Question
Answer
$$\dfrac{3\sqrt{7}}{2-\sqrt{7}}=-2\sqrt{7}-7$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{7}}{2-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{7}}{2-\sqrt{7}}\frac{2+\sqrt{7}}{2+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{7}+21}{4+2\sqrt{7}-2\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{7}+21}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2\sqrt{7}+7}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{2\sqrt{7}+7}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(2\sqrt{7}+7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-2\sqrt{7}-7\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{7} } \cdot \left( 2 + \sqrt{7}\right) = \color{blue}{ 3 \sqrt{7}} \cdot2+\color{blue}{ 3 \sqrt{7}} \cdot \sqrt{7} = \\ = 6 \sqrt{7} + 21 $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{7}\right) } \cdot \left( 2 + \sqrt{7}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot2\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 4 + 2 \sqrt{7}- 2 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Place a negative sign in front of a fraction. |
| ⑥ | Remove the parenthesis by changing the sign of each term within them. |
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