Question
Answer
$$\dfrac{4\sqrt{3}}{\sqrt{6}-2}=\dfrac{12\sqrt{2}+8\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4\sqrt{3}}{\sqrt{6}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4\sqrt{3}}{\sqrt{6}-2}\frac{\sqrt{6}+2}{\sqrt{6}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{2}+8\sqrt{3}}{6+2\sqrt{6}-2\sqrt{6}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{2}+8\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 \sqrt{3} } \cdot \left( \sqrt{6} + 2\right) = \color{blue}{ 4 \sqrt{3}} \cdot \sqrt{6}+\color{blue}{ 4 \sqrt{3}} \cdot2 = \\ = 12 \sqrt{2} + 8 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6}-2\right) } \cdot \left( \sqrt{6} + 2\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot2\color{blue}{-2} \cdot \sqrt{6}\color{blue}{-2} \cdot2 = \\ = 6 + 2 \sqrt{6}- 2 \sqrt{6}-4 $$ |
| ③ | Simplify numerator and denominator |
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