$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, 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?

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 $b=1$.
