Question
Answer
$$\dfrac{6}{-2+\sqrt{15}}=\dfrac{12+6\sqrt{15}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{-2+\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{-2+\sqrt{15}}\frac{-2-\sqrt{15}}{-2-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-12-6\sqrt{15}}{4+2\sqrt{15}-2\sqrt{15}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-12-6\sqrt{15}}{-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12+6\sqrt{15}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -2- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( -2- \sqrt{15}\right) = \color{blue}{6} \cdot-2+\color{blue}{6} \cdot- \sqrt{15} = \\ = -12- 6 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( -2 + \sqrt{15}\right) } \cdot \left( -2- \sqrt{15}\right) = \color{blue}{-2} \cdot-2\color{blue}{-2} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot-2+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 4 + 2 \sqrt{15}- 2 \sqrt{15}-15 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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