Difference between revisions of "2021 AMC 10B Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
− | Note that the square root of any | + | 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 <math>3</math> and <math>2\sqrt{3}</math>, we see that <math>2\sqrt{3}>3</math>, thus <math>3-2\sqrt{3}</math> is negative, so we must take the absolute value of <math>3-2\sqrt{3}</math>, which is just <math>2\sqrt{3}-3</math>. Knowing this, the first term in the expression equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>, and summing the two gives <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math>. |
~bjc, abhinavg0627 and JackBocresion | ~bjc, abhinavg0627 and JackBocresion | ||
==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 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 |
==Video Solution== | ==Video Solution== | ||
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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
https://youtu.be/Df3AIGD78xM | https://youtu.be/Df3AIGD78xM | ||
+ | |||
+ | ~pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
− | https://youtu.be/DvpN56Ob6Zw?t= | + | https://youtu.be/DvpN56Ob6Zw?t=1 |
~Interstigation | ~Interstigation | ||
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==Video Solution by Mathematical Dexterity (50 Seconds)== | ==Video Solution by Mathematical Dexterity (50 Seconds)== | ||
https://www.youtube.com/watch?v=ScZ5VK7QTpY | https://www.youtube.com/watch?v=ScZ5VK7QTpY | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/3GHD62FK0xY | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:49, 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.