Question
Answer
$$\dfrac{\sqrt{5}-3\sqrt{3}}{\sqrt{5}+3\sqrt{3}}=\dfrac{-16+3\sqrt{15}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}-3\sqrt{3}}{\sqrt{5}+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}-3\sqrt{3}}{\sqrt{5}+3\sqrt{3}}\frac{\sqrt{5}-3\sqrt{3}}{\sqrt{5}-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5-3\sqrt{15}-3\sqrt{15}+27}{5-3\sqrt{15}+3\sqrt{15}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{32-6\sqrt{15}}{-22} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{16-3\sqrt{15}}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-16+3\sqrt{15}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5}- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{5}- 3 \sqrt{3}\right) } \cdot \left( \sqrt{5}- 3 \sqrt{3}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot- 3 \sqrt{3}\color{blue}{- 3 \sqrt{3}} \cdot \sqrt{5}\color{blue}{- 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 5- 3 \sqrt{15}- 3 \sqrt{15} + 27 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5} + 3 \sqrt{3}\right) } \cdot \left( \sqrt{5}- 3 \sqrt{3}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot \sqrt{5}+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 5- 3 \sqrt{15} + 3 \sqrt{15}-27 $$ |
| 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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