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