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