Question
Answer
$$\dfrac{3+\sqrt{2}}{2-\sqrt{5}}=-6-3\sqrt{5}-2\sqrt{2}-\sqrt{10}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{2}}{2-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{2}}{2-\sqrt{5}}\frac{2+\sqrt{5}}{2+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+3\sqrt{5}+2\sqrt{2}+\sqrt{10}}{4+2\sqrt{5}-2\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+3\sqrt{5}+2\sqrt{2}+\sqrt{10}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{6+3\sqrt{5}+2\sqrt{2}+\sqrt{10}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(6+3\sqrt{5}+2\sqrt{2}+\sqrt{10}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-6-3\sqrt{5}-2\sqrt{2}-\sqrt{10}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{2}\right) } \cdot \left( 2 + \sqrt{5}\right) = \color{blue}{3} \cdot2+\color{blue}{3} \cdot \sqrt{5}+\color{blue}{ \sqrt{2}} \cdot2+\color{blue}{ \sqrt{2}} \cdot \sqrt{5} = \\ = 6 + 3 \sqrt{5} + 2 \sqrt{2} + \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{5}\right) } \cdot \left( 2 + \sqrt{5}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot2\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 4 + 2 \sqrt{5}- 2 \sqrt{5}-5 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
| ⑤ | Remove the parenthesis by changing the sign of each term within them. |
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