Question
Answer
$$\dfrac{\sqrt{12}}{\sqrt{3}+3}=-1+\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{12}}{\sqrt{3}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{12}}{\sqrt{3}+3}\frac{\sqrt{3}-3}{\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-6\sqrt{3}}{3-3\sqrt{3}+3\sqrt{3}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6-6\sqrt{3}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1-\sqrt{3}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-1+\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-1+\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{12} } \cdot \left( \sqrt{3}-3\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{3}+\color{blue}{ \sqrt{12}} \cdot-3 = \\ = 6- 6 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 3\right) } \cdot \left( \sqrt{3}-3\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-3+\color{blue}{3} \cdot \sqrt{3}+\color{blue}{3} \cdot-3 = \\ = 3- 3 \sqrt{3} + 3 \sqrt{3}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
| ⑤ | Multiply both numerator and denominator by -1. |
| ⑥ | Remove 1 from denominator. |
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