Question
Answer
$$\dfrac{1}{1+\sqrt{3}\cdot\sqrt{5}}=\dfrac{-1+\sqrt{15}}{14}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{1+\sqrt{3}\cdot\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{1+\sqrt{3}\cdot\sqrt{5}}\frac{1-\sqrt{15}}{1-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1-\sqrt{15}}{1-\sqrt{15}+\sqrt{15}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1-\sqrt{15}}{-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-1+\sqrt{15}}{14}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 1- \sqrt{15}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{15} = \\ = 1- \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + \sqrt{15}\right) } \cdot \left( 1- \sqrt{15}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot1+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 1- \sqrt{15} + \sqrt{15}-15 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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