Question
Answer
$$\dfrac{2\sqrt{2}}{2\sqrt{3}-\sqrt{8}}=\sqrt{6}+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}}{2\sqrt{3}-\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{2}}{2\sqrt{3}-\sqrt{8}}\frac{2\sqrt{3}+\sqrt{8}}{2\sqrt{3}+\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{6}+8}{12+4\sqrt{6}-4\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{6}+8}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{6}+2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{6}+2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3} + \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{2} } \cdot \left( 2 \sqrt{3} + \sqrt{8}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{2}} \cdot \sqrt{8} = \\ = 4 \sqrt{6} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3}- \sqrt{8}\right) } \cdot \left( 2 \sqrt{3} + \sqrt{8}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot 2 \sqrt{3}\color{blue}{- \sqrt{8}} \cdot \sqrt{8} = \\ = 12 + 4 \sqrt{6}- 4 \sqrt{6}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Remove 1 from denominator. |
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