Question
Answer
$$\dfrac{32+\sqrt{2}}{4+\sqrt{2}}=9-2\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{32+\sqrt{2}}{4+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{32+\sqrt{2}}{4+\sqrt{2}}\frac{4-\sqrt{2}}{4-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{128-32\sqrt{2}+4\sqrt{2}-2}{16-4\sqrt{2}+4\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{126-28\sqrt{2}}{14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9-2\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}9-2\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 32 + \sqrt{2}\right) } \cdot \left( 4- \sqrt{2}\right) = \color{blue}{32} \cdot4+\color{blue}{32} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot4+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 128- 32 \sqrt{2} + 4 \sqrt{2}-2 $$ Simplify denominator. $$ \color{blue}{ \left( 4 + \sqrt{2}\right) } \cdot \left( 4- \sqrt{2}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot4+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 16- 4 \sqrt{2} + 4 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 14. |
| ⑤ | Remove 1 from denominator. |
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