Difference between revisions of "2019 AMC 10A Problems/Problem 19"

Revision as of 15:33, 18 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 1

Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ , which can be simplified to $(x^2+5x+5)^2-1+2019$ . Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$ , the answer is $\boxed{\textbf{(B) } 2018}$ .

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

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 $2019-1=\boxed{\textbf{(B) }2018}$ .

Solution 3 (using calculus)

Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$ .

Letting $y=x^2+5x$ , we get the expression $(y+4)(y+6)+2019$ . Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:

$\frac{d}{dx}(y^2+10y+24)=0$

$2y+10=0$

$2y(y+5)=0$

$y=-5,0$

To minimize the result, we use $y=-5$ . Hence, the minimum is $(-5+4)(-5+6)=-1$ , so $-1+2019 = \boxed{\textbf{(B) }2018}$ .

Note: The minimum/maximum point of a parabola $y = ax^2 + bx + c$ occurs at $x=-\frac{b}{2a}$ .

Solution 4

The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x -3$ . Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$ , which is very close to $-1$ . $-1 + 2019 = \boxed{\textbf{(B) }2018}$ .

Solution 5 (using the answer choices)

Using the answer choices, we know that $\textbf{(C)}$ , $\textbf{(D)}$ , and $\textbf{(E)}$ are impossible since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when $x = -\frac{3}{2}$ ). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$ , so round this to $\boxed{\textbf{(B) }2018}$ .

Video Solution

For those who want a video solution: https://www.youtube.com/watch?v=Mfa7j2BoNjI

@@ Line 16: / Line 16: @@
 We can use the difference of two 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^2+\tfrac{9}{16}</math>.
-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>.
+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{\textbf{(B) }2018}</math>.
 ==Solution 3 (using calculus)==

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
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