Question
Answer
$$\dfrac{(\sqrt{3}+2)^2}{\sqrt{3}-2}=\dfrac{21+7\sqrt{2}+12\sqrt{3}+4\sqrt{6}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(\sqrt{3}+2)^2}{\sqrt{3}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+2\sqrt{3}+2\sqrt{3}+4}{3-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7+4\sqrt{3}}{3-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7+4\sqrt{3}}{3-\sqrt{2}}\frac{3+\sqrt{2}}{3+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{21+7\sqrt{2}+12\sqrt{3}+4\sqrt{6}}{9+3\sqrt{2}-3\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{21+7\sqrt{2}+12\sqrt{3}+4\sqrt{6}}{7}\end{aligned} $$ | |
| ① | $$ (\sqrt{3}+2)^2 = \left( \sqrt{3} + 2 \right) \cdot \left( \sqrt{3} + 2 \right) = 3 + 2 \sqrt{3} + 2 \sqrt{3} + 4 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ \left( 7 + 4 \sqrt{3}\right) } \cdot \left( 3 + \sqrt{2}\right) = \color{blue}{7} \cdot3+\color{blue}{7} \cdot \sqrt{2}+\color{blue}{ 4 \sqrt{3}} \cdot3+\color{blue}{ 4 \sqrt{3}} \cdot \sqrt{2} = \\ = 21 + 7 \sqrt{2} + 12 \sqrt{3} + 4 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 3- \sqrt{2}\right) } \cdot \left( 3 + \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot3\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 9 + 3 \sqrt{2}- 3 \sqrt{2}-2 $$ |
| ⑤ | Simplify numerator and denominator |
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