Question
Answer
$$\dfrac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}=\dfrac{-\sqrt{6}+3\sqrt{2}}{8}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}-6\sqrt{2}}{8-8\sqrt{3}+8\sqrt{3}-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}-6\sqrt{2}}{-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{6}-3\sqrt{2}}{-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-\sqrt{6}+3\sqrt{2}}{8}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2}- 2 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 2 \sqrt{2}- 2 \sqrt{6}\right) = \color{blue}{ \sqrt{3}} \cdot 2 \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot- 2 \sqrt{6} = \\ = 2 \sqrt{6}- 6 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2} + 2 \sqrt{6}\right) } \cdot \left( 2 \sqrt{2}- 2 \sqrt{6}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{6}+\color{blue}{ 2 \sqrt{6}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{6}} \cdot- 2 \sqrt{6} = \\ = 8- 8 \sqrt{3} + 8 \sqrt{3}-24 $$ |
| 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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