Question
Answer
$$\dfrac{2\sqrt{12}-3}{\sqrt{12}+4}=\dfrac{-18+11\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{12}-3}{\sqrt{12}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{12}-3}{\sqrt{12}+4}\frac{\sqrt{12}-4}{\sqrt{12}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{24-16\sqrt{3}-6\sqrt{3}+12}{12-8\sqrt{3}+8\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{36-22\sqrt{3}}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{18-11\sqrt{3}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-18+11\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{12}-3\right) } \cdot \left( \sqrt{12}-4\right) = \color{blue}{ 2 \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ 2 \sqrt{12}} \cdot-4\color{blue}{-3} \cdot \sqrt{12}\color{blue}{-3} \cdot-4 = \\ = 24- 16 \sqrt{3}- 6 \sqrt{3} + 12 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12} + 4\right) } \cdot \left( \sqrt{12}-4\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot-4+\color{blue}{4} \cdot \sqrt{12}+\color{blue}{4} \cdot-4 = \\ = 12- 8 \sqrt{3} + 8 \sqrt{3}-16 $$ |
| 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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