Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sqrt{3}} + \frac{b}{2 - \sqrt{3}} = \sqrt{3}+1$ which results in $2(a+b)+\sqrt{3}(b-a) = \sqrt{3}+1$, I'm not sure how to proceed from that.

Let's use the completion of square as @Kay K: did.
$\sqrt{ 4 + 2 \sqrt{3}} = \sqrt{ \sqrt {3} ^ 2 + 2 \sqrt{3} + 1} = \sqrt{ (\sqrt{3} + 1)^2} = \sqrt{3} + 1$
$\sqrt{ 7 + 4 \sqrt{3}}= \sqrt{ 2^2 + 2\cdot 2 \sqrt{3} + \sqrt{3}^2} = \sqrt{ (2+ \sqrt{3})^2}= 2+ \sqrt{3}$
$\sqrt{ 7 - 4 \sqrt{3}}= \sqrt{ 2^2 - 2\cdot 2 \sqrt{3} + \sqrt{3}^2} = \sqrt{ (2- \sqrt{3})^2}= 2- \sqrt{3}$ since $2- \sqrt{3}>0$.
Note now that $(2+ \sqrt{3})(2- \sqrt{3} ) = 1$. Thus we need
$$a(2- \sqrt{3})+ b (2+ \sqrt{3}) = \sqrt{3} + 1$$ that is
$$2 a + 2 b = 1\\
b-a =1$$
with solution $a = -1/4, b = 3/4$
