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