Difference between revisions of "2015 AMC 10A Problems/Problem 23"

Latest revision as of 09:07, 10 July 2024

Problem

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution 1

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.

Because the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields $(a - 4)^2 = k^2 + 16.$ Therefore $(a-4)^2 - k^2 = 16$ and $((a-4) - k)((a-4) + k) = 16.$ Let $(a-4) - k = u$ and $(a-4) + k = v$ ; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$ . Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$ ), $(2, 8), (4, 4), (-2, -8), (-4, -4)$ , yields $a = 9, 8, -1, 0$ . These $a$ sum to $16$ , so our answer is $\boxed{\textbf{(C) }16}$ .

Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. , By Vieta's Formulas, $r_1 + r_2 = a\text{ and }r_1r_2 = 2a.$

Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2).$

Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.$

These factors $(f_1,f_2)$ (ignoring order, because we want the sum of factors), can be $(1, 4), (-1, -4), (2, 2),$ or $(-2, -2)$ .

The sum of distinct $a = r_1 + r_2 = (f_1+2) + (f_2+2)$ , and these factors give $\sum_a a = (5+4) + (-5+4) + (4+4) + (0+4) = \boxed{\textbf{(C) }16}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc10a/397

Video Solution

https://youtu.be/RQ4ZCttwmA4

~savannahsolver

@@ Line 15: / Line 15: @@
 ==Solution 2==
-Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. Since the coefficient of the <math>x^2</math> term is <math>1</math>, the quadratic can be written as <cmath>(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2</cmath>
+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>
-By comparing this with <math>x^2 - ax + 2a</math>, <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>
-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 can be<math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),</math> or <math>(-2, -2).</math>
+Rearranging gives
+<cmath>r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.</cmath>
-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>.
+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 = (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>.
+=== Video Solution by Richard Rusczyk ===
+https://artofproblemsolving.com/videos/amc/2015amc10a/397
+==Video Solution==
+https://youtu.be/RQ4ZCttwmA4
+~savannahsolver
 ==See Also==

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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