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

Given an expression like $\sqrt{a\pm\sqrt{b}}$, can we simplify it to an expression of the form $\left|c\pm\sqrt{d}\right|$?
Well, suppose we could.  Then we would find that
$$
(c\pm\sqrt{d})^2 = a\pm\sqrt{b}
$$
Carrying out the square, we get
$$
c^2+d\pm2\sqrt{4c^2d} = a\pm\sqrt{b}
$$
We might therefore try to equate $c^2+d$ with $a$, and $c^2d$ with $b/4$.  In this case, we have $a = 4$ and $b = 12$.  Can we find two numbers whose sum is $4$, and whose product is $12/4 = 3$?  The obvious answer is $1$ and $3$, and in this case, $1$ is a perfect square, so it makes a good choice for $c^2$.  Then $d = 3$, and we have
$$
\sqrt{4-\sqrt{12}} = \left|1-\sqrt{3}\right| = \sqrt{3}-1
$$
Since the value desired is not restricted to the positive value, then $1-\sqrt{3}$ will also work.  (We could also have had $c = -1$, but that just yields $-1+\sqrt{3}$ again.)

If you want an algorithmic way of finding two values whose sum is $s$ and whose product is $p$, note that these two values are the roots to the quadratic equation
$$
x^2-sx+p = 0
$$
Solve for the roots, and $c^2$ can be any positive root, if one exists, and $d$ is then the other root.
