So we have a problem, find \(\sqrt{16 + 8\sqrt{3}}\)
Clearly it is \(2 + 2\sqrt{3}\)

Let us now begin our journey to generalize this problem!

So given that \(a,b,c,d,e \isin Q\) with \(\sqrt{e}, \sqrt{b}\) being irrational
and \(\sqrt{c + d\sqrt{e}} = a + \sqrt{b}\)
\(\implies c + d\sqrt{e} = a^2 + b + 2a\sqrt{b}\)
\(\implies c = a^2 + b\) and \(d\sqrt{e}= 2a\sqrt{b}\)
\(\implies a^2 = \tfrac{d^2e}{4b}\)
\(\implies c = \tfrac{d^2e}{4b} + b\)
\(\implies 4b^2 - 4bc + d^2e = 0\)
or we have, b = \(\tfrac{c \pm \sqrt{c^2 - d^2e}}{2}\)

And we get an important property that \(c^2 - d^2e \ge 0\)
or \(\tfrac{c^2}{d^2} \ge e\)

\(\because a = \pm \sqrt{c-b}\)
\(\implies a = \pm \sqrt{c- \tfrac{c \pm \sqrt{c^2 - d^2e}}{2}}\)

\(\implies a = \pm \sqrt{\tfrac{c \mp \sqrt{c^2 - d^2e}}{2}}\)

Hence we have now generalized and only one thing is left now!!
Let \(B_1 = \tfrac{c + \sqrt{c^2 - d^2e}}{2}\)
and Let \(B_2 = \tfrac{c - \sqrt{c^2 - d^2e}}{2}\)

Now let \(A_1, A_2 = \pm \sqrt{\tfrac{c + \sqrt{c^2 - d^2e}}{2}}\) respectively
and let \(A_3, A_4 = \pm \sqrt{\tfrac{c - \sqrt{c^2 - d^2e}}{2}}\) respectively

Then from the two pairs either \((A_1,B_2)\) or \((A_2,B_2)\) satisfy the pair (a,b)
and from these two pairs, either \((A_3,B_1)\) or \((A_4,B_1)\) satisfy the pair (a,b)

Also if \(d \gt 0\), we have
\((A_1,B_2)\) and \((A_3,B_1)\) will satisfy the pairs (a,b)

and if \(d \lt 0\), we have
\((A_2,B_2)\) and \((A_4,B_1)\) will satisfy the pairs (a,b)