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Does $(\sqrt3-1)^2=3+1$ or $3-2\sqrt3+1$?

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Hint: Remember that $$\begin{align} \color{red}{(a+b)^2}&=(a+b)(a+b) \\ &=a(a+b)+b(a+b)\\ &=a^2+ab+ba+b^2 \\ &=\color{red}{a^2+2ab+b^2}. \end{align}$$

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Note that: $(a-b)^2=a^2-2ab+b^2,(a^b)^c=a^{bc}$

Let $a=3^{1/2}$ and let $b=1$ in both sides of the above result, we get:

$(3^{1/2}-1)^2=(3^{1/2})^2-2(3^{1/2})(1)+(1)^2=3-2(3^{1/2})+1=4-2(3^{1/2})=4-2\sqrt{3}=2(2-\sqrt{3})$.

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  • $\begingroup$ Use $$\begin{align} A&=B \\ &=C \\ &=D.\end{align}$$ for $$\begin{align} A&=B \\ &=C \\ &=D.\end{align}$$ $\endgroup$ – Shaun Sep 14 at 22:02
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$(\sqrt 3 -1 )^2 = (\sqrt 3)^2 + 1^2 - 2 \cdot 1 \cdot \sqrt 3 = 4-2\sqrt 3$ , by applying $$(a-b)^2 = a^2 + b^2 -2ab$$

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It is equal to $3 - 2 \sqrt{3} +1$

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    $\begingroup$ Welcome to MSE. Please use MathJax to format your posts. $\endgroup$ – saulspatz Sep 14 at 21:32

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