Question
Answer
$$\dfrac{2}{\sqrt{11}-\sqrt{2}}=\dfrac{2\sqrt{11}+2\sqrt{2}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{\sqrt{11}-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{11}-\sqrt{2}}\frac{\sqrt{11}+\sqrt{2}}{\sqrt{11}+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{11}+2\sqrt{2}}{11+\sqrt{22}-\sqrt{22}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{11}+2\sqrt{2}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{11} + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{11} + \sqrt{2}\right) = \color{blue}{2} \cdot \sqrt{11}+\color{blue}{2} \cdot \sqrt{2} = \\ = 2 \sqrt{11} + 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{11}- \sqrt{2}\right) } \cdot \left( \sqrt{11} + \sqrt{2}\right) = \color{blue}{ \sqrt{11}} \cdot \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot \sqrt{11}\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 11 + \sqrt{22}- \sqrt{22}-2 $$ |
| ③ | Simplify numerator and denominator |
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