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Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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Therefore <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>
 
<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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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==
 
==Solution 2==
  
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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>
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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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By comparing this with <math>x^2 - ax + 2a</math>, <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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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>
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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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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>.
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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>.
  
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==Solution 3==
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The sum of distinct <math>a = r_1 + r_2 = (f_1+2) + (f_2+2)</math>, and these factors give
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Let <math>r_1</math> and <math>r_2</math> be the roots of the equation, then by vietas we have that <math>r_1</math>+<math>r_2</math> = a and  <math>r_1</math><math>r_2</math> = 2a. Writing 2a/a in terms of the roots (and cleaning up some algebra) we get that 0 =  2<math>r_1</math>+2<math>r_2</math>-<math>r_1</math><math>r_2</math>. We can now use Simon's Favorite Factoring trick to nicely factor this into (<math>r_1</math>-2)(<math>r_2</math>-2) = 4. We now try these possible pairs for the binomial factors: (1,4),(-1,-4),(2,2), and (-2,-2). After computing all the possible roots from these pairs we get that the sum of all possible values of a  is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>.  
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<cmath>\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}</cmath>.
  
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Note: This problem, or at least the method I used to solve this, is really just asking what is the sum of possible sums of the roots given in a quadratic form
 
  
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~triggod.
 
  
 
=== Video Solution by Richard Rusczyk ===
 
=== Video Solution by Richard Rusczyk ===

Latest revision as of 23:44, 23 December 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 \[\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}\].


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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