Question
Answer
$$\dfrac{16}{\sqrt{41}-5}=\sqrt{41}+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{16}{\sqrt{41}-5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{16}{\sqrt{41}-5}\frac{\sqrt{41}+5}{\sqrt{41}+5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16\sqrt{41}+80}{41+5\sqrt{41}-5\sqrt{41}-25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{16\sqrt{41}+80}{16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{41}+5}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{41}+5\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{41} + 5} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 16 } \cdot \left( \sqrt{41} + 5\right) = \color{blue}{16} \cdot \sqrt{41}+\color{blue}{16} \cdot5 = \\ = 16 \sqrt{41} + 80 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{41}-5\right) } \cdot \left( \sqrt{41} + 5\right) = \color{blue}{ \sqrt{41}} \cdot \sqrt{41}+\color{blue}{ \sqrt{41}} \cdot5\color{blue}{-5} \cdot \sqrt{41}\color{blue}{-5} \cdot5 = \\ = 41 + 5 \sqrt{41}- 5 \sqrt{41}-25 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 16. |
| ⑤ | Remove 1 from denominator. |
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