Question
Answer
$$\dfrac{3\sqrt{2}}{2\sqrt{7}-\sqrt{2}}=\dfrac{6\sqrt{14}+6}{26}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{2}}{2\sqrt{7}-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{2}}{2\sqrt{7}-\sqrt{2}}\frac{2\sqrt{7}+\sqrt{2}}{2\sqrt{7}+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{14}+6}{28+2\sqrt{14}-2\sqrt{14}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{14}+6}{26}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{7} + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{2} } \cdot \left( 2 \sqrt{7} + \sqrt{2}\right) = \color{blue}{ 3 \sqrt{2}} \cdot 2 \sqrt{7}+\color{blue}{ 3 \sqrt{2}} \cdot \sqrt{2} = \\ = 6 \sqrt{14} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{7}- \sqrt{2}\right) } \cdot \left( 2 \sqrt{7} + \sqrt{2}\right) = \color{blue}{ 2 \sqrt{7}} \cdot 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot 2 \sqrt{7}\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 28 + 2 \sqrt{14}- 2 \sqrt{14}-2 $$ |
| ③ | Simplify numerator and denominator |
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