Question
Answer
$$\dfrac{6-3\sqrt{3}}{6+4\sqrt{3}}=\dfrac{-12+7\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6-3\sqrt{3}}{6+4\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6-3\sqrt{3}}{6+4\sqrt{3}}\frac{6-4\sqrt{3}}{6-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36-24\sqrt{3}-18\sqrt{3}+36}{36-24\sqrt{3}+24\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{72-42\sqrt{3}}{-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12-7\sqrt{3}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-12+7\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- 4 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 6- 3 \sqrt{3}\right) } \cdot \left( 6- 4 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 4 \sqrt{3}\color{blue}{- 3 \sqrt{3}} \cdot6\color{blue}{- 3 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 36- 24 \sqrt{3}- 18 \sqrt{3} + 36 $$ Simplify denominator. $$ \color{blue}{ \left( 6 + 4 \sqrt{3}\right) } \cdot \left( 6- 4 \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot6+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 36- 24 \sqrt{3} + 24 \sqrt{3}-48 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
| ⑤ | Multiply both numerator and denominator by -1. |
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