Question
Answer
$$\dfrac{3+\sqrt{7}}{6+\sqrt{2}}=\dfrac{18-3\sqrt{2}+6\sqrt{7}-\sqrt{14}}{34}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{7}}{6+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{7}}{6+\sqrt{2}}\frac{6-\sqrt{2}}{6-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18-3\sqrt{2}+6\sqrt{7}-\sqrt{14}}{36-6\sqrt{2}+6\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{18-3\sqrt{2}+6\sqrt{7}-\sqrt{14}}{34}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{7}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{3} \cdot6+\color{blue}{3} \cdot- \sqrt{2}+\color{blue}{ \sqrt{7}} \cdot6+\color{blue}{ \sqrt{7}} \cdot- \sqrt{2} = \\ = 18- 3 \sqrt{2} + 6 \sqrt{7}- \sqrt{14} $$ Simplify denominator. $$ \color{blue}{ \left( 6 + \sqrt{2}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot6+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 36- 6 \sqrt{2} + 6 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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