Question
Answer
$$\dfrac{\sqrt{2}-\sqrt{11}}{\sqrt{11}-\sqrt{8}}=\dfrac{-\sqrt{22}-7}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}-\sqrt{11}}{\sqrt{11}-\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}-\sqrt{11}}{\sqrt{11}-\sqrt{8}}\frac{\sqrt{11}+\sqrt{8}}{\sqrt{11}+\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{22}+4-11-2\sqrt{22}}{11+2\sqrt{22}-2\sqrt{22}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-\sqrt{22}-7}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{11} + \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2}- \sqrt{11}\right) } \cdot \left( \sqrt{11} + \sqrt{8}\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{11}+\color{blue}{ \sqrt{2}} \cdot \sqrt{8}\color{blue}{- \sqrt{11}} \cdot \sqrt{11}\color{blue}{- \sqrt{11}} \cdot \sqrt{8} = \\ = \sqrt{22} + 4-11- 2 \sqrt{22} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{11}- \sqrt{8}\right) } \cdot \left( \sqrt{11} + \sqrt{8}\right) = \color{blue}{ \sqrt{11}} \cdot \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot \sqrt{11}\color{blue}{- \sqrt{8}} \cdot \sqrt{8} = \\ = 11 + 2 \sqrt{22}- 2 \sqrt{22}-8 $$ |
| ③ | Simplify numerator and denominator |
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