Difference between revisions of "2002 AMC 12A Problems/Problem 12"
| Line 43: | Line 43: | ||
=== Note === | === Note === | ||
| − | The solution in this video may be | + | The solution in this video may be similar to others. |
==See Also== | ==See Also== | ||
Latest revision as of 08:55, 1 September 2024
- The following problem is from both the 2002 AMC 12A #12 and 2002 AMC 10A #14, so both problems redirect to this page.
Contents
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
![\[a+b=63\]](http://latex.artofproblemsolving.com/f/7/8/f78cca460ad1c36974bce73c317a853717e1571e.png)
![\[ab=k\]](http://latex.artofproblemsolving.com/e/7/c/e7c29e60814388974641c8bc4c3abc8ff4561422.png)
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
Note
All these solutions are principled in Vieta's Formulas, and can be viewed as identical.
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Note
The solution in this video may be similar to others.
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. 
