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

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<cmath>((a-4) - k)((a-4) + k) = 16.</cmath>
 
<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>. However, <math>a = 8</math> and <math>a=0</math> do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only <math>a</math> that work are <math>a=9</math> and <math>a=-1</math>. These <math>a</math> sum to <math>8</math>, so our answer is <math>\textbf{(B)}</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>. However, <math>a = 8</math> and <math>a=0</math> do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only <math>a</math> that work are <math>a=9</math> and <math>a=-1</math>. These <math>a</math> sum to <math>8</math>, so our answer is <math>\textbf{(B)}</math>.
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FOR THE WRITER OF THIS SOLUTION: Generally, roots that are double roots are still counted as "two" roots by multiplicity. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3735612&#p3735612 (the forum discussion for this problem). The answer should be C as shown below. I won't edit your solution; instead I'll let you edit your own solution how you want it.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 12:00, 5 February 2015

Problem

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

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. Unknown error_msg)


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 \[(a - 4)^2 = k^2 + 16.\] Hence $(a-4)^2 - k^2 = 16$ 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 $a = 9, 8, -1, 0$. However, $a = 8$ and $a=0$ do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only $a$ that work are $a=9$ and $a=-1$. These $a$ sum to $8$, so our answer is $\textbf{(B)}$.

FOR THE WRITER OF THIS SOLUTION: Generally, roots that are double roots are still counted as "two" roots by multiplicity. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3735612&#p3735612 (the forum discussion for this problem). The answer should be C as shown below. I won't edit your solution; instead I'll let you edit your own solution how you want it.

Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic.

Since the coefficent of the $x^2$ term is $1$, the quadratic can be written as $(x - r_1)(x - r_2)$ or $x^2 - (r_1 + r_2)x + r_1r_2)$.

By comparing this with $x^2 - ax + 2a$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$.

Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2)$. Rearranging gives $r_1r_2 - 2r_1 - 2r_2 = 0$.

This can be factored as $(r_1 - 2)(r_2 - 2) = 4$.

These factors can be: $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$.

We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = -1, 0, 8, 9$.

So the answer is $-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}$.

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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