Answer
$$2\sqrt{27}=6\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2\sqrt{27}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2\cdot \sqrt{ 9 \cdot 3 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2\cdot \sqrt{ 9 } \cdot \sqrt{ 3 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2\cdot3 \sqrt{ 3 } \xlongequal{ } \\[1 em] & \xlongequal{ }6\sqrt{3}\end{aligned} $$ | |
| ① | Factor out the largest perfect square of 27. ( in this example we factored out $ 9 $ ) |
| ② | Rewrite $ \sqrt{ 9 \cdot 3 } $ as the product of two radicals. |
| ③ | The square root of $ 9 $ is $ 3 $. |
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