Question
Answer
$$\dfrac{1}{-4+\sqrt{14}}=\dfrac{-4-\sqrt{14}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{-4+\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{-4+\sqrt{14}}\frac{-4-\sqrt{14}}{-4-\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-4-\sqrt{14}}{16+4\sqrt{14}-4\sqrt{14}-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-4-\sqrt{14}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -4- \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( -4- \sqrt{14}\right) = \color{blue}{1} \cdot-4+\color{blue}{1} \cdot- \sqrt{14} = \\ = -4- \sqrt{14} $$ Simplify denominator. $$ \color{blue}{ \left( -4 + \sqrt{14}\right) } \cdot \left( -4- \sqrt{14}\right) = \color{blue}{-4} \cdot-4\color{blue}{-4} \cdot- \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot-4+\color{blue}{ \sqrt{14}} \cdot- \sqrt{14} = \\ = 16 + 4 \sqrt{14}- 4 \sqrt{14}-14 $$ |
| ③ | Simplify numerator and denominator |
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