2021 AMC 10B Problems/Problem 2: Difference between revisions
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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 | ||
==Solution 3 (Memorizing Roots)== | |||
Memorizing your square roots from 1 - 10 are really important for cheesing some AMC problems, so try to memorize them. | |||
\( \sqrt{3} \) is about 1.7. | |||
We then substitute \( \sqrt{3} \) for 1.7 to solve this. | |||
We get | |||
<cmath> | |||
\sqrt{(3-2 \cdot 1.7)^2} + \sqrt{(3+2 \cdot 1.7)^2} | |||
</cmath> | |||
<cmath> | |||
= \sqrt{(-0.4)^2} + \sqrt{(6.4)^2} | |||
</cmath> | |||
<cmath> | |||
= \sqrt{0.16} + \sqrt{6.4^2} | |||
</cmath> | |||
<cmath> | |||
= 0.4 + 6.4 | |||
</cmath> | |||
<cmath> | |||
= 6.8 | |||
</cmath> | |||
Looking at the answer choices, we see that <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math> gives 6.8. | |||
==Video Solution== | ==Video Solution== | ||
Revision as of 11:53, 5 September 2025
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
Solution 3 (Memorizing Roots)
Memorizing your square roots from 1 - 10 are really important for cheesing some AMC problems, so try to memorize them.
\( \sqrt{3} \) is about 1.7.
We then substitute \( \sqrt{3} \) for 1.7 to solve this.
We get
Looking at the answer choices, we see that
gives 6.8.
Video Solution
https://youtu.be/HHVdPTLQsLc ~Math Python
Video Solution (EASY TO UNDERSTAND📈)
https://www.youtube.com/watch?v=A1Li_jciTZY
~CalculaCore
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 | ||
These problems are copyrighted © by the Mathematical Association of America.