# What is $(\sqrt3-1)^2$?

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

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}

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:

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

• Use \begin{align} A&=B \\ &=C \\ &=D.\end{align} for \begin{align} A&=B \\ &=C \\ &=D.\end{align} – Shaun Sep 14 at 22:02

$$(\sqrt 3 -1 )^2 = (\sqrt 3)^2 + 1^2 - 2 \cdot 1 \cdot \sqrt 3 = 4-2\sqrt 3$$ , by applying $$(a-b)^2 = a^2 + b^2 -2ab$$

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

• Welcome to MSE. Please use MathJax to format your posts. – saulspatz Sep 14 at 21:32