Question
Answer
$$\dfrac{5}{\sqrt{21}-4}=\sqrt{21}+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5}{\sqrt{21}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{\sqrt{21}-4}\frac{\sqrt{21}+4}{\sqrt{21}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{21}+20}{21+4\sqrt{21}-4\sqrt{21}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5\sqrt{21}+20}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{21}+4}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{21}+4\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{21} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 5 } \cdot \left( \sqrt{21} + 4\right) = \color{blue}{5} \cdot \sqrt{21}+\color{blue}{5} \cdot4 = \\ = 5 \sqrt{21} + 20 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{21}-4\right) } \cdot \left( \sqrt{21} + 4\right) = \color{blue}{ \sqrt{21}} \cdot \sqrt{21}+\color{blue}{ \sqrt{21}} \cdot4\color{blue}{-4} \cdot \sqrt{21}\color{blue}{-4} \cdot4 = \\ = 21 + 4 \sqrt{21}- 4 \sqrt{21}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
| ⑤ | Remove 1 from denominator. |
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