Question
Answer
$$\dfrac{1}{\sqrt{17}+8}=\dfrac{-\sqrt{17}+8}{47}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{17}+8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{17}+8}\frac{\sqrt{17}-8}{\sqrt{17}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{17}-8}{17-8\sqrt{17}+8\sqrt{17}-64} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{17}-8}{-47} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{17}+8}{47}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{17}-8} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{17}-8\right) = \color{blue}{1} \cdot \sqrt{17}+\color{blue}{1} \cdot-8 = \\ = \sqrt{17}-8 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{17} + 8\right) } \cdot \left( \sqrt{17}-8\right) = \color{blue}{ \sqrt{17}} \cdot \sqrt{17}+\color{blue}{ \sqrt{17}} \cdot-8+\color{blue}{8} \cdot \sqrt{17}+\color{blue}{8} \cdot-8 = \\ = 17- 8 \sqrt{17} + 8 \sqrt{17}-64 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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