Question
Answer
$$\dfrac{1}{\sqrt{2}+5}=\dfrac{-\sqrt{2}+5}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{2}+5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{2}+5}\frac{\sqrt{2}-5}{\sqrt{2}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{2}-5}{2-5\sqrt{2}+5\sqrt{2}-25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{2}-5}{-23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{2}+5}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}-5} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{2}-5\right) = \color{blue}{1} \cdot \sqrt{2}+\color{blue}{1} \cdot-5 = \\ = \sqrt{2}-5 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2} + 5\right) } \cdot \left( \sqrt{2}-5\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot-5+\color{blue}{5} \cdot \sqrt{2}+\color{blue}{5} \cdot-5 = \\ = 2- 5 \sqrt{2} + 5 \sqrt{2}-25 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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