Question
Answer
$$\dfrac{2}{4-\sqrt{15}}=8+2\sqrt{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{4-\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{4-\sqrt{15}}\frac{4+\sqrt{15}}{4+\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8+2\sqrt{15}}{16+4\sqrt{15}-4\sqrt{15}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8+2\sqrt{15}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}8+2\sqrt{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( 4 + \sqrt{15}\right) = \color{blue}{2} \cdot4+\color{blue}{2} \cdot \sqrt{15} = \\ = 8 + 2 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 4- \sqrt{15}\right) } \cdot \left( 4 + \sqrt{15}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot \sqrt{15}\color{blue}{- \sqrt{15}} \cdot4\color{blue}{- \sqrt{15}} \cdot \sqrt{15} = \\ = 16 + 4 \sqrt{15}- 4 \sqrt{15}-15 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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