Question
Answer
$$\sqrt{27}^2-4\sqrt{4}^2=11$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{27}^2-4\sqrt{4}^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3\sqrt{3})^2-4\cdot2^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}27-4\cdot4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}11\end{aligned} $$ | |
| ① | $$ \sqrt{27} =
\sqrt{ 3 ^2 \cdot 3 } =
\sqrt{ 3 ^2 } \, \sqrt{ 3 } =
3 \sqrt{ 3 }$$ |
| ② | $$ \sqrt{4} = 2 $$ |
| ③ | $$ (3\sqrt{3})^2 =
3^{ 2 } \cdot \sqrt{3} ^ { 2 } =
3^{ 2 } \sqrt{3} ^2 =
3^{ 2 } \lvert 3 \rvert =
27 $$$ 2 ^ 2 = 4 $ |
| ④ | Combine like terms |
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