Question
Answer
$$\dfrac{6-\sqrt{14}}{3+\sqrt{14}}=\dfrac{-32+9\sqrt{14}}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6-\sqrt{14}}{3+\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6-\sqrt{14}}{3+\sqrt{14}}\frac{3-\sqrt{14}}{3-\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18-6\sqrt{14}-3\sqrt{14}+14}{9-3\sqrt{14}+3\sqrt{14}-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{32-9\sqrt{14}}{-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-32+9\sqrt{14}}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 6- \sqrt{14}\right) } \cdot \left( 3- \sqrt{14}\right) = \color{blue}{6} \cdot3+\color{blue}{6} \cdot- \sqrt{14}\color{blue}{- \sqrt{14}} \cdot3\color{blue}{- \sqrt{14}} \cdot- \sqrt{14} = \\ = 18- 6 \sqrt{14}- 3 \sqrt{14} + 14 $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{14}\right) } \cdot \left( 3- \sqrt{14}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot3+\color{blue}{ \sqrt{14}} \cdot- \sqrt{14} = \\ = 9- 3 \sqrt{14} + 3 \sqrt{14}-14 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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