Question
Answer
$$\dfrac{3-\sqrt{2}}{4+2\sqrt{2}}=\dfrac{8-5\sqrt{2}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3-\sqrt{2}}{4+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3-\sqrt{2}}{4+2\sqrt{2}}\frac{4-2\sqrt{2}}{4-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12-6\sqrt{2}-4\sqrt{2}+4}{16-8\sqrt{2}+8\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{16-10\sqrt{2}}{8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{8-5\sqrt{2}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3- \sqrt{2}\right) } \cdot \left( 4- 2 \sqrt{2}\right) = \color{blue}{3} \cdot4+\color{blue}{3} \cdot- 2 \sqrt{2}\color{blue}{- \sqrt{2}} \cdot4\color{blue}{- \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 12- 6 \sqrt{2}- 4 \sqrt{2} + 4 $$ Simplify denominator. $$ \color{blue}{ \left( 4 + 2 \sqrt{2}\right) } \cdot \left( 4- 2 \sqrt{2}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot4+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 16- 8 \sqrt{2} + 8 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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