Question
Answer
$$\dfrac{3+\sqrt{5}}{2\sqrt{5}+3}=\dfrac{3\sqrt{5}+1}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{5}}{2\sqrt{5}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{5}}{2\sqrt{5}+3}\frac{2\sqrt{5}-3}{2\sqrt{5}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{5}-9+10-3\sqrt{5}}{20-6\sqrt{5}+6\sqrt{5}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{5}+1}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{5}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{5}\right) } \cdot \left( 2 \sqrt{5}-3\right) = \color{blue}{3} \cdot 2 \sqrt{5}+\color{blue}{3} \cdot-3+\color{blue}{ \sqrt{5}} \cdot 2 \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot-3 = \\ = 6 \sqrt{5}-9 + 10- 3 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{5} + 3\right) } \cdot \left( 2 \sqrt{5}-3\right) = \color{blue}{ 2 \sqrt{5}} \cdot 2 \sqrt{5}+\color{blue}{ 2 \sqrt{5}} \cdot-3+\color{blue}{3} \cdot 2 \sqrt{5}+\color{blue}{3} \cdot-3 = \\ = 20- 6 \sqrt{5} + 6 \sqrt{5}-9 $$ |
| ③ | Simplify numerator and denominator |
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