Question
Answer
$$\dfrac{6}{\sqrt{12}-\sqrt{19}}=-\dfrac{12\sqrt{3}+6\sqrt{19}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{\sqrt{12}-\sqrt{19}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{\sqrt{12}-\sqrt{19}}\frac{\sqrt{12}+\sqrt{19}}{\sqrt{12}+\sqrt{19}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12\sqrt{3}+6\sqrt{19}}{12+2\sqrt{57}-2\sqrt{57}-19} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{3}+6\sqrt{19}}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{12\sqrt{3}+6\sqrt{19}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{12} + \sqrt{19}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( \sqrt{12} + \sqrt{19}\right) = \color{blue}{6} \cdot \sqrt{12}+\color{blue}{6} \cdot \sqrt{19} = \\ = 12 \sqrt{3} + 6 \sqrt{19} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{12}- \sqrt{19}\right) } \cdot \left( \sqrt{12} + \sqrt{19}\right) = \color{blue}{ \sqrt{12}} \cdot \sqrt{12}+\color{blue}{ \sqrt{12}} \cdot \sqrt{19}\color{blue}{- \sqrt{19}} \cdot \sqrt{12}\color{blue}{- \sqrt{19}} \cdot \sqrt{19} = \\ = 12 + 2 \sqrt{57}- 2 \sqrt{57}-19 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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