Answer
$$\frac{-8}{4+\sqrt{13}}=\frac{-32+8\sqrt{13}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-8}{4+\sqrt{13}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-8}{4+\sqrt{13}}\frac{4-\sqrt{13}}{4-\sqrt{13}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-32+8\sqrt{13}}{16-4\sqrt{13}+4\sqrt{13}-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-32+8\sqrt{13}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- \sqrt{13}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -8 } \cdot \left( 4- \sqrt{13}\right) = \color{blue}{-8} \cdot4\color{blue}{-8} \cdot- \sqrt{13} = \\ = -32 + 8 \sqrt{13} $$ Simplify denominator. $$ \color{blue}{ \left( 4 + \sqrt{13}\right) } \cdot \left( 4- \sqrt{13}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot4+\color{blue}{ \sqrt{13}} \cdot- \sqrt{13} = \\ = 16- 4 \sqrt{13} + 4 \sqrt{13}-13 $$ |
| ③ | Simplify numerator and denominator |
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