Question
Answer
$$\dfrac{2\sqrt{2}}{2-\sqrt{8}}=-\sqrt{2}-2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}}{2-\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{2}}{2-\sqrt{8}}\frac{2+\sqrt{8}}{2+\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{2}+8}{4+4\sqrt{2}-4\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{2}+8}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{2}+2}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{2}+2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(\sqrt{2}+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\sqrt{2}-2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{2} } \cdot \left( 2 + \sqrt{8}\right) = \color{blue}{ 2 \sqrt{2}} \cdot2+\color{blue}{ 2 \sqrt{2}} \cdot \sqrt{8} = \\ = 4 \sqrt{2} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{8}\right) } \cdot \left( 2 + \sqrt{8}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot2\color{blue}{- \sqrt{8}} \cdot \sqrt{8} = \\ = 4 + 4 \sqrt{2}- 4 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Place a negative sign in front of a fraction. |
| ⑥ | Remove the parenthesis by changing the sign of each term within them. |
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