Question
Answer
$$\dfrac{(6+\sqrt{2})\cdot(6-\sqrt{2})}{4\sqrt{17}}=\dfrac{\sqrt{17}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(6+\sqrt{2})\cdot(6-\sqrt{2})}{4\sqrt{17}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{36-6\sqrt{2}+6\sqrt{2}-2}{4\sqrt{17}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{34}{4\sqrt{17}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{34}{4\sqrt{17}}\frac{\sqrt{17}}{\sqrt{17}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{34\sqrt{17}}{68} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{17}}{2}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 6 + \sqrt{2}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot6+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 36- 6 \sqrt{2} + 6 \sqrt{2}-2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{17}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 34 } \cdot \sqrt{17} = 34 \sqrt{17} $$ Simplify denominator. $$ \color{blue}{ 4 \sqrt{17} } \cdot \sqrt{17} = 68 $$ |
| ⑤ | Divide both numerator and denominator by 34. |
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