Question
Answer
$$\dfrac{-7}{\sqrt{12}+\sqrt{3}}=\dfrac{-7\sqrt{3}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-7}{\sqrt{12}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-7}{\sqrt{12}+\sqrt{3}}\frac{\sqrt{12}-\sqrt{3}}{\sqrt{12}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-14\sqrt{3}+7\sqrt{3}}{12-6+6-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-7\sqrt{3}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12}- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -7 } \cdot \left( \sqrt{12}- \sqrt{3}\right) = \color{blue}{-7} \cdot \sqrt{12}\color{blue}{-7} \cdot- \sqrt{3} = \\ = - 14 \sqrt{3} + 7 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12} + \sqrt{3}\right) } \cdot \left( \sqrt{12}- \sqrt{3}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{12}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 12-6 + 6-3 $$ |
| ③ | Simplify numerator and denominator |
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