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
 A: 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.
A: $4-\sqrt{12}=3-2\sqrt3+1=(\sqrt3-1)^2$
A: 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.
A: If you want to deduce without guessing:
$$ (x^2-4)^2 = 12  \Rightarrow  x^4-8x^2+16=12 \Leftrightarrow  x^4-8x^2+4=0$$
A quartic polynomial without odd degree terms has a unique factorization as
$$ (x^2+ax+b)(x^2-ax+b) =x^4 +(2b -a^2) x^2 + b^2$$ 
You find either $b=2, a=\pm \sqrt{12}$ or $b=-2,a=\pm 2$. The first doesn't give a radical of the wanted form but $x^2-2x\pm 2=0$ does, with solutions $x=1-\sqrt{3}$ or $x=-1+\sqrt{3}$ satisfying the required condition.
A: $$x^2=4-\sqrt{12}$$
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).
