Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. | ||
− | Since the | + | Since the coefficient of the <math>x^2</math> term is <math>1</math>, the quadratic can be written as <math>(x - r_1)(x - r_2)</math> or <math>x^2 - (r_1 + r_2)x + r_1r_2</math>. |
By comparing this with <math>x^2 - ax + 2a</math>, <math>r_1 + r_2 = a</math> and <math>r_1r_2 = 2a</math>. | By comparing this with <math>x^2 - ax + 2a</math>, <math>r_1 + r_2 = a</math> and <math>r_1r_2 = 2a</math>. |
Revision as of 23:21, 19 February 2016
Contents
Problem
The zeroes of the function are integers .What is the sum of the possible values of a?
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 Hence 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 or .
By comparing this with , and .
Plugging the first equation in the second, . Rearranging gives .
This can be factored as .
These factors can be: .
We want the number of distinct , and these factors gives .
So the answer is .
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 |
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