2021 AMC 10B Problems/Problem 2
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Memorizing Roots)
- 5 Video Solution
- 6 Video Solution (EASY TO UNDERSTAND📈)
- 7 Video Solution by OmegaLearn
- 8 Video Solution
- 9 Video Solution by TheBeautyofMath
- 10 Video Solution by Interstigation
- 11 Video Solution by Mathematical Dexterity (50 Seconds)
- 12 Video Solution
- 13 See Also
Problem
What is the value of ?
Solution 1
Note that the square root of any number squared is always the absolute value of the squared number because the square root function will only return a nonnegative number. By squaring both and
, we see that
, thus
is negative, so we must take the absolute value of
, which is just
. Knowing this, the first term in the expression equals
and the second term is
, and summing the two gives
.
~bjc, abhinavg0627 and JackBocresion
Solution 2
Let , then
. The
term is there due to difference of squares when you simplify
from
. Simplifying the expression gives us
, so
~ shrungpatel
Solution 3 (Memorizing Roots)
Memorizing your square roots from 1 - 10 are really important for cheesing some AMC problems, so try to memorize them.
\( \sqrt{3} \) is about 1.7.
We then substitute \( \sqrt{3} \) for 1.7 to solve this.
We get
Looking at the answer choices, we see that gives
for when \( \sqrt{3} = 1.7 \).
~Pinotation
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
Video Solution (EASY TO UNDERSTAND📈)
https://www.youtube.com/watch?v=A1Li_jciTZY
~CalculaCore
Video Solution by OmegaLearn
~pi_is_3.14
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=154
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=1
~Interstigation
Video Solution by Mathematical Dexterity (50 Seconds)
https://www.youtube.com/watch?v=ScZ5VK7QTpY
Video Solution
~Education, the Study of Everything
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.