2019 AMC 10A Problems/Problem 19
Contents
Problem
What is the least possible value of
where
is a real number?
Solution 1
Grouping the first and last terms and two middle terms gives , which can be simplified as
. Since squares are nonnegative, the answer is
.
Solution 2
Let . Then
becomes
.
We can use difference of squares to get , and expand this to get
.
Refactor this by completing the square to get , which has a minimum value of
. The answer is thus
.
-WannabeCharmander
Solution 3 (using calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get .
Letting , we get the expression
. Now, we can find the critical points of
to minimize the function:
To minimize the result, we use . Hence, the minimum is
, so
.
(inspired by solution by oO8715_alexOo)
Note: The minimum/maximum of a parabola occurs at .
Solution 4
The expression is negative when an odd number of the factors are negative. This happens when or
. Plugging in
or
yields
, which is very close to
.
.
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=Mfa7j2BoNjI
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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