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.
$$x^2=4-\sqrt{12}$$
Rewrite the square root:
$$x^2=4-\sqrt{4\cdot3}$$
$$x^2=4-2\sqrt{3}$$
Rewrite (using binomial formula):
$$x^2=3-2\sqrt{3}+1=(\sqrt{3}-1)^2$$
Take square root:
$$x_1=(\sqrt{3}-1)$$
$$x_2=(1-\sqrt{3})$$

In general you should first look for square roots with non-squarefree radicals, because you can always factor out the square number.
Also if you take the square root of a sum try rewriting it as a product (binomial formula).