Question
Answer
$$\dfrac{4+2\sqrt{7}}{3-\sqrt{7}}=13+5\sqrt{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+2\sqrt{7}}{3-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+2\sqrt{7}}{3-\sqrt{7}}\frac{3+\sqrt{7}}{3+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12+4\sqrt{7}+6\sqrt{7}+14}{9+3\sqrt{7}-3\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{26+10\sqrt{7}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{13+5\sqrt{7}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}13+5\sqrt{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + 2 \sqrt{7}\right) } \cdot \left( 3 + \sqrt{7}\right) = \color{blue}{4} \cdot3+\color{blue}{4} \cdot \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot3+\color{blue}{ 2 \sqrt{7}} \cdot \sqrt{7} = \\ = 12 + 4 \sqrt{7} + 6 \sqrt{7} + 14 $$ Simplify denominator. $$ \color{blue}{ \left( 3- \sqrt{7}\right) } \cdot \left( 3 + \sqrt{7}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot3\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 9 + 3 \sqrt{7}- 3 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Remove 1 from denominator. |
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