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