Question
Answer
$$\dfrac{13}{3+\sqrt{23}}=\dfrac{-39+13\sqrt{23}}{14}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{13}{3+\sqrt{23}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{13}{3+\sqrt{23}}\frac{3-\sqrt{23}}{3-\sqrt{23}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{39-13\sqrt{23}}{9-3\sqrt{23}+3\sqrt{23}-23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{39-13\sqrt{23}}{-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-39+13\sqrt{23}}{14}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{23}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 13 } \cdot \left( 3- \sqrt{23}\right) = \color{blue}{13} \cdot3+\color{blue}{13} \cdot- \sqrt{23} = \\ = 39- 13 \sqrt{23} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{23}\right) } \cdot \left( 3- \sqrt{23}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{23}+\color{blue}{ \sqrt{23}} \cdot3+\color{blue}{ \sqrt{23}} \cdot- \sqrt{23} = \\ = 9- 3 \sqrt{23} + 3 \sqrt{23}-23 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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