Question
Answer
$$\dfrac{4+\sqrt{3}}{2\sqrt{2}-\sqrt{7}}=8\sqrt{2}+4\sqrt{7}+2\sqrt{6}+\sqrt{21}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+\sqrt{3}}{2\sqrt{2}-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+\sqrt{3}}{2\sqrt{2}-\sqrt{7}}\frac{2\sqrt{2}+\sqrt{7}}{2\sqrt{2}+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{2}+4\sqrt{7}+2\sqrt{6}+\sqrt{21}}{8+2\sqrt{14}-2\sqrt{14}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{2}+4\sqrt{7}+2\sqrt{6}+\sqrt{21}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}8\sqrt{2}+4\sqrt{7}+2\sqrt{6}+\sqrt{21}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2} + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + \sqrt{3}\right) } \cdot \left( 2 \sqrt{2} + \sqrt{7}\right) = \color{blue}{4} \cdot 2 \sqrt{2}+\color{blue}{4} \cdot \sqrt{7}+\color{blue}{ \sqrt{3}} \cdot 2 \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot \sqrt{7} = \\ = 8 \sqrt{2} + 4 \sqrt{7} + 2 \sqrt{6} + \sqrt{21} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2}- \sqrt{7}\right) } \cdot \left( 2 \sqrt{2} + \sqrt{7}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot 2 \sqrt{2}\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 8 + 2 \sqrt{14}- 2 \sqrt{14}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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