Using Vieta's formulas, we see that $$x_1^2 + x_2^2 = (x_1 + x_2)^2 -2x_1 x_2 = (-a)^2 -2(b) = a^2 -2b.$$The problem tells us that we also have $x_1^2 + x_2^2 = 1$, so $a^2 -2b = 1$, that is, $1 + 2b = a^2$. Regardless of the value of $a$, we have that $a^2 \geq 0$, so $1 + 2b \geq 0$.