Question
Answer
$$2\sqrt{27}-\sqrt{48}+4\sqrt{12}=10\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2\sqrt{27}-\sqrt{48}+4\sqrt{12}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6\sqrt{3}-4\sqrt{3}+8\sqrt{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}10\sqrt{3}\end{aligned} $$ | |
| ① | $$ 2 \sqrt{27} =
2 \sqrt{ 3 ^2 \cdot 3 } =
2 \sqrt{ 3 ^2 } \, \sqrt{ 3 } =
2 \cdot 3 \sqrt{ 3 } =
6 \sqrt{ 3 } $$ |
| ② | $$ - \sqrt{48} =
- \sqrt{ 4 ^2 \cdot 3 } =
- \sqrt{ 4 ^2 } \, \sqrt{ 3 } =
- 4 \sqrt{ 3 }$$ |
| ③ | $$ 4 \sqrt{12} =
4 \sqrt{ 2 ^2 \cdot 3 } =
4 \sqrt{ 2 ^2 } \, \sqrt{ 3 } =
4 \cdot 2 \sqrt{ 3 } =
8 \sqrt{ 3 } $$ |
| ④ | Combine like terms |
This page was created using
Simplify Radical Expression