Question
Answer
$$\dfrac{3\sqrt{2}+\sqrt{5}}{4\sqrt{3}-2\sqrt{2}}=\dfrac{6\sqrt{6}+6+2\sqrt{15}+\sqrt{10}}{20}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{2}+\sqrt{5}}{4\sqrt{3}-2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{2}+\sqrt{5}}{4\sqrt{3}-2\sqrt{2}}\frac{4\sqrt{3}+2\sqrt{2}}{4\sqrt{3}+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{6}+12+4\sqrt{15}+2\sqrt{10}}{48+8\sqrt{6}-8\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{6}+12+4\sqrt{15}+2\sqrt{10}}{40} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{6\sqrt{6}+6+2\sqrt{15}+\sqrt{10}}{20}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 \sqrt{3} + 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 \sqrt{2} + \sqrt{5}\right) } \cdot \left( 4 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{ 3 \sqrt{2}} \cdot 4 \sqrt{3}+\color{blue}{ 3 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ \sqrt{5}} \cdot 4 \sqrt{3}+\color{blue}{ \sqrt{5}} \cdot 2 \sqrt{2} = \\ = 12 \sqrt{6} + 12 + 4 \sqrt{15} + 2 \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( 4 \sqrt{3}- 2 \sqrt{2}\right) } \cdot \left( 4 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{ 4 \sqrt{3}} \cdot 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot 2 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot 4 \sqrt{3}\color{blue}{- 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 48 + 8 \sqrt{6}- 8 \sqrt{6}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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