Problem - 2600
If one root of $x^2 + \sqrt{2}x + a = 0$ is $1-\sqrt{2}$, find the other root as well as the value of $a$.
By the Vieta's theorem, the sum of the two roots are $-\sqrt{2}$. Therefoer the other root is $\boxed{-1}$.
Applying the Vieta's theorem again leads $a=(-1)\times(1-\sqrt{2})=\boxed{\sqrt{2}-1}$.