Question
Answer
$$\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{15}}=\dfrac{-2\sqrt{3}+3\sqrt{10}}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{6}}{\sqrt{2}+\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{6}}{\sqrt{2}+\sqrt{15}}\frac{\sqrt{2}-\sqrt{15}}{\sqrt{2}-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{3}-3\sqrt{10}}{2-\sqrt{30}+\sqrt{30}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{3}-3\sqrt{10}}{-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{3}+3\sqrt{10}}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{6} } \cdot \left( \sqrt{2}- \sqrt{15}\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{2}+\color{blue}{ \sqrt{6}} \cdot- \sqrt{15} = \\ = 2 \sqrt{3}- 3 \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2} + \sqrt{15}\right) } \cdot \left( \sqrt{2}- \sqrt{15}\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot \sqrt{2}+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 2- \sqrt{30} + \sqrt{30}-15 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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