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