Question
Answer
$$\dfrac{5\sqrt{5}+4}{-2+4\sqrt{3}}=\dfrac{5\sqrt{5}+10\sqrt{15}+4+8\sqrt{3}}{22}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5\sqrt{5}+4}{-2+4\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5\sqrt{5}+4}{-2+4\sqrt{3}}\frac{-2-4\sqrt{3}}{-2-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-10\sqrt{5}-20\sqrt{15}-8-16\sqrt{3}}{4+8\sqrt{3}-8\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-10\sqrt{5}-20\sqrt{15}-8-16\sqrt{3}}{-44} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-5\sqrt{5}-10\sqrt{15}-4-8\sqrt{3}}{-22} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{5\sqrt{5}+10\sqrt{15}+4+8\sqrt{3}}{22}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -2- 4 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 \sqrt{5} + 4\right) } \cdot \left( -2- 4 \sqrt{3}\right) = \color{blue}{ 5 \sqrt{5}} \cdot-2+\color{blue}{ 5 \sqrt{5}} \cdot- 4 \sqrt{3}+\color{blue}{4} \cdot-2+\color{blue}{4} \cdot- 4 \sqrt{3} = \\ = - 10 \sqrt{5}- 20 \sqrt{15}-8- 16 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( -2 + 4 \sqrt{3}\right) } \cdot \left( -2- 4 \sqrt{3}\right) = \color{blue}{-2} \cdot-2\color{blue}{-2} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot-2+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 4 + 8 \sqrt{3}- 8 \sqrt{3}-48 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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