How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge of the answer?
Solution to a problem

From $x^2=4-2\sqrt3$, we can take the square root to obtain$$x=\pm\sqrt{4-2\sqrt3}\tag1$$
However, we can factor $(1)$ into$$x^2=4-2\sqrt3\iff (x-1+\sqrt3)(x+1-\sqrt3)=0\tag2$$
Which implies the roots are $x=\pm\sqrt3\mp1$. To prove $(2)$, we can employ a general formula for $\sqrt{X\pm Y}$ for $X,Y\in\mathbb{R}$ and $X>Y$:

General Denesting Identity:$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag3$$

Here, we have $X=4,Y=2\sqrt3$. Hence,$$\sqrt{4-2\sqrt3}=\sqrt{\dfrac {4+\sqrt{16-12}}2}-\sqrt{\dfrac {4-\sqrt{16-12}}2}=\sqrt3-1\tag4$$
Therefore, $(1)$ can be expressed as$$x=\pm\sqrt{4-2\sqrt3}=\pm(\sqrt3-1)\tag5$$
Can you continue from here?

Another way is to assume$$\sqrt{4-2\sqrt{3}}=a-\sqrt b\iff 4-2\sqrt3=a^2+b-2a\sqrt b\tag6$$And solve the resulting system that follows.
