Question
Answer
$$\dfrac{4}{(1+\sqrt{2})^2}=12-8\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{(1+\sqrt{2})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{1+\sqrt{2}+\sqrt{2}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4}{3+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4}{3+2\sqrt{2}}\frac{3-2\sqrt{2}}{3-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12-8\sqrt{2}}{9-6\sqrt{2}+6\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{12-8\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}12-8\sqrt{2}\end{aligned} $$ | |
| ① | $$ (1+\sqrt{2})^2 = \left( 1 + \sqrt{2} \right) \cdot \left( 1 + \sqrt{2} \right) = 1 + \sqrt{2} + \sqrt{2} + 2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- 2 \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 3- 2 \sqrt{2}\right) = \color{blue}{4} \cdot3+\color{blue}{4} \cdot- 2 \sqrt{2} = \\ = 12- 8 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + 2 \sqrt{2}\right) } \cdot \left( 3- 2 \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot3+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 9- 6 \sqrt{2} + 6 \sqrt{2}-8 $$ |
| ⑤ | Simplify numerator and denominator |
| ⑥ | Remove 1 from denominator. |
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