Question
Answer
$$\dfrac{8+\sqrt{12}}{2+\sqrt{12}}=\dfrac{-1+3\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{8+\sqrt{12}}{2+\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8+\sqrt{12}}{2+\sqrt{12}}\frac{2-\sqrt{12}}{2-\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16-16\sqrt{3}+4\sqrt{3}-12}{4-4\sqrt{3}+4\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-12\sqrt{3}}{-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1-3\sqrt{3}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-1+3\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 8 + \sqrt{12}\right) } \cdot \left( 2- \sqrt{12}\right) = \color{blue}{8} \cdot2+\color{blue}{8} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot2+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 16- 16 \sqrt{3} + 4 \sqrt{3}-12 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{12}\right) } \cdot \left( 2- \sqrt{12}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot2+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 4- 4 \sqrt{3} + 4 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Multiply both numerator and denominator by -1. |
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