Question
Answer
$$\dfrac{4}{\sqrt{20}-\sqrt{12}}=\sqrt{5}+\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{\sqrt{20}-\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{\sqrt{20}-\sqrt{12}}\frac{\sqrt{20}+\sqrt{12}}{\sqrt{20}+\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{5}+8\sqrt{3}}{20+4\sqrt{15}-4\sqrt{15}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{5}+8\sqrt{3}}{8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{5}+\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{5}+\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{20} + \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( \sqrt{20} + \sqrt{12}\right) = \color{blue}{4} \cdot \sqrt{20}+\color{blue}{4} \cdot \sqrt{12} = \\ = 8 \sqrt{5} + 8 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{20}- \sqrt{12}\right) } \cdot \left( \sqrt{20} + \sqrt{12}\right) = \color{blue}{ \sqrt{20}} \cdot \sqrt{20}+\color{blue}{ \sqrt{20}} \cdot \sqrt{12}\color{blue}{- \sqrt{12}} \cdot \sqrt{20}\color{blue}{- \sqrt{12}} \cdot \sqrt{12} = \\ = 20 + 4 \sqrt{15}- 4 \sqrt{15}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 8. |
| ⑤ | Remove 1 from denominator. |
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