Question
Answer
$$\dfrac{7}{\sqrt{15}-1}=\dfrac{\sqrt{15}+1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{\sqrt{15}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{\sqrt{15}-1}\frac{\sqrt{15}+1}{\sqrt{15}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7\sqrt{15}+7}{15+\sqrt{15}-\sqrt{15}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7\sqrt{15}+7}{14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{15}+1}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( \sqrt{15} + 1\right) = \color{blue}{7} \cdot \sqrt{15}+\color{blue}{7} \cdot1 = \\ = 7 \sqrt{15} + 7 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15}-1\right) } \cdot \left( \sqrt{15} + 1\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot1\color{blue}{-1} \cdot \sqrt{15}\color{blue}{-1} \cdot1 = \\ = 15 + \sqrt{15}- \sqrt{15}-1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 7. |
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