Question
Answer
$$\dfrac{1}{2\sqrt{3}+8}=\dfrac{-\sqrt{3}+4}{26}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{2\sqrt{3}+8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{2\sqrt{3}+8}\frac{2\sqrt{3}-8}{2\sqrt{3}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{3}-8}{12-16\sqrt{3}+16\sqrt{3}-64} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{3}-8}{-52} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}-4}{-26} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-\sqrt{3}+4}{26}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3}-8} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 2 \sqrt{3}-8\right) = \color{blue}{1} \cdot 2 \sqrt{3}+\color{blue}{1} \cdot-8 = \\ = 2 \sqrt{3}-8 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3} + 8\right) } \cdot \left( 2 \sqrt{3}-8\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot-8+\color{blue}{8} \cdot 2 \sqrt{3}+\color{blue}{8} \cdot-8 = \\ = 12- 16 \sqrt{3} + 16 \sqrt{3}-64 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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