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