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