Question
Answer
$$\dfrac{-9}{3+\sqrt{15}}=\dfrac{27-9\sqrt{15}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-9}{3+\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-9}{3+\sqrt{15}}\frac{3-\sqrt{15}}{3-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-27+9\sqrt{15}}{9-3\sqrt{15}+3\sqrt{15}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-27+9\sqrt{15}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{27-9\sqrt{15}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -9 } \cdot \left( 3- \sqrt{15}\right) = \color{blue}{-9} \cdot3\color{blue}{-9} \cdot- \sqrt{15} = \\ = -27 + 9 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{15}\right) } \cdot \left( 3- \sqrt{15}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot3+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 9- 3 \sqrt{15} + 3 \sqrt{15}-15 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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