2015 AMC 10A Problems/Problem 23: Difference between revisions

Latest revision as of 10:55, 26 October 2025

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 $\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}$ .

Video Solution by Richard Rusczyk

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

Solution 3

Let the roots be $r$ and $s$ . By Vieta's we have that $r+s = a$ , and $rs = 2a$ . So by Simon's Favorite Factoring Trick: $2r+2s = rs \Longrightarrow rs-2r-2s = 0 \Longrightarrow (r-2)(s-2) = 4.$ Now, we test out values of $r$ and $s$ to see which work. Lets say that $r-2 = 1$ , and $s-2 = 4$ . This implies that $(r,s) = (3,6)$ , so $a = 9$ . Now, we let $r-2 = 2$ , and $s-2 = 2$ , this gives us $(r,s) = (4,4)$ , which gives us a value for $a$ of $8$ . Now, we circle an answer of $17$ and move onto the next problem. Yike! They didn't say that $a$ can't be negative! This is where many people would go wrong. Let $r-2 = -1$ , and $s-2 = -4$ . This gives us that $(r,s) = (1,-2)$ . This results in a value of $-1$ for $a$ . Now, we let $r-2 = -2$ , and $s-2 = -2$ . This gives $a = 0$ . Now, we are done. Our total sum is $9+8-1 = \boxed{\textbf{(C) }16}$ .

-jb2015007 - minor edits by mathlover1205 and PojoDotCom

Video Solution

https://youtu.be/RQ4ZCttwmA4

~savannahsolver

@@ Line 1: / Line 1: @@
 ==Problem==
-The zeroes of the function <math>f(x)=x^2-ax+2a</math> are integers .What is the sum of the possible values of a?
+The zeroes of the function <math>f(x)=x^2-ax+2a</math> are integers. What is the sum of the possible values of <math>a?</math>
-<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18</math>
+<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math>
 ==Solution 1==
@@ Line 10: / Line 9: @@
 Because the zeros are integral, the discriminant of the function, <math>a^2 - 8a</math>, is a perfect square, say <math>k^2</math>. Then adding 16 to both sides and completing the square yields
 <cmath>(a - 4)^2 = k^2 + 16.</cmath>
-Hence <math>(a-4)^2 - k^2 = 16</math> and
+Therefore <math>(a-4)^2 - k^2 = 16</math> and
 <cmath>((a-4) - k)((a-4) + k) = 16.</cmath>
-Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect <math>u + v</math>), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to 16, so our answer is <math>\textbf{(C)}</math>.
+Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect (<math>u + v</math>)), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>.
 ==Solution 2==
 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 = (f_1+2) + (f_2+2)</math>, and these factors give
+<cmath>\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}</cmath>.
-Since the coefficent 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>.
+=== Video Solution by Richard Rusczyk ===
-By comparing this with <math>x^2 - ax + a^2</math>, <math>r_1 + r_2 = a</math> and <math>r_1r_2 = 2a</math>.
+https://artofproblemsolving.com/videos/amc/2015amc10a/397
-Plugging the first equation in the second, <math>r_1r_2 = 2 (r_1 + r_2)</math>. Rearranging gives <math>r_1r_2 - 2r_1 - 2r_2 = 0</math>.
+==Solution 3==
-This can be factored as <math>(r_1 - 2)(r_2 - 2) = 4</math>.
+Let the roots be <math>r</math> and <math>s</math>. By Vieta's we have that <math>r+s = a</math>, and <math>rs = 2a</math>. So by Simon's Favorite Factoring Trick: <cmath>2r+2s = rs \Longrightarrow rs-2r-2s = 0 \Longrightarrow (r-2)(s-2) = 4.</cmath> Now, we test out values of <math>r</math> and <math>s</math> to see which work. Lets say that <math>r-2 = 1</math>, and <math>s-2 = 4</math>. This implies that <math>(r,s) = (3,6)</math>, so <math>a = 9</math>. Now, we let <math>r-2 = 2</math>, and <math>s-2 = 2</math>, this gives us <math>(r,s) = (4,4)</math>, which gives us a value for <math>a</math> of <math>8</math>. Now, we circle an answer of <math>17</math> and move onto the next problem. Yike! They didn't say that <math>a</math> can't be negative! This is where many people would go wrong. Let <math>r-2 = -1</math>, and <math>s-2 = -4</math>. This gives us that <math>(r,s) = (1,-2)</math>. This results in a value of <math>-1</math> for <math>a</math>. Now, we let <math>r-2 = -2</math>, and <math>s-2 = -2</math>. This gives <math>a = 0</math>. Now, we are done. Our total sum is <math>9+8-1 = \boxed{\textbf{(C) }16}</math>.
-These factors can be: <math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)</math>.
+-jb2015007 - minor edits by mathlover1205 and PojoDotCom
-We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>.
+==Video Solution==
+https://youtu.be/RQ4ZCttwmA4
-So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>.
+~savannahsolver
 ==See Also==
-{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
+{{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}}
 {{MAA Notice}}
+[[Category: Intermediate Number Theory Problems]]

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