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