$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, 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?
 A: 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.
A: We can start with $\sqrt{4-2\sqrt{3}} = a + b\sqrt{3}$ and square both sides. This gives us
$$ 4-2\sqrt{3} = a^2 + 2ab\sqrt{3} + 3b^2 $$
Because $a$ and $b$ are said to be rational, we know that the only term which will be a multiple of $\sqrt{3}$ is $2ab\sqrt{3}$ From this, we can split our system into two equations:
$$ 2ab = -2 $$
and
$$ a^2 + 3b^2 = 4$$
Let's use the first equation to sub $b = -1/a$. This gives us
$$ a^2 + 3/a^2 = 4 \Rightarrow a^4 -4a^2 + 3 = 0 $$
The quadratic formula then gives us $a = 1$ and $a = -1$, which would make $b = -1$ and $b = 1$, respectively. Because $\sqrt{3} > 1$ and the square root is always defined as positive, $\sqrt{4-2\sqrt{3}} = 1 - \sqrt{3}$ is an extraneous solution. Therefor, the only solution is $a = -1$ and $b = 1$.
A: Using formula
$$\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}$$
you will get
$$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt{12}}=\sqrt{\dfrac{4+\sqrt{16-12}}{2}}-\sqrt{\dfrac{4-\sqrt{16-12}}{2}}=-1+\sqrt3$$
So, $a=-1$ and $b=1$.
