Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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− | We can use difference of squares to get <math>(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})</math>, and expand this to get <math>a^4-\tfrac{5}{2}a+\ | + | We can use difference of squares to get <math>(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})</math>, and expand this to get <math>a^4-\tfrac{5}{2}a+\tfrac{9}{16}</math>. |
Revision as of 19:13, 9 February 2019
Contents
Problem
What is the least possible value of
where
is a real number?
Solution
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
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.