Question
Answer
$$(4-2\sqrt{2})\cdot(3+\sqrt{8})=4+2\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4-2\sqrt{2})\cdot(3+\sqrt{8})& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(4-2\sqrt{2})\cdot(3+2\sqrt{2}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12+8\sqrt{2}-6\sqrt{2}-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4+2\sqrt{2}\end{aligned} $$ | |
| ① | $$ \sqrt{8} =
\sqrt{ 2 ^2 \cdot 2 } =
\sqrt{ 2 ^2 } \, \sqrt{ 2 } =
2 \sqrt{ 2 }$$ |
| ② | $$ \color{blue}{ \left( 4- 2 \sqrt{2}\right) } \cdot \left( 3 + 2 \sqrt{2}\right) = \color{blue}{4} \cdot3+\color{blue}{4} \cdot 2 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot3\color{blue}{- 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 12 + 8 \sqrt{2}- 6 \sqrt{2}-8 $$ |
| ③ | Combine like terms |
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