Question
Answer
$$\dfrac{2-\sqrt{20}}{5+\sqrt{5}}=\dfrac{5-3\sqrt{5}}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-\sqrt{20}}{5+\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-\sqrt{20}}{5+\sqrt{5}}\frac{5-\sqrt{5}}{5-\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10-2\sqrt{5}-10\sqrt{5}+10}{25-5\sqrt{5}+5\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{20-12\sqrt{5}}{20} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5-3\sqrt{5}}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5- \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- \sqrt{20}\right) } \cdot \left( 5- \sqrt{5}\right) = \color{blue}{2} \cdot5+\color{blue}{2} \cdot- \sqrt{5}\color{blue}{- \sqrt{20}} \cdot5\color{blue}{- \sqrt{20}} \cdot- \sqrt{5} = \\ = 10- 2 \sqrt{5}- 10 \sqrt{5} + 10 $$ Simplify denominator. $$ \color{blue}{ \left( 5 + \sqrt{5}\right) } \cdot \left( 5- \sqrt{5}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot- \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot5+\color{blue}{ \sqrt{5}} \cdot- \sqrt{5} = \\ = 25- 5 \sqrt{5} + 5 \sqrt{5}-5 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
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