Question
Answer
$$\dfrac{2+\sqrt{3}}{8+\sqrt{3}}=\dfrac{13+6\sqrt{3}}{61}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+\sqrt{3}}{8+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2+\sqrt{3}}{8+\sqrt{3}}\frac{8-\sqrt{3}}{8-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16-2\sqrt{3}+8\sqrt{3}-3}{64-8\sqrt{3}+8\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{13+6\sqrt{3}}{61}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 8- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 + \sqrt{3}\right) } \cdot \left( 8- \sqrt{3}\right) = \color{blue}{2} \cdot8+\color{blue}{2} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot8+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 16- 2 \sqrt{3} + 8 \sqrt{3}-3 $$ 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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