Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | + | 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> | |
==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 18:31, 4 February 2015
Problem
The zeroes of the function 
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 
. We immediately see that 
must be a perfect square in order for the solution to be rational. Thus, 
is a perfect square. Note that if 
is more than 
or less than 
, thus value cannot possibly be a perfect square. Trying all the values in between, 
, 
, and 
work. Their sum yeilds 
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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