Question
Answer
$$\dfrac{\sqrt{75}-2}{1+2\sqrt{3}}=\dfrac{-9\sqrt{3}+32}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{75}-2}{1+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{75}-2}{1+2\sqrt{3}}\frac{1-2\sqrt{3}}{1-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{3}-30-2+4\sqrt{3}}{1-2\sqrt{3}+2\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{3}-32}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-9\sqrt{3}+32}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{75}-2\right) } \cdot \left( 1- 2 \sqrt{3}\right) = \color{blue}{ \sqrt{75}} \cdot1+\color{blue}{ \sqrt{75}} \cdot- 2 \sqrt{3}\color{blue}{-2} \cdot1\color{blue}{-2} \cdot- 2 \sqrt{3} = \\ = 5 \sqrt{3}-30-2 + 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + 2 \sqrt{3}\right) } \cdot \left( 1- 2 \sqrt{3}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot1+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 1- 2 \sqrt{3} + 2 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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