Difference between revisions of "2021 AMC 10B Problems/Problem 2"
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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
https://youtu.be/Df3AIGD78xM | https://youtu.be/Df3AIGD78xM | ||
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+ | ==Video Solution 3== | ||
+ | https://youtu.be/v71C6cFbErQ | ||
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+ | ~savannahsolver | ||
==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{(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 | ||
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}} |
Revision as of 20:39, 16 February 2021
Contents
Problem
What is the value of
Solution
Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that is actually negative, thus the absolute value is not but . So the first term equals and the second term is .
Summed up you get ~bjc and abhinavg0627
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
Video Solution by OmegaLearn
Video Solution 3
~savannahsolver
Solution 2
Let , then . The term is there due to difference of squares. Simplifying the expression gives us , so ~ shrungpatel
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 |