Question
Answer
$$\dfrac{16}{2\sqrt{3}+\sqrt{10}}=\dfrac{32\sqrt{3}-16\sqrt{10}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{16}{2\sqrt{3}+\sqrt{10}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{16}{2\sqrt{3}+\sqrt{10}}\frac{2\sqrt{3}-\sqrt{10}}{2\sqrt{3}-\sqrt{10}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{32\sqrt{3}-16\sqrt{10}}{12-2\sqrt{30}+2\sqrt{30}-10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{32\sqrt{3}-16\sqrt{10}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3}- \sqrt{10}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 16 } \cdot \left( 2 \sqrt{3}- \sqrt{10}\right) = \color{blue}{16} \cdot 2 \sqrt{3}+\color{blue}{16} \cdot- \sqrt{10} = \\ = 32 \sqrt{3}- 16 \sqrt{10} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3} + \sqrt{10}\right) } \cdot \left( 2 \sqrt{3}- \sqrt{10}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot- \sqrt{10}+\color{blue}{ \sqrt{10}} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{10}} \cdot- \sqrt{10} = \\ = 12- 2 \sqrt{30} + 2 \sqrt{30}-10 $$ |
| ③ | Simplify numerator and denominator |
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