Question
Answer
$$\dfrac{4}{-\sqrt{2}+\sqrt{14}}=\dfrac{\sqrt{2}+\sqrt{14}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{-\sqrt{2}+\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{-\sqrt{2}+\sqrt{14}}\frac{-\sqrt{2}-\sqrt{14}}{-\sqrt{2}-\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-4\sqrt{2}-4\sqrt{14}}{2+2\sqrt{7}-2\sqrt{7}-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-4\sqrt{2}-4\sqrt{14}}{-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{2}-\sqrt{14}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{2}+\sqrt{14}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ - \sqrt{2}- \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( - \sqrt{2}- \sqrt{14}\right) = \color{blue}{4} \cdot- \sqrt{2}+\color{blue}{4} \cdot- \sqrt{14} = \\ = - 4 \sqrt{2}- 4 \sqrt{14} $$ Simplify denominator. $$ \color{blue}{ \left( - \sqrt{2} + \sqrt{14}\right) } \cdot \left( - \sqrt{2}- \sqrt{14}\right) = \color{blue}{- \sqrt{2}} \cdot- \sqrt{2}\color{blue}{- \sqrt{2}} \cdot- \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot- \sqrt{2}+\color{blue}{ \sqrt{14}} \cdot- \sqrt{14} = \\ = 2 + 2 \sqrt{7}- 2 \sqrt{7}-14 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Multiply both numerator and denominator by -1. |
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