Question
Answer
$$\dfrac{2-5\sqrt{2}}{2+5\sqrt{2}}=\dfrac{-27+10\sqrt{2}}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-5\sqrt{2}}{2+5\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-5\sqrt{2}}{2+5\sqrt{2}}\frac{2-5\sqrt{2}}{2-5\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-10\sqrt{2}-10\sqrt{2}+50}{4-10\sqrt{2}+10\sqrt{2}-50} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{54-20\sqrt{2}}{-46} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{27-10\sqrt{2}}{-23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-27+10\sqrt{2}}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 5 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- 5 \sqrt{2}\right) } \cdot \left( 2- 5 \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 5 \sqrt{2}\color{blue}{- 5 \sqrt{2}} \cdot2\color{blue}{- 5 \sqrt{2}} \cdot- 5 \sqrt{2} = \\ = 4- 10 \sqrt{2}- 10 \sqrt{2} + 50 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 5 \sqrt{2}\right) } \cdot \left( 2- 5 \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 5 \sqrt{2}+\color{blue}{ 5 \sqrt{2}} \cdot2+\color{blue}{ 5 \sqrt{2}} \cdot- 5 \sqrt{2} = \\ = 4- 10 \sqrt{2} + 10 \sqrt{2}-50 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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