How do you find a polynomial with integer coefficients with this sum of radicals as a root? Let $$x=\sqrt{a + \sqrt{b}} + \sqrt{c + \sqrt{d}}$$
How do you find the polynomial with this value as a root?  Where a, b, c, and d are integers.
 A: For the sake of defending my honor, here's most of the solution I alluded to involving squaring. It is somewhat messy, and the solution alluded to by Patrick Da Silva is nicer. 
First square: 
$$x^2 = a + \sqrt{b} + 2 \sqrt{(a + \sqrt{b})(c + \sqrt{d})} + c + \sqrt{d}.$$
Second square:
$$(x^2 - a - \sqrt{b} - c - \sqrt{d})^2 = 4 (a + \sqrt{b})(c + \sqrt{d}).$$
Expand:
$$x^4 - 2x^2 (a + \sqrt{b} + c + \sqrt{d}) + (a^2 + 2a \sqrt{b} + b) + 2(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd}) + c^2 + 2c \sqrt{d} + d = 4(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd})$$
Rearrange:
$$x^4 - 2x^2 (a + c) + (-2x^2 + 2a - 2c) \sqrt{b} + (a^2 + b - 2ac + c^2 + d) = (2x + 2a - 2c) \sqrt{d} + 2 \sqrt{bd}.$$
I won't write everything out from here (I've probably already made a mistake). Squaring a third time removes all of the radicals except those of the form $\sqrt{b}$. Rearranging and squaring a fourth time removes these. 
This method does not generalize; putting too many additional square roots into the problem will cause squaring to keep giving you more terms regardless of how cleverly you rearrange. 
