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

Latest revision as of 00:28, 27 September 2021

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

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

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 can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$

We want the number of distinct $a = r_1 + r_2$ , and these factors gives $a = -1, 0, 8, 9$ . So the answer is $-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}$ .

Video Solution by Richard Rusczyk

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

Video Solution

https://youtu.be/RQ4ZCttwmA4

~savannahsolver

@@ Line 23: / Line 23: @@
 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>.
-==Solution 3==
-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>.
-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
-~triggod.
 === Video Solution by Richard Rusczyk ===

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