Difference between revisions of "2021 AMC 10B Problems/Problem 2"
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~bjc, abhinavg0627 and JackBocresion | ~bjc, abhinavg0627 and JackBocresion | ||
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+ | ==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 | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 13:18, 22 February 2021
Contents
Problem
What is the value of
Solution
Note that the square root of any square is always the absolute value of the squared number because the square root function will only return a positive 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. Simplifying the expression gives us , so ~ shrungpatel
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
Video Solution by OmegaLearn
Video Solution 3
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=154
~IceMatrix
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 |