Question
Answer
$$\dfrac{\sqrt{2}-2}{3\sqrt{2}+4}=7-5\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}-2}{3\sqrt{2}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}-2}{3\sqrt{2}+4}\frac{3\sqrt{2}-4}{3\sqrt{2}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-4\sqrt{2}-6\sqrt{2}+8}{18-12\sqrt{2}+12\sqrt{2}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14-10\sqrt{2}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7-5\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}7-5\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{2}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2}-2\right) } \cdot \left( 3 \sqrt{2}-4\right) = \color{blue}{ \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot-4\color{blue}{-2} \cdot 3 \sqrt{2}\color{blue}{-2} \cdot-4 = \\ = 6- 4 \sqrt{2}- 6 \sqrt{2} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{2} + 4\right) } \cdot \left( 3 \sqrt{2}-4\right) = \color{blue}{ 3 \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot-4+\color{blue}{4} \cdot 3 \sqrt{2}+\color{blue}{4} \cdot-4 = \\ = 18- 12 \sqrt{2} + 12 \sqrt{2}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Remove 1 from denominator. |
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