Question
Answer
$$\dfrac{5}{\sqrt{14}-2}=\dfrac{\sqrt{14}+2}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5}{\sqrt{14}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{\sqrt{14}-2}\frac{\sqrt{14}+2}{\sqrt{14}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{14}+10}{14+2\sqrt{14}-2\sqrt{14}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5\sqrt{14}+10}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{14}+2}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{14} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 5 } \cdot \left( \sqrt{14} + 2\right) = \color{blue}{5} \cdot \sqrt{14}+\color{blue}{5} \cdot2 = \\ = 5 \sqrt{14} + 10 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{14}-2\right) } \cdot \left( \sqrt{14} + 2\right) = \color{blue}{ \sqrt{14}} \cdot \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot2\color{blue}{-2} \cdot \sqrt{14}\color{blue}{-2} \cdot2 = \\ = 14 + 2 \sqrt{14}- 2 \sqrt{14}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
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