Question
Answer
$$\dfrac{5+\sqrt{3}}{7-2\sqrt{3}}=\dfrac{41+17\sqrt{3}}{37}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+\sqrt{3}}{7-2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+\sqrt{3}}{7-2\sqrt{3}}\frac{7+2\sqrt{3}}{7+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{35+10\sqrt{3}+7\sqrt{3}+6}{49+14\sqrt{3}-14\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{41+17\sqrt{3}}{37}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7 + 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + \sqrt{3}\right) } \cdot \left( 7 + 2 \sqrt{3}\right) = \color{blue}{5} \cdot7+\color{blue}{5} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot7+\color{blue}{ \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 35 + 10 \sqrt{3} + 7 \sqrt{3} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 7- 2 \sqrt{3}\right) } \cdot \left( 7 + 2 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot7\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 49 + 14 \sqrt{3}- 14 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
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