$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$
Then what is the value of $a$? The answer is $-1$.
$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$ 
$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,
$$\sqrt{u^2 - \sqrt{3}u} = a + b\sqrt{3}$$
$$u^2 - \sqrt{3}u = u(u - \sqrt{3})$$
$$a + b\sqrt{3} = \sqrt{u}\sqrt{u - \sqrt{3}}$$
What should I do?

Such square roots can  be computed by a Simple Denesting Rule:
Here $\ 4-2\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\  2-2\sqrt 3\:$
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\ \  \rm \color{brown}{Dividing\ it\ out}\ $ of the above  yields $\ \ 1-\sqrt 3$ 
Finally, since the result is negative, we need to negate it, which yields $\ \sqrt 3 - 1$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
