Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
− | 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. , |
+ | By [[Vieta's Formulas]], <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath> | ||
− | + | Plugging the first equation in the second, | |
+ | <cmath>r_1r_2 = 2 (r_1 + r_2).</cmath> | ||
− | + | Rearranging gives | |
+ | <cmath>r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.</cmath> | ||
− | + | 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</math> = (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}$. | ||
Revision as of 10:06, 10 July 2024
Contents
Problem
The zeroes of the function are integers. What is the sum of the possible values of
Solution 1
By Vieta's Formula, is the sum of the integral zeros of the function, and so
is integral.
Because the zeros are integral, the discriminant of the function, , is a perfect square, say
. Then adding 16 to both sides and completing the square yields
Therefore
and
Let
and
; then,
and so
. Listing all possible
pairs (not counting transpositions because this does not affect (
),
, yields
. These
sum to
, so our answer is
.
Solution 2
Let and
be the integer zeroes of the quadratic. ,
By Vieta's Formulas,
Plugging the first equation in the second,
Rearranging gives
These factors (ignoring order, because we want the sum of factors), can be
or
.
The sum of distinct = (f_1+2) + (f_2+2)
\sum_a a = (5+4) + (-5+4) + (4+4) + (0+4) = \boxed{\textbf{(C) }16}$.
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.