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