Difference between revisions of "2015 AMC 10A Problems/Problem 12"
(→Solution) |
m (→Alternate Solution) |
||
Line 35: | Line 35: | ||
<math>x^2-y=1</math> and <math>x^2-y=-1</math>. | <math>x^2-y=1</math> and <math>x^2-y=-1</math>. | ||
− | One of these <math>y</math>s is <math>a</math> and one is <math>b</math>.Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>. | + | One of these <math>y</math>'s is <math>a</math> and one is <math>b</math>. Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. |
+ | |||
+ | So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>. | ||
The answer is <math>\boxed{\textbf{(C) }2}</math> | The answer is <math>\boxed{\textbf{(C) }2}</math> |
Revision as of 14:39, 30 January 2016
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
Alternate Solution
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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.