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Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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==Solution==
 
==Solution==
  
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We use quadratic formula, yeilding <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{(D)}16}</math>
+
We use quadratic formula, yeilding <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>
  
 
==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 18:32, 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

We use quadratic formula, yeilding $x=\frac{a\pm \sqrt{a^2-8a}}{2}$. We immediately see that $a^2-8a$ must be a perfect square in order for the solution to be rational. Thus, $a(a-8)$ is a perfect square. Note that if $a$ is more than $16$ or less than $-8$, thus value cannot possibly be a perfect square. Trying all the values in between, $-1$, $8$, and $9$ work. Their sum yeilds $\boxed{\textbf{(C)}\ 16}$

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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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