Question
Answer
$$\dfrac{1}{\sqrt{2}-3}=-\dfrac{\sqrt{2}+3}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{2}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{2}-3}\frac{\sqrt{2}+3}{\sqrt{2}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{2}+3}{2+3\sqrt{2}-3\sqrt{2}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{2}+3}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{\sqrt{2}+3}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{2} + 3\right) = \color{blue}{1} \cdot \sqrt{2}+\color{blue}{1} \cdot3 = \\ = \sqrt{2} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{2}-3\right) } \cdot \left( \sqrt{2} + 3\right) = \color{blue}{ \sqrt{2}} \cdot \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot3\color{blue}{-3} \cdot \sqrt{2}\color{blue}{-3} \cdot3 = \\ = 2 + 3 \sqrt{2}- 3 \sqrt{2}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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