Question
Answer
$$\dfrac{2}{\sqrt{7}+4}=\dfrac{-2\sqrt{7}+8}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{\sqrt{7}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{\sqrt{7}+4}\frac{\sqrt{7}-4}{\sqrt{7}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{7}-8}{7-4\sqrt{7}+4\sqrt{7}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{7}-8}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{7}+8}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 } \cdot \left( \sqrt{7}-4\right) = \color{blue}{2} \cdot \sqrt{7}+\color{blue}{2} \cdot-4 = \\ = 2 \sqrt{7}-8 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7} + 4\right) } \cdot \left( \sqrt{7}-4\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot-4+\color{blue}{4} \cdot \sqrt{7}+\color{blue}{4} \cdot-4 = \\ = 7- 4 \sqrt{7} + 4 \sqrt{7}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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