Question
Answer
$$\dfrac{7\sqrt{3}}{3\sqrt{5}-2\sqrt{6}}=\sqrt{15}+2\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7\sqrt{3}}{3\sqrt{5}-2\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7\sqrt{3}}{3\sqrt{5}-2\sqrt{6}}\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{21\sqrt{15}+42\sqrt{2}}{45+6\sqrt{30}-6\sqrt{30}-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{21\sqrt{15}+42\sqrt{2}}{21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{15}+2\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{15}+2\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{5} + 2 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 \sqrt{3} } \cdot \left( 3 \sqrt{5} + 2 \sqrt{6}\right) = \color{blue}{ 7 \sqrt{3}} \cdot 3 \sqrt{5}+\color{blue}{ 7 \sqrt{3}} \cdot 2 \sqrt{6} = \\ = 21 \sqrt{15} + 42 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{5}- 2 \sqrt{6}\right) } \cdot \left( 3 \sqrt{5} + 2 \sqrt{6}\right) = \color{blue}{ 3 \sqrt{5}} \cdot 3 \sqrt{5}+\color{blue}{ 3 \sqrt{5}} \cdot 2 \sqrt{6}\color{blue}{- 2 \sqrt{6}} \cdot 3 \sqrt{5}\color{blue}{- 2 \sqrt{6}} \cdot 2 \sqrt{6} = \\ = 45 + 6 \sqrt{30}- 6 \sqrt{30}-24 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 21. |
| ⑤ | Remove 1 from denominator. |
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