Answer
$$\frac{1}{7+4\sqrt{3}}=7-4\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{7+4\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{7+4\sqrt{3}}\frac{7-4\sqrt{3}}{7-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7-4\sqrt{3}}{49-28\sqrt{3}+28\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7-4\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}7-4\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 4 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{1} \cdot7+\color{blue}{1} \cdot- 4 \sqrt{3} = \\ = 7- 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 4 \sqrt{3}\right) } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot7+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 49- 28 \sqrt{3} + 28 \sqrt{3}-48 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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