Question
Answer
$$\dfrac{\sqrt{2}}{4-\sqrt{7}}=\dfrac{4\sqrt{2}+\sqrt{14}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}}{4-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}}{4-\sqrt{7}}\frac{4+\sqrt{7}}{4+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4\sqrt{2}+\sqrt{14}}{16+4\sqrt{7}-4\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{2}+\sqrt{14}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{2} } \cdot \left( 4 + \sqrt{7}\right) = \color{blue}{ \sqrt{2}} \cdot4+\color{blue}{ \sqrt{2}} \cdot \sqrt{7} = \\ = 4 \sqrt{2} + \sqrt{14} $$ Simplify denominator. $$ \color{blue}{ \left( 4- \sqrt{7}\right) } \cdot \left( 4 + \sqrt{7}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot4\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 16 + 4 \sqrt{7}- 4 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
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