC 10B #7 and 2022 AMC 12B #4, so both problems redirect to this page.
Contents
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
It follows 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 ~
Solution 2
Note that must be an integer. Using the quadratic formula, Since divides evenly, and have the same parity, so is an integer if and only if is a perfect square.
Let Then, Since is an integer and is even, and must both be even. Assuming that is positive, we get possible values of , namely , which will give distinct positive values of , but gives and , giving identical integer roots. Therefore, there are distinct positive values of Multiplying that by to take the negative values into account, we get values of .
~pianoboy
Solution 3 (Pythagorean Triples)
Proceed similar to Solution 2 and deduce that the discriminant of must be a perfect square greater than to satisfy all given conditions. Seeing something like might remind us of a right triangle, where is the hypotenuse, and is a leg. There are four ways we could have this: a -- triangle, a -- triangle, a -- triangle, and a -- triangle.
Multiply by to account for negative values (since is being squared), and our answer is .
Solution 4
Since , that means there are possible factors of . Since is one possible factor, subtract from to get . Note that we do not need to take into consideration overcounting in this case since we also include negatives.
~songmath20 Edited 5.1.2023
Video Solution 1
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=679
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.