Question
Answer
$$\dfrac{1}{6-\sqrt{35}}=6+\sqrt{35}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{6-\sqrt{35}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{6-\sqrt{35}}\frac{6+\sqrt{35}}{6+\sqrt{35}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+\sqrt{35}}{36+6\sqrt{35}-6\sqrt{35}-35} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+\sqrt{35}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6+\sqrt{35}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 + \sqrt{35}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 6 + \sqrt{35}\right) = \color{blue}{1} \cdot6+\color{blue}{1} \cdot \sqrt{35} = \\ = 6 + \sqrt{35} $$ Simplify denominator. $$ \color{blue}{ \left( 6- \sqrt{35}\right) } \cdot \left( 6 + \sqrt{35}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot \sqrt{35}\color{blue}{- \sqrt{35}} \cdot6\color{blue}{- \sqrt{35}} \cdot \sqrt{35} = \\ = 36 + 6 \sqrt{35}- 6 \sqrt{35}-35 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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