Answer
$$\frac{12}{\sqrt{3}+\sqrt{7}}=\frac{-12\sqrt{3}+12\sqrt{7}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{12}{\sqrt{3}+\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{12}{\sqrt{3}+\sqrt{7}}\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}-\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{3}-12\sqrt{7}}{3-\sqrt{21}+\sqrt{21}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{3}-12\sqrt{7}}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-12\sqrt{3}+12\sqrt{7}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}- \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 12 } \cdot \left( \sqrt{3}- \sqrt{7}\right) = \color{blue}{12} \cdot \sqrt{3}+\color{blue}{12} \cdot- \sqrt{7} = \\ = 12 \sqrt{3}- 12 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3} + \sqrt{7}\right) } \cdot \left( \sqrt{3}- \sqrt{7}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{3}+\color{blue}{ \sqrt{7}} \cdot- \sqrt{7} = \\ = 3- \sqrt{21} + \sqrt{21}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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