Question
Answer
$$\dfrac{\sqrt{6}}{2\sqrt{5}-3}=\dfrac{2\sqrt{30}+3\sqrt{6}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{6}}{2\sqrt{5}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{6}}{2\sqrt{5}-3}\frac{2\sqrt{5}+3}{2\sqrt{5}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{30}+3\sqrt{6}}{20+6\sqrt{5}-6\sqrt{5}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{30}+3\sqrt{6}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{5} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{6} } \cdot \left( 2 \sqrt{5} + 3\right) = \color{blue}{ \sqrt{6}} \cdot 2 \sqrt{5}+\color{blue}{ \sqrt{6}} \cdot3 = \\ = 2 \sqrt{30} + 3 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{5}-3\right) } \cdot \left( 2 \sqrt{5} + 3\right) = \color{blue}{ 2 \sqrt{5}} \cdot 2 \sqrt{5}+\color{blue}{ 2 \sqrt{5}} \cdot3\color{blue}{-3} \cdot 2 \sqrt{5}\color{blue}{-3} \cdot3 = \\ = 20 + 6 \sqrt{5}- 6 \sqrt{5}-9 $$ |
| ③ | Simplify numerator and denominator |
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