Question
Answer
$$\dfrac{1}{6+\sqrt{8}}=\dfrac{3-\sqrt{2}}{14}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{6+\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{6+\sqrt{8}}\frac{6-\sqrt{8}}{6-\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-2\sqrt{2}}{36-12\sqrt{2}+12\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6-2\sqrt{2}}{28} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3-\sqrt{2}}{14}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 6- \sqrt{8}\right) = \color{blue}{1} \cdot6+\color{blue}{1} \cdot- \sqrt{8} = \\ = 6- 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 6 + \sqrt{8}\right) } \cdot \left( 6- \sqrt{8}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot6+\color{blue}{ \sqrt{8}} \cdot- \sqrt{8} = \\ = 36- 12 \sqrt{2} + 12 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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