Question
Answer
$$\dfrac{6}{\sqrt{20}-\sqrt{8}}=\sqrt{5}+\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{\sqrt{20}-\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{\sqrt{20}-\sqrt{8}}\frac{\sqrt{20}+\sqrt{8}}{\sqrt{20}+\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{5}+12\sqrt{2}}{20+4\sqrt{10}-4\sqrt{10}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{5}+12\sqrt{2}}{12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{5}+\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{5}+\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{20} + \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( \sqrt{20} + \sqrt{8}\right) = \color{blue}{6} \cdot \sqrt{20}+\color{blue}{6} \cdot \sqrt{8} = \\ = 12 \sqrt{5} + 12 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{20}- \sqrt{8}\right) } \cdot \left( \sqrt{20} + \sqrt{8}\right) = \color{blue}{ \sqrt{20}} \cdot \sqrt{20}+\color{blue}{ \sqrt{20}} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot \sqrt{20}\color{blue}{- \sqrt{8}} \cdot \sqrt{8} = \\ = 20 + 4 \sqrt{10}- 4 \sqrt{10}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 12. |
| ⑤ | Remove 1 from denominator. |
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