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