Question
Answer
$$\dfrac{7}{2-\sqrt{3}}=14+7\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{2-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{2-\sqrt{3}}\frac{2+\sqrt{3}}{2+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14+7\sqrt{3}}{4+2\sqrt{3}-2\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14+7\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}14+7\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 2 + \sqrt{3}\right) = \color{blue}{7} \cdot2+\color{blue}{7} \cdot \sqrt{3} = \\ = 14 + 7 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{3}\right) } \cdot \left( 2 + \sqrt{3}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot2\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 4 + 2 \sqrt{3}- 2 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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