2015 AMC 10A Problems/Problem 12
Contents
[hide]Problem
Points and are distinct points on the graph of . What is ?
Solution
Since points on the graph make the equation true, substitute in to the equation and then solve to find and .
There are only two solutions to the equation, so one of them is the value of and the other is . The order does not matter because of the absolute value sign.
The answer is
Solution #2
This solution is very related to Solution #1 but just simplifies the problem earlier to make it easier.
can be written as . Recognizing that this is a binomial square, simplify this to . This gives us two equations:
and .
One of these 's is and one is . Substituting for , we get and .
So, .
The answer is
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AMC 10 Problems and Solutions |
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