If $b, c$ are integers, and $\sqrt{2} + \sqrt{3}$ is a root of the equation, $x^4 + bx^2 + c = 0$, which is greater, $b$ or $c$; where $b, c$ are both integers.
Since $\sqrt{2} + \sqrt{3}$ is a root, sub it in gives:
$ (5 + 2\sqrt{6})b + c = -49 - 20\sqrt{6}$
This gives:
$$c = -49 - 20\sqrt{6} - (5 + 2\sqrt{6})b$$
$$b = \frac{-49 - 20\sqrt{6} - c}{5 + 2\sqrt{6}}$$
If $c > 0$ then $b < 0$ and if $b > 0$ then $ c < 0$.
So how to go about this?
I know that $\sum \text{roots} = 0$ hence, WLOG, $r_2 + r_3 + r_4 = -\sqrt{2} - \sqrt{3}$, but that doesn't help.