Question
Answer
$$\dfrac{3}{\sqrt{11}+7}=\dfrac{-3\sqrt{11}+21}{38}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{\sqrt{11}+7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{\sqrt{11}+7}\frac{\sqrt{11}-7}{\sqrt{11}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{11}-21}{11-7\sqrt{11}+7\sqrt{11}-49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{11}-21}{-38} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-3\sqrt{11}+21}{38}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{11}-7} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( \sqrt{11}-7\right) = \color{blue}{3} \cdot \sqrt{11}+\color{blue}{3} \cdot-7 = \\ = 3 \sqrt{11}-21 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{11} + 7\right) } \cdot \left( \sqrt{11}-7\right) = \color{blue}{ \sqrt{11}} \cdot \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot-7+\color{blue}{7} \cdot \sqrt{11}+\color{blue}{7} \cdot-7 = \\ = 11- 7 \sqrt{11} + 7 \sqrt{11}-49 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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