Question
Answer
$$\dfrac{1}{20-\sqrt{399}}=20+\sqrt{399}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{20-\sqrt{399}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{20-\sqrt{399}}\frac{20+\sqrt{399}}{20+\sqrt{399}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{20+\sqrt{399}}{400+20\sqrt{399}-20\sqrt{399}-399} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{20+\sqrt{399}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}20+\sqrt{399}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 20 + \sqrt{399}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 20 + \sqrt{399}\right) = \color{blue}{1} \cdot20+\color{blue}{1} \cdot \sqrt{399} = \\ = 20 + \sqrt{399} $$ Simplify denominator. $$ \color{blue}{ \left( 20- \sqrt{399}\right) } \cdot \left( 20 + \sqrt{399}\right) = \color{blue}{20} \cdot20+\color{blue}{20} \cdot \sqrt{399}\color{blue}{- \sqrt{399}} \cdot20\color{blue}{- \sqrt{399}} \cdot \sqrt{399} = \\ = 400 + 20 \sqrt{399}- 20 \sqrt{399}-399 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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