Question
Answer
$$\dfrac{\sqrt{3}\cdot\sqrt{3}\cdot9}{\sqrt{3}-4}=-\dfrac{27\sqrt{3}+108}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}\cdot\sqrt{3}\cdot9}{\sqrt{3}-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}\cdot\sqrt{3}\cdot9}{\sqrt{3}-4}\frac{\sqrt{3}+4}{\sqrt{3}+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27\sqrt{3}+108}{3+4\sqrt{3}-4\sqrt{3}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{27\sqrt{3}+108}{-13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{27\sqrt{3}+108}{13}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 9 \sqrt{9} } \cdot \left( \sqrt{3} + 4\right) = \color{blue}{ 9 \sqrt{9}} \cdot \sqrt{3}+\color{blue}{ 9 \sqrt{9}} \cdot4 = \\ = 27 \sqrt{3} + 108 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}-4\right) } \cdot \left( \sqrt{3} + 4\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot4\color{blue}{-4} \cdot \sqrt{3}\color{blue}{-4} \cdot4 = \\ = 3 + 4 \sqrt{3}- 4 \sqrt{3}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
This page was created using
Rationalize Denominator Calculator