Question
Answer
$$\dfrac{\sqrt{15}+2\sqrt{3}}{\sqrt{15}-2\sqrt{3}}=9+4\sqrt{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{15}+2\sqrt{3}}{\sqrt{15}-2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{15}+2\sqrt{3}}{\sqrt{15}-2\sqrt{3}}\frac{\sqrt{15}+2\sqrt{3}}{\sqrt{15}+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{15+6\sqrt{5}+6\sqrt{5}+12}{15+6\sqrt{5}-6\sqrt{5}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{27+12\sqrt{5}}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9+4\sqrt{5}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}9+4\sqrt{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15} + 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{15} + 2 \sqrt{3}\right) } \cdot \left( \sqrt{15} + 2 \sqrt{3}\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot \sqrt{15}+\color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 15 + 6 \sqrt{5} + 6 \sqrt{5} + 12 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15}- 2 \sqrt{3}\right) } \cdot \left( \sqrt{15} + 2 \sqrt{3}\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot \sqrt{15}\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 15 + 6 \sqrt{5}- 6 \sqrt{5}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Remove 1 from denominator. |
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