Difference between revisions of "2021 AMC 10B Problems/Problem 2"
(→Solution 1) |
|||
| (42 intermediate revisions by 22 users not shown) | |||
| Line 1: | Line 1: | ||
| − | What is the value of < | + | ==Problem== |
| + | What is the value of <math>\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}</math>? | ||
<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6</math> | <math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6</math> | ||
| + | |||
| + | ==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 <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 | ||
| + | |||
| + | ==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 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/HHVdPTLQsLc | ||
| + | ~Math Python | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/Df3AIGD78xM | ||
| + | |||
| + | ~pi_is_3.14 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/v71C6cFbErQ | ||
| + | |||
| + | ~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== | ||
| + | https://youtu.be/3GHD62FK0xY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:09, 2 October 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. 
