Difference between revisions of "2015 AMC 10A Problems/Problem 23"
Line 22: | Line 22: | ||
We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>. So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>. So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>r_1</math> and <math>r_2</math> be the roots of the equation, then by vietas we have that <math>r_1</math>+<math>r_2</math> = a and <math>r_1</math><math>r_2</math> = 2a. Writing 2a/a in terms of the roots (and cleaning up some algebra) we get that 0 = 2<math>r_1</math>+2<math>r_2</math>-<math>r_1</math><math>r_2</math>. We can now use Simon's Favorite Factoring trick to nicely factor this into (<math>r_1</math>-2)(<math>r_2</math>-2) = 4. We now try these possible pairs for the binomial factors: (1,4),(-1,-4),(2,2), and (-2,-2). After computing all the possible roots from these pairs we get that the sum of all possible values of a is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | ||
+ | |||
+ | Note: This problem, or at least the method I used to solve this, is really just asking what is the sum of possible sums of the roots given in a quadratic form | ||
+ | |||
+ | ~triggod. | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Revision as of 23:17, 26 September 2021
Contents
Problem
The zeroes of the function are integers. What is the sum of the possible values of
Solution 1
By Vieta's Formula, is the sum of the integral zeros of the function, and so is integral.
Because the zeros are integral, the discriminant of the function, , is a perfect square, say . Then adding 16 to both sides and completing the square yields Therefore and Let and ; then, and so . Listing all possible pairs (not counting transpositions because this does not affect (), , yields . These sum to , so our answer is .
Solution 2
Let and be the integer zeroes of the quadratic. Since the coefficient of the term is , the quadratic can be written as
By comparing this with ,
Plugging the first equation in the second, Rearranging gives These factors can be or
We want the number of distinct , and these factors gives . So the answer is .
Solution 3
Let and be the roots of the equation, then by vietas we have that + = a and = 2a. Writing 2a/a in terms of the roots (and cleaning up some algebra) we get that 0 = 2+2-. We can now use Simon's Favorite Factoring trick to nicely factor this into (-2)(-2) = 4. We now try these possible pairs for the binomial factors: (1,4),(-1,-4),(2,2), and (-2,-2). After computing all the possible roots from these pairs we get that the sum of all possible values of a is .
Note: This problem, or at least the method I used to solve this, is really just asking what is the sum of possible sums of the roots given in a quadratic form
~triggod.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/397
Video Solution
~savannahsolver
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.