Difference between revisions of "2015 AMC 10A Problems/Problem 12"
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Thus, the answer is <math>\boxed{\textbf{(C) }2}</math> | Thus, the answer is <math>\boxed{\textbf{(C) }2}</math> | ||
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+ | -DuckDuckGooseGoose | ||
==Video Solution== | ==Video Solution== |
Revision as of 22:46, 3 November 2021
Problem
Points and
are distinct points on the graph of
. What is
?
Solution 1
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 closely related to Solution #1 and 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
Solution 3
This solution is similar to Solution #1 but uses a different way to find at the end.
Just like Solution #1, we arrive at the conclusion that .
Simplifying we get:
We now can factor this quadratic. We must find two terms that multiply to and add to
.
These terms are and
.
Subtracting one from the other, we get .
Thus, the answer is
-DuckDuckGooseGoose
Video Solution
~savannahsolver
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.