Question
Answer
$$\dfrac{2}{\sqrt{3}-4}=-\dfrac{2\sqrt{3}+8}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{\sqrt{3}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{3}-4}\frac{\sqrt{3}+4}{\sqrt{3}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{3}+8}{3+4\sqrt{3}-4\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{3}+8}{-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{2\sqrt{3}+8}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{3} + 4\right) = \color{blue}{2} \cdot \sqrt{3}+\color{blue}{2} \cdot4 = \\ = 2 \sqrt{3} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}-4\right) } \cdot \left( \sqrt{3} + 4\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot4\color{blue}{-4} \cdot \sqrt{3}\color{blue}{-4} \cdot4 = \\ = 3 + 4 \sqrt{3}- 4 \sqrt{3}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
This page was created using
Rationalize Denominator Calculator