<s>[INST] 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
 [/INST] 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.
</s>