Question
Answer
$$\dfrac{4+\sqrt{12}}{4-\sqrt{12}}=7+4\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+\sqrt{12}}{4-\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+\sqrt{12}}{4-\sqrt{12}}\frac{4+\sqrt{12}}{4+\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16+8\sqrt{3}+8\sqrt{3}+12}{16+8\sqrt{3}-8\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{28+16\sqrt{3}}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7+4\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}7+4\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + \sqrt{12}\right) } \cdot \left( 4 + \sqrt{12}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot4+\color{blue}{ \sqrt{12}} \cdot \sqrt{12} = \\ = 16 + 8 \sqrt{3} + 8 \sqrt{3} + 12 $$ Simplify denominator. $$ \color{blue}{ \left( 4- \sqrt{12}\right) } \cdot \left( 4 + \sqrt{12}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot \sqrt{12}\color{blue}{- \sqrt{12}} \cdot4\color{blue}{- \sqrt{12}} \cdot \sqrt{12} = \\ = 16 + 8 \sqrt{3}- 8 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Remove 1 from denominator. |
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