Question
Answer
$$\dfrac{\sqrt{-1-\sqrt{3}}^2}{4}+\dfrac{(1-\sqrt{3})^2}{4}=\dfrac{\sqrt{-1-\sqrt{3}}^2}{4}+\dfrac{2-\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{-1-\sqrt{3}}^2}{4}+\frac{(1-\sqrt{3})^2}{4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{-1-\sqrt{3}}^2}{4}+\frac{1-\sqrt{3}-\sqrt{3}+3}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{-1-\sqrt{3}}^2}{4}+\frac{4-2\sqrt{3}}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{-1-\sqrt{3}}^2}{4}+\frac{2-\sqrt{3}}{2}\end{aligned} $$ | |
| ① | $$ (1-\sqrt{3})^2 = \left( 1- \sqrt{3} \right) \cdot \left( 1- \sqrt{3} \right) = 1- \sqrt{3}- \sqrt{3} + 3 $$ |
| ② | Combine like terms |
| ③ | Divide both numerator and denominator by 2. |
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