Question
Answer
$$\dfrac{3}{-2+\sqrt{35}}=\dfrac{6+3\sqrt{35}}{31}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{-2+\sqrt{35}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{-2+\sqrt{35}}\frac{-2-\sqrt{35}}{-2-\sqrt{35}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-6-3\sqrt{35}}{4+2\sqrt{35}-2\sqrt{35}-35} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-6-3\sqrt{35}}{-31} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{6+3\sqrt{35}}{31}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -2- \sqrt{35}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( -2- \sqrt{35}\right) = \color{blue}{3} \cdot-2+\color{blue}{3} \cdot- \sqrt{35} = \\ = -6- 3 \sqrt{35} $$ Simplify denominator. $$ \color{blue}{ \left( -2 + \sqrt{35}\right) } \cdot \left( -2- \sqrt{35}\right) = \color{blue}{-2} \cdot-2\color{blue}{-2} \cdot- \sqrt{35}+\color{blue}{ \sqrt{35}} \cdot-2+\color{blue}{ \sqrt{35}} \cdot- \sqrt{35} = \\ = 4 + 2 \sqrt{35}- 2 \sqrt{35}-35 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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