Answer
$$\frac{\sqrt{3}-7}{\sqrt{3}+7}=\frac{-26+7\sqrt{3}}{23}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-7}{\sqrt{3}+7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-7}{\sqrt{3}+7}\frac{\sqrt{3}-7}{\sqrt{3}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-7\sqrt{3}-7\sqrt{3}+49}{3-7\sqrt{3}+7\sqrt{3}-49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{52-14\sqrt{3}}{-46} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{26-7\sqrt{3}}{-23} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-26+7\sqrt{3}}{23}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}-7} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3}-7\right) } \cdot \left( \sqrt{3}-7\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-7\color{blue}{-7} \cdot \sqrt{3}\color{blue}{-7} \cdot-7 = \\ = 3- 7 \sqrt{3}- 7 \sqrt{3} + 49 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + 7\right) } \cdot \left( \sqrt{3}-7\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot-7+\color{blue}{7} \cdot \sqrt{3}+\color{blue}{7} \cdot-7 = \\ = 3- 7 \sqrt{3} + 7 \sqrt{3}-49 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Multiply both numerator and denominator by -1. |
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