Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{2}{\sqrt{3}-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{3}-\sqrt{5}}\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{3}+2\sqrt{5}}{3+\sqrt{15}-\sqrt{15}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{3}+2\sqrt{5}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}+\sqrt{5}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{3}+\sqrt{5}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(\sqrt{3}+\sqrt{5}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\sqrt{3}-\sqrt{5}\end{aligned} $$ | |
① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + \sqrt{5}} $$. |
② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{3} + \sqrt{5}\right) = \color{blue}{2} \cdot \sqrt{3}+\color{blue}{2} \cdot \sqrt{5} = \\ = 2 \sqrt{3} + 2 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}- \sqrt{5}\right) } \cdot \left( \sqrt{3} + \sqrt{5}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot \sqrt{3}\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 3 + \sqrt{15}- \sqrt{15}-5 $$ |
③ | Simplify numerator and denominator |
④ | Divide both numerator and denominator by 2. |
⑤ | Place a negative sign in front of a fraction. |
⑥ | Remove the parenthesis by changing the sign of each term within them. |