Answer
$$\frac{12}{\sqrt{15}+3}=\frac{12\sqrt{15}-36}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{12}{\sqrt{15}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{12}{\sqrt{15}+3}\frac{\sqrt{15}-3}{\sqrt{15}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{15}-36}{15-3\sqrt{15}+3\sqrt{15}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{15}-36}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 12 } \cdot \left( \sqrt{15}-3\right) = \color{blue}{12} \cdot \sqrt{15}+\color{blue}{12} \cdot-3 = \\ = 12 \sqrt{15}-36 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15} + 3\right) } \cdot \left( \sqrt{15}-3\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot-3+\color{blue}{3} \cdot \sqrt{15}+\color{blue}{3} \cdot-3 = \\ = 15- 3 \sqrt{15} + 3 \sqrt{15}-9 $$ |
| ③ | Simplify numerator and denominator |
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