Question
Answer
$$\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}=\dfrac{43-24\sqrt{3}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{3}-4}{3\sqrt{3}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{3}-4}{3\sqrt{3}+4}\frac{3\sqrt{3}-4}{3\sqrt{3}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27-12\sqrt{3}-12\sqrt{3}+16}{27-12\sqrt{3}+12\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{43-24\sqrt{3}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 \sqrt{3}-4\right) } \cdot \left( 3 \sqrt{3}-4\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot-4\color{blue}{-4} \cdot 3 \sqrt{3}\color{blue}{-4} \cdot-4 = \\ = 27- 12 \sqrt{3}- 12 \sqrt{3} + 16 $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3} + 4\right) } \cdot \left( 3 \sqrt{3}-4\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot-4+\color{blue}{4} \cdot 3 \sqrt{3}+\color{blue}{4} \cdot-4 = \\ = 27- 12 \sqrt{3} + 12 \sqrt{3}-16 $$ |
| ③ | Simplify numerator and denominator |
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