Answer
$$\frac{6}{\sqrt{3}+4}=\frac{-6\sqrt{3}+24}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{\sqrt{3}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{\sqrt{3}+4}\frac{\sqrt{3}-4}{\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{3}-24}{3-4\sqrt{3}+4\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{3}-24}{-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-6\sqrt{3}+24}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( \sqrt{3}-4\right) = \color{blue}{6} \cdot \sqrt{3}+\color{blue}{6} \cdot-4 = \\ = 6 \sqrt{3}-24 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 4\right) } \cdot \left( \sqrt{3}-4\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-4+\color{blue}{4} \cdot \sqrt{3}+\color{blue}{4} \cdot-4 = \\ = 3- 4 \sqrt{3} + 4 \sqrt{3}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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