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
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
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Solution 2
Note that must be an integer. By the quadratic formula, Since is a multiple of , 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
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Solution 3 (Pythagorean Triples)
Proceed similar to Solution 2 and deduce that the discriminant of must be a perfect square greater than 0 to satisfy all given conditions. Seeing something like might remind us of a right triangle where k is the hypotenuse, and 12 is a leg. There are four ways we could have this: a triangle, a triangle, a triangle, and a triangle. Multiply by two to account for negative k values (since k is being squared) and our answer is .
Solution 4 (Similar to Solution 1)
Let and be the roots of the given polynomial. By Vieta's, , and .
The positive pairs of that satisfy the second constraint are: and
We multiply this number by to account for the negative numbers that will multiply together to give us , and subtract , since is both a double root and is counted twice.
Therefore, the answer is pairs.
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 |
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