Question
Answer
$$\dfrac{2}{\sqrt{13}-\sqrt{6}}=\dfrac{2\sqrt{13}+2\sqrt{6}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{\sqrt{13}-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{13}-\sqrt{6}}\frac{\sqrt{13}+\sqrt{6}}{\sqrt{13}+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{13}+2\sqrt{6}}{13+\sqrt{78}-\sqrt{78}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{13}+2\sqrt{6}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{13} + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{13} + \sqrt{6}\right) = \color{blue}{2} \cdot \sqrt{13}+\color{blue}{2} \cdot \sqrt{6} = \\ = 2 \sqrt{13} + 2 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{13}- \sqrt{6}\right) } \cdot \left( \sqrt{13} + \sqrt{6}\right) = \color{blue}{ \sqrt{13}} \cdot \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot \sqrt{13}\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 13 + \sqrt{78}- \sqrt{78}-6 $$ |
| ③ | Simplify numerator and denominator |
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