Question
Answer
$$\dfrac{5+\sqrt{3}}{7-3\sqrt{3}}=2+\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+\sqrt{3}}{7-3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+\sqrt{3}}{7-3\sqrt{3}}\frac{7+3\sqrt{3}}{7+3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{35+15\sqrt{3}+7\sqrt{3}+9}{49+21\sqrt{3}-21\sqrt{3}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{44+22\sqrt{3}}{22} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2+\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2+\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 + 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + \sqrt{3}\right) } \cdot \left( 7 + 3 \sqrt{3}\right) = \color{blue}{5} \cdot7+\color{blue}{5} \cdot 3 \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot7+\color{blue}{ \sqrt{3}} \cdot 3 \sqrt{3} = \\ = 35 + 15 \sqrt{3} + 7 \sqrt{3} + 9 $$ Simplify denominator. $$ \color{blue}{ \left( 7- 3 \sqrt{3}\right) } \cdot \left( 7 + 3 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot 3 \sqrt{3}\color{blue}{- 3 \sqrt{3}} \cdot7\color{blue}{- 3 \sqrt{3}} \cdot 3 \sqrt{3} = \\ = 49 + 21 \sqrt{3}- 21 \sqrt{3}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 22. |
| ⑤ | Remove 1 from denominator. |
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