Question
Answer
$$\dfrac{4}{2+5\sqrt{3}}=\dfrac{-8+20\sqrt{3}}{71}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{2+5\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{2+5\sqrt{3}}\frac{2-5\sqrt{3}}{2-5\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8-20\sqrt{3}}{4-10\sqrt{3}+10\sqrt{3}-75} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8-20\sqrt{3}}{-71} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-8+20\sqrt{3}}{71}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 5 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 2- 5 \sqrt{3}\right) = \color{blue}{4} \cdot2+\color{blue}{4} \cdot- 5 \sqrt{3} = \\ = 8- 20 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 5 \sqrt{3}\right) } \cdot \left( 2- 5 \sqrt{3}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 5 \sqrt{3}+\color{blue}{ 5 \sqrt{3}} \cdot2+\color{blue}{ 5 \sqrt{3}} \cdot- 5 \sqrt{3} = \\ = 4- 10 \sqrt{3} + 10 \sqrt{3}-75 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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