Question
Answer
$$\dfrac{2}{2\sqrt{3}-4}=-\sqrt{3}-2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{2\sqrt{3}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{2\sqrt{3}-4}\frac{2\sqrt{3}+4}{2\sqrt{3}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{3}+8}{12+8\sqrt{3}-8\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{3}+8}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}+2}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{3}+2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(\sqrt{3}+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\sqrt{3}-2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( 2 \sqrt{3} + 4\right) = \color{blue}{2} \cdot 2 \sqrt{3}+\color{blue}{2} \cdot4 = \\ = 4 \sqrt{3} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3}-4\right) } \cdot \left( 2 \sqrt{3} + 4\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot4\color{blue}{-4} \cdot 2 \sqrt{3}\color{blue}{-4} \cdot4 = \\ = 12 + 8 \sqrt{3}- 8 \sqrt{3}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Place a negative sign in front of a fraction. |
| ⑥ | Remove the parenthesis by changing the sign of each term within them. |
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