Question
Answer
$$\dfrac{3+\sqrt{5}}{2-\sqrt{3}}=6+3\sqrt{3}+2\sqrt{5}+\sqrt{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{5}}{2-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{5}}{2-\sqrt{3}}\frac{2+\sqrt{3}}{2+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{4+2\sqrt{3}-2\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+3\sqrt{3}+2\sqrt{5}+\sqrt{15}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6+3\sqrt{3}+2\sqrt{5}+\sqrt{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{5}\right) } \cdot \left( 2 + \sqrt{3}\right) = \color{blue}{3} \cdot2+\color{blue}{3} \cdot \sqrt{3}+\color{blue}{ \sqrt{5}} \cdot2+\color{blue}{ \sqrt{5}} \cdot \sqrt{3} = \\ = 6 + 3 \sqrt{3} + 2 \sqrt{5} + \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{3}\right) } \cdot \left( 2 + \sqrt{3}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot2\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 4 + 2 \sqrt{3}- 2 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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