Question
Answer
$$\dfrac{2}{\sqrt{6}+7}=\dfrac{-2\sqrt{6}+14}{43}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{\sqrt{6}+7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{6}+7}\frac{\sqrt{6}-7}{\sqrt{6}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}-14}{6-7\sqrt{6}+7\sqrt{6}-49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}-14}{-43} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{6}+14}{43}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6}-7} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{6}-7\right) = \color{blue}{2} \cdot \sqrt{6}+\color{blue}{2} \cdot-7 = \\ = 2 \sqrt{6}-14 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6} + 7\right) } \cdot \left( \sqrt{6}-7\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot-7+\color{blue}{7} \cdot \sqrt{6}+\color{blue}{7} \cdot-7 = \\ = 6- 7 \sqrt{6} + 7 \sqrt{6}-49 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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