Question
Answer
$$\dfrac{2}{-4+\sqrt{3}}=\dfrac{-8-2\sqrt{3}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{-4+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{-4+\sqrt{3}}\frac{-4-\sqrt{3}}{-4-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-8-2\sqrt{3}}{16+4\sqrt{3}-4\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-8-2\sqrt{3}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -4- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( -4- \sqrt{3}\right) = \color{blue}{2} \cdot-4+\color{blue}{2} \cdot- \sqrt{3} = \\ = -8- 2 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( -4 + \sqrt{3}\right) } \cdot \left( -4- \sqrt{3}\right) = \color{blue}{-4} \cdot-4\color{blue}{-4} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-4+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 16 + 4 \sqrt{3}- 4 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
This page was created using
Rationalize Denominator Calculator