Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{2\sqrt{3}}{7-4\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}}{7-4\sqrt{3}}\frac{7+4\sqrt{3}}{7+4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14\sqrt{3}+24}{49+28\sqrt{3}-28\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14\sqrt{3}+24}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}14\sqrt{3}+24\end{aligned} $$ | |
① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 + 4 \sqrt{3}} $$. |
② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{3} } \cdot \left( 7 + 4 \sqrt{3}\right) = \color{blue}{ 2 \sqrt{3}} \cdot7+\color{blue}{ 2 \sqrt{3}} \cdot 4 \sqrt{3} = \\ = 14 \sqrt{3} + 24 $$ Simplify denominator. $$ \color{blue}{ \left( 7- 4 \sqrt{3}\right) } \cdot \left( 7 + 4 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot 4 \sqrt{3}\color{blue}{- 4 \sqrt{3}} \cdot7\color{blue}{- 4 \sqrt{3}} \cdot 4 \sqrt{3} = \\ = 49 + 28 \sqrt{3}- 28 \sqrt{3}-48 $$ |
③ | Simplify numerator and denominator |
④ | Remove 1 from denominator. |