Question
Answer
$$\dfrac{2}{-\sqrt{3}+\sqrt{15}}=\dfrac{\sqrt{3}+\sqrt{15}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{-\sqrt{3}+\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{-\sqrt{3}+\sqrt{15}}\frac{-\sqrt{3}-\sqrt{15}}{-\sqrt{3}-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-2\sqrt{3}-2\sqrt{15}}{3+3\sqrt{5}-3\sqrt{5}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2\sqrt{3}-2\sqrt{15}}{-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{3}-\sqrt{15}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{3}+\sqrt{15}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ - \sqrt{3}- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( - \sqrt{3}- \sqrt{15}\right) = \color{blue}{2} \cdot- \sqrt{3}+\color{blue}{2} \cdot- \sqrt{15} = \\ = - 2 \sqrt{3}- 2 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( - \sqrt{3} + \sqrt{15}\right) } \cdot \left( - \sqrt{3}- \sqrt{15}\right) = \color{blue}{- \sqrt{3}} \cdot- \sqrt{3}\color{blue}{- \sqrt{3}} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 3 + 3 \sqrt{5}- 3 \sqrt{5}-15 $$ |
| 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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