Question
Answer
$$\dfrac{4+\sqrt{8}}{\sqrt{2}-1}=6\sqrt{2}+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+\sqrt{8}}{\sqrt{2}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+\sqrt{8}}{\sqrt{2}-1}\frac{\sqrt{2}+1}{\sqrt{2}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{2}+4+4+2\sqrt{2}}{2+\sqrt{2}-\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{2}+8}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6\sqrt{2}+8\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + \sqrt{8}\right) } \cdot \left( \sqrt{2} + 1\right) = \color{blue}{4} \cdot \sqrt{2}+\color{blue}{4} \cdot1+\color{blue}{ \sqrt{8}} \cdot \sqrt{2}+\color{blue}{ \sqrt{8}} \cdot1 = \\ = 4 \sqrt{2} + 4 + 4 + 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2}-1\right) } \cdot \left( \sqrt{2} + 1\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot1\color{blue}{-1} \cdot \sqrt{2}\color{blue}{-1} \cdot1 = \\ = 2 + \sqrt{2}- \sqrt{2}-1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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