Question
Answer
$$\dfrac{3\sqrt{3}}{\sqrt{3}-2}=-9-6\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{3}}{\sqrt{3}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{3}}{\sqrt{3}-2}\frac{\sqrt{3}+2}{\sqrt{3}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9+6\sqrt{3}}{3+2\sqrt{3}-2\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9+6\sqrt{3}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{9+6\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(9+6\sqrt{3}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-9-6\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{3} } \cdot \left( \sqrt{3} + 2\right) = \color{blue}{ 3 \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot2 = \\ = 9 + 6 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}-2\right) } \cdot \left( \sqrt{3} + 2\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot2\color{blue}{-2} \cdot \sqrt{3}\color{blue}{-2} \cdot2 = \\ = 3 + 2 \sqrt{3}- 2 \sqrt{3}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
| ⑤ | Remove the parenthesis by changing the sign of each term within them. |
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