Question
Answer
$$\dfrac{1-\sqrt{2}}{2+3\sqrt{2}}=\dfrac{-8+5\sqrt{2}}{14}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-\sqrt{2}}{2+3\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-\sqrt{2}}{2+3\sqrt{2}}\frac{2-3\sqrt{2}}{2-3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2-3\sqrt{2}-2\sqrt{2}+6}{4-6\sqrt{2}+6\sqrt{2}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8-5\sqrt{2}}{-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-8+5\sqrt{2}}{14}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 3 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1- \sqrt{2}\right) } \cdot \left( 2- 3 \sqrt{2}\right) = \color{blue}{1} \cdot2+\color{blue}{1} \cdot- 3 \sqrt{2}\color{blue}{- \sqrt{2}} \cdot2\color{blue}{- \sqrt{2}} \cdot- 3 \sqrt{2} = \\ = 2- 3 \sqrt{2}- 2 \sqrt{2} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 3 \sqrt{2}\right) } \cdot \left( 2- 3 \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot2+\color{blue}{ 3 \sqrt{2}} \cdot- 3 \sqrt{2} = \\ = 4- 6 \sqrt{2} + 6 \sqrt{2}-18 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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