Difference between revisions of "2021 AMC 10B Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
| − | Let <math>x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}</math>, then <math>x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2</math>. The <math>2\sqrt{(-3)^2}</math> term is there due to difference of squares. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel | + | Let <math>x = \sqrt{\left(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3}\right)^2}</math>, then <math>x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2</math>. The <math>2\sqrt{(-3)^2}</math> term is there due to difference of squares. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel |
==Video Solution== | ==Video Solution== | ||
Revision as of 21:35, 21 October 2021
Contents
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 $x = \sqrt{\left(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3}\right)^2}$ (Error compiling LaTeX. Unknown error_msg), then 
. The 
term is there due to difference of squares. Simplifying the expression gives us 
, so 
~ shrungpatel
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
Video Solution by OmegaLearn
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=154
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=101
~Interstigation
Video Solution by Mathematical Dexterity (50 Seconds)
https://www.youtube.com/watch?v=ScZ5VK7QTpY
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
