Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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==Solution== | ==Solution== | ||
Grouping the first and last terms and two middle terms gives <math>(x^2+5x+4)(x^2+5x+6)+2019</math> which can be simplified as <math>(x^2+5x+5)^2-1+2019</math>. Since squares are nonnegative, the answer is <math>\boxed{(B) 2018}</math> | Grouping the first and last terms and two middle terms gives <math>(x^2+5x+4)(x^2+5x+6)+2019</math> which can be simplified as <math>(x^2+5x+5)^2-1+2019</math>. Since squares are nonnegative, the answer is <math>\boxed{(B) 2018}</math> | ||
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+ | ==Solution 2== | ||
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+ | Let <math>a=x+\tfrac{5}{2}</math>. Then <math>(x+1)(x+2)(x+3)(x+4)</math> becomes <math>(a-\tfrac{3}{2})(a-\tfrac{1}{2})(a+\tfrac{1}{2})(a+\tfrac{3}{2})</math> | ||
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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+\frac{9}{16}</math>. | ||
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+ | Refactor this by completing the square to get <math>(a^2-\tfrac{5}{4})^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{2018}</math> | ||
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+ | -WannabeCharmander | ||
==See Also== | ==See Also== |
Revision as of 18:13, 9 February 2019
Contents
[hide]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 |
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