Question
Answer
$$\dfrac{\sqrt{2}-8}{\sqrt{2}-2}=7+3\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}-8}{\sqrt{2}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}-8}{\sqrt{2}-2}\frac{\sqrt{2}+2}{\sqrt{2}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2+2\sqrt{2}-8\sqrt{2}-16}{2+2\sqrt{2}-2\sqrt{2}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-14-6\sqrt{2}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-7-3\sqrt{2}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{7+3\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}7+3\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2}-8\right) } \cdot \left( \sqrt{2} + 2\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot2\color{blue}{-8} \cdot \sqrt{2}\color{blue}{-8} \cdot2 = \\ = 2 + 2 \sqrt{2}- 8 \sqrt{2}-16 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2}-2\right) } \cdot \left( \sqrt{2} + 2\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot2\color{blue}{-2} \cdot \sqrt{2}\color{blue}{-2} \cdot2 = \\ = 2 + 2 \sqrt{2}- 2 \sqrt{2}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
| ⑥ | Remove 1 from denominator. |
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