Question
Answer
$$\dfrac{6}{-9+\sqrt{3}}=\dfrac{-9-\sqrt{3}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{-9+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{-9+\sqrt{3}}\frac{-9-\sqrt{3}}{-9-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-54-6\sqrt{3}}{81+9\sqrt{3}-9\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-54-6\sqrt{3}}{78} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-9-\sqrt{3}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -9- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( -9- \sqrt{3}\right) = \color{blue}{6} \cdot-9+\color{blue}{6} \cdot- \sqrt{3} = \\ = -54- 6 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( -9 + \sqrt{3}\right) } \cdot \left( -9- \sqrt{3}\right) = \color{blue}{-9} \cdot-9\color{blue}{-9} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-9+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 81 + 9 \sqrt{3}- 9 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
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