Question
Answer
$$\dfrac{\sqrt{15}+\sqrt{6}}{\sqrt{15}-\sqrt{6}}=\dfrac{7+2\sqrt{10}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{15}+\sqrt{6}}{\sqrt{15}-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{15}+\sqrt{6}}{\sqrt{15}-\sqrt{6}}\frac{\sqrt{15}+\sqrt{6}}{\sqrt{15}+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{15+3\sqrt{10}+3\sqrt{10}+6}{15+3\sqrt{10}-3\sqrt{10}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{21+6\sqrt{10}}{9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7+2\sqrt{10}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15} + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{15} + \sqrt{6}\right) } \cdot \left( \sqrt{15} + \sqrt{6}\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot \sqrt{15}+\color{blue}{ \sqrt{6}} \cdot \sqrt{6} = \\ = 15 + 3 \sqrt{10} + 3 \sqrt{10} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15}- \sqrt{6}\right) } \cdot \left( \sqrt{15} + \sqrt{6}\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot \sqrt{15}\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 15 + 3 \sqrt{10}- 3 \sqrt{10}-6 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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