Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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These factors <math>(f_1,f_2)</math> (ignoring order, because we want the sum of factors), can be <math>(1, 4), (-1, -4), (2, 2),</math> or <math>(-2, -2)</math>. | These factors <math>(f_1,f_2)</math> (ignoring order, because we want the sum of factors), can be <math>(1, 4), (-1, -4), (2, 2),</math> or <math>(-2, -2)</math>. | ||
− | The sum of distinct <math>a = r_1 + r_2 | + | The sum of distinct <math>a = r_1 + r_2 = (f_1+2) + (f_2+2)</math>, and these factors give <math>\sum_a a = (5+4) + (-5+4) + (4+4) + (0+4) = \boxed{\textbf{(C) }16}</math>. |
Revision as of 09:07, 10 July 2024
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. , By Vieta's Formulas,
Plugging the first equation in the second,
Rearranging gives
These factors (ignoring order, because we want the sum of factors), can be or .
The sum of distinct , and these factors give .
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.