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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

Problem

What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\]where $x$ is a real number?

$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ which can be simplified as $(x^2+5x+5)^2-1+2019$. Since squares are nonnegative, the answer is $\boxed{(B) 2018}$

Solution 2

Let $a=x+\tfrac{5}{2}$. Then $(x+1)(x+2)(x+3)(x+4)$ becomes $(a-\tfrac{3}{2})(a-\tfrac{1}{2})(a+\tfrac{1}{2})(a+\tfrac{3}{2})$


We can use difference of squares to get $(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})$, and expand this to get $a^4-\tfrac{5}{2}a+\frac{9}{16}$.


Refactor this by completing the square to get $(a^2-\tfrac{5}{4})^2-1$, which has a minimum value of $-1$. The answer is thus $2019-1=\boxed{2018}$

-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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