Question
Answer
$$\dfrac{6}{3-\sqrt{12}}=-\dfrac{18+12\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{3-\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{3-\sqrt{12}}\frac{3+\sqrt{12}}{3+\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18+12\sqrt{3}}{9+6\sqrt{3}-6\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{18+12\sqrt{3}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{18+12\sqrt{3}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( 3 + \sqrt{12}\right) = \color{blue}{6} \cdot3+\color{blue}{6} \cdot \sqrt{12} = \\ = 18 + 12 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3- \sqrt{12}\right) } \cdot \left( 3 + \sqrt{12}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot \sqrt{12}\color{blue}{- \sqrt{12}} \cdot3\color{blue}{- \sqrt{12}} \cdot \sqrt{12} = \\ = 9 + 6 \sqrt{3}- 6 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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