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

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==Problem==
 
==Problem==
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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?
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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 <math>a?</math>
  
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<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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<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 1==
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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.
  
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==Solution==
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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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Therefore <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 <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>.
  
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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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==Solution 2==
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Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. ,
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By [[Vieta's Formulas]], <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath>
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Plugging the first equation in the second,
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<cmath>r_1r_2 = 2 (r_1 + r_2).</cmath>  
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Rearranging gives
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<cmath>r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.</cmath>
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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>.
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=== Video Solution by Richard Rusczyk ===
 +
 
 +
https://artofproblemsolving.com/videos/amc/2015amc10a/397
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==Video Solution==
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https://youtu.be/RQ4ZCttwmA4
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
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{{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Intermediate Number Theory Problems]]

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

See Also

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

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