2002 AMC 12A Problems/Problem 12
- The following problem is from both the 2002 AMC 12A #12 and 2002 AMC 10A #14, so both problems redirect to this page.
Problem
Both roots of the quadratic equation are prime numbers. The number of possible values of is
Solution
Consider a general quadratic with the coefficient of being and the roots being and . It can be factored as which is just . Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas).
We now have that the sum of the two roots is while the product is . Since both roots are primes, one must be , otherwise the sum would be even. That means the other root is and the product must be . Hence, our answer is .
Solution 2
Let the quadratic be factored as . Then and . We know that both and are prime, and all primes but are odd. The sum of two odd numbers is also always even, but in this case the sum of two primes is odd, which means that one of the primes is . This leaves only one possible value for the other root and one possible value for their product .
Solution 3
By Vieta's you have Since we know odd + even = odd, we must have either or equal to and the other equal to Both of these are prime so they satisfy the restraints. Thus there is solution.
Video Solution
https://youtu.be/5QdPQ3__a7I?t=130
~ pi_is_3.14
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.