Question
Answer
$$\dfrac{\sqrt{8}-\sqrt{11}}{\sqrt{2}+\sqrt{11}}=\dfrac{-5+\sqrt{22}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{8}-\sqrt{11}}{\sqrt{2}+\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{8}-\sqrt{11}}{\sqrt{2}+\sqrt{11}}\frac{\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-2\sqrt{22}-\sqrt{22}+11}{2-\sqrt{22}+\sqrt{22}-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{15-3\sqrt{22}}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5-\sqrt{22}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-5+\sqrt{22}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}- \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{8}- \sqrt{11}\right) } \cdot \left( \sqrt{2}- \sqrt{11}\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{2}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{11}\color{blue}{- \sqrt{11}} \cdot \sqrt{2}\color{blue}{- \sqrt{11}} \cdot- \sqrt{11} = \\ = 4- 2 \sqrt{22}- \sqrt{22} + 11 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2} + \sqrt{11}\right) } \cdot \left( \sqrt{2}- \sqrt{11}\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot- \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot \sqrt{2}+\color{blue}{ \sqrt{11}} \cdot- \sqrt{11} = \\ = 2- \sqrt{22} + \sqrt{22}-11 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Multiply both numerator and denominator by -1. |
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