Question
Answer
$$\dfrac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{12}}=\dfrac{-2+\sqrt{6}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{12}}\frac{\sqrt{2}-\sqrt{12}}{\sqrt{2}-\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-4\sqrt{6}-\sqrt{6}+6}{2-2\sqrt{6}+2\sqrt{6}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10-5\sqrt{6}}{-10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2-\sqrt{6}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-2+\sqrt{6}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}- \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{8}- \sqrt{3}\right) } \cdot \left( \sqrt{2}- \sqrt{12}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{2}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{12}\color{blue}{- \sqrt{3}} \cdot \sqrt{2}\color{blue}{- \sqrt{3}} \cdot- \sqrt{12} = \\ = 4- 4 \sqrt{6}- \sqrt{6} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2} + \sqrt{12}\right) } \cdot \left( \sqrt{2}- \sqrt{12}\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot- \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot \sqrt{2}+\color{blue}{ \sqrt{12}} \cdot- \sqrt{12} = \\ = 2- 2 \sqrt{6} + 2 \sqrt{6}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
| ⑤ | Multiply both numerator and denominator by -1. |
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