Question
Answer
$$\dfrac{5+2\sqrt{7}}{\sqrt{7}+3}=\dfrac{\sqrt{7}+1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+2\sqrt{7}}{\sqrt{7}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+2\sqrt{7}}{\sqrt{7}+3}\frac{\sqrt{7}-3}{\sqrt{7}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{7}-15+14-6\sqrt{7}}{7-3\sqrt{7}+3\sqrt{7}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-\sqrt{7}-1}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{7}+1}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + 2 \sqrt{7}\right) } \cdot \left( \sqrt{7}-3\right) = \color{blue}{5} \cdot \sqrt{7}+\color{blue}{5} \cdot-3+\color{blue}{ 2 \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot-3 = \\ = 5 \sqrt{7}-15 + 14- 6 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7} + 3\right) } \cdot \left( \sqrt{7}-3\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot-3+\color{blue}{3} \cdot \sqrt{7}+\color{blue}{3} \cdot-3 = \\ = 7- 3 \sqrt{7} + 3 \sqrt{7}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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