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

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==Problem==
 
==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?
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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?
  
 
<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>
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==Solution==
 
==Solution==
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By Vieta's Formula, <math>a</math> is the sum of the integral zeros of the function, and so <math>a</math> is integral.
  
We use quadratic formula, yielding <math>x=\frac{a\pm \sqrt{a^2-8a}}{2}</math>. We immediately see that <math>a^2-8a</math> must be a perfect square in order for the solution to be rational. Thus, <math>a(a-8)</math> is a perfect square. Note that if <math>a</math> is more than <math>16</math> or less than <math>-8</math>, thus value cannot possibly be a perfect square. Trying all the values in between, <math>-1</math>, <math>8</math>, and <math>9</math> work. Their sum yeilds <math>\boxed{\textbf{(C)}\ 16}</math>
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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
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<cmath>(a - 4)^2 = k^2 + 16.</cmath>
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Hence <math>(a-4)^2 - k^2 = 16</math> and
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<cmath>((a-4) - k)((a-4) + k) = 16.</cmath>
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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>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:33, 4 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

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 $a = 9, 8, -1, 0$. These $a$ sum to 16, so our answer is $\textbf{(C)}$.

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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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