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 when you simplify <math>2ab</math> from <math>(a + b)^2</math>. 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{(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 when you simplify <math>2ab</math> from <math>(a + b)^2</math>. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel | ||
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==Video Solution== | ==Video Solution== |
Revision as of 11:49, 26 May 2024
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 , then . The term is there due to difference of squares when you simplify from . Simplifying the expression gives us , so ~ shrungpatel
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.