Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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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 to <math>(x^2+5x+5)^2-1+2019</math>. Noting that squares are nonnegative, and verifying that <math>x^2+5x+5=0</math> for some real <math>x</math>, the answer is <math>\boxed{\textbf{(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 to <math>(x^2+5x+5)^2-1+2019</math>. Noting that squares are nonnegative, and verifying that <math>x^2+5x+5=0</math> for some real <math>x</math>, the answer is <math>\boxed{\textbf{(B) } 2018}</math>. | ||
+ | |||
+ | Better to understand solution by Williamgolly: | ||
+ | After we factor into <math>(x^2+5x+4)(x^2+5+6)+2019</math>, let <math>y=x^2+5x+5</math> and the forner expression turns to <math>(y-1)(y+1)+2019=y^2+2018</math>. Clearly, <math>y^2</math> is non negative, so the minimum is achieved when <math>y=0</math> and the expression equals 2018 | ||
==Solution 2== | ==Solution 2== |
Revision as of 13:00, 2 January 2020
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 to
. Noting that squares are nonnegative, and verifying that
for some real
, the answer is
.
Better to understand solution by Williamgolly:
After we factor into , let
and the forner expression turns to
. Clearly,
is non negative, so the minimum is achieved when
and the expression equals 2018
Solution 2
Let . Then the expression
becomes
.
We can now use the difference of two 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
.
Solution 3 (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
.
Note: We could also have used the result that minimum/maximum point 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
. Thus the answer is
.
Solution 5 (using the answer choices)
Answer choices ,
, and
are impossible, since
can be negative (as seen when e.g.
). Plug in
to see that it becomes
, so round this to
.
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
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 | |
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.