Question
Answer
$$\dfrac{\sqrt{3}}{4+4\sqrt{3}}=\dfrac{-\sqrt{3}+3}{8}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{4+4\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{4+4\sqrt{3}}\frac{4-4\sqrt{3}}{4-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{3}-12}{16-16\sqrt{3}+16\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{3}-12}{-32} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}-3}{-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-\sqrt{3}+3}{8}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- 4 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 4- 4 \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot4+\color{blue}{ \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 4 \sqrt{3}-12 $$ Simplify denominator. $$ \color{blue}{ \left( 4 + 4 \sqrt{3}\right) } \cdot \left( 4- 4 \sqrt{3}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot4+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 16- 16 \sqrt{3} + 16 \sqrt{3}-48 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Multiply both numerator and denominator by -1. |
This page was created using
Rationalize Denominator Calculator