Answer
$$\frac{4}{8-\sqrt{3}}=\frac{32+4\sqrt{3}}{61}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{8-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{8-\sqrt{3}}\frac{8+\sqrt{3}}{8+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{32+4\sqrt{3}}{64+8\sqrt{3}-8\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{32+4\sqrt{3}}{61}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8 + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 8 + \sqrt{3}\right) = \color{blue}{4} \cdot8+\color{blue}{4} \cdot \sqrt{3} = \\ = 32 + 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 8- \sqrt{3}\right) } \cdot \left( 8 + \sqrt{3}\right) = \color{blue}{8} \cdot8+\color{blue}{8} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot8\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 64 + 8 \sqrt{3}- 8 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
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