(a) Yes, they do; take for example $(a,b)=(3,3)$. The polynomial $x^2+3x+3$ has discriminant $3^2-4\cdot 3<0$ and hence no real root; but $x=-\frac43$ is a root of $\lfloor x^2\rfloor+3x+3=0$. $\blacksquare$ (b) No. For $x^2+2ax+b=0$ to have no real roots, we need $(2a)^2-4\cdot b<0\iff b>a^2\implies b\ge a^2+1$. So $$\lfloor x^2\rfloor +2ax+b\ge (x^2-1)+2ax+(a^2+1)=(x+a)^2\ge 0.$$For $\lfloor x^2\rfloor +2ax+b=0$ to hold, we must have equality everywhere in the above line, so $x=-a$ and $b=a^2+1$, which doesn't work since $\lfloor (-a)^2\rfloor+2a\cdot a(-a)+(a^2+1)=1\ne 0.$ $\blacksquare$