Question
Answer
$$\dfrac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2}}=2+\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2}}\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18+4\sqrt{6}+15\sqrt{6}+20}{27+6\sqrt{6}-6\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{38+19\sqrt{6}}{19} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2+\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2+\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3} + 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{3} + 5 \sqrt{2}\right) } \cdot \left( 3 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{2}+\color{blue}{ 5 \sqrt{2}} \cdot 3 \sqrt{3}+\color{blue}{ 5 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 18 + 4 \sqrt{6} + 15 \sqrt{6} + 20 $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3}- 2 \sqrt{2}\right) } \cdot \left( 3 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot 2 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot 3 \sqrt{3}\color{blue}{- 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 27 + 6 \sqrt{6}- 6 \sqrt{6}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 19. |
| ⑤ | Remove 1 from denominator. |
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