Question
Answer
$$\dfrac{\sqrt{3}}{14+6\sqrt{3}}=\dfrac{7\sqrt{3}-9}{44}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{14+6\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{14+6\sqrt{3}}\frac{14-6\sqrt{3}}{14-6\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14\sqrt{3}-18}{196-84\sqrt{3}+84\sqrt{3}-108} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14\sqrt{3}-18}{88} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7\sqrt{3}-9}{44}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 14- 6 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 14- 6 \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot14+\color{blue}{ \sqrt{3}} \cdot- 6 \sqrt{3} = \\ = 14 \sqrt{3}-18 $$ Simplify denominator. $$ \color{blue}{ \left( 14 + 6 \sqrt{3}\right) } \cdot \left( 14- 6 \sqrt{3}\right) = \color{blue}{14} \cdot14+\color{blue}{14} \cdot- 6 \sqrt{3}+\color{blue}{ 6 \sqrt{3}} \cdot14+\color{blue}{ 6 \sqrt{3}} \cdot- 6 \sqrt{3} = \\ = 196- 84 \sqrt{3} + 84 \sqrt{3}-108 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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