Question
Answer
$$\dfrac{9}{-2+3\sqrt{3}}=\dfrac{18+27\sqrt{3}}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9}{-2+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{-2+3\sqrt{3}}\frac{-2-3\sqrt{3}}{-2-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-18-27\sqrt{3}}{4+6\sqrt{3}-6\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-18-27\sqrt{3}}{-23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{18+27\sqrt{3}}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -2- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 9 } \cdot \left( -2- 3 \sqrt{3}\right) = \color{blue}{9} \cdot-2+\color{blue}{9} \cdot- 3 \sqrt{3} = \\ = -18- 27 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( -2 + 3 \sqrt{3}\right) } \cdot \left( -2- 3 \sqrt{3}\right) = \color{blue}{-2} \cdot-2\color{blue}{-2} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot-2+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 4 + 6 \sqrt{3}- 6 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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