Question
Answer
$$\dfrac{4}{1+2\sqrt{3}}=\dfrac{-4+8\sqrt{3}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{1+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{1+2\sqrt{3}}\frac{1-2\sqrt{3}}{1-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-8\sqrt{3}}{1-2\sqrt{3}+2\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-8\sqrt{3}}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-4+8\sqrt{3}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 1- 2 \sqrt{3}\right) = \color{blue}{4} \cdot1+\color{blue}{4} \cdot- 2 \sqrt{3} = \\ = 4- 8 \sqrt{3} $$ 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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