\[x^{2} + (a + 3)x + 1 = 0\]
\[Уравнение\ не\ имеет\ корней\ при\ D < 0.\]
\[D = (a + 3)^{2} - 4 \cdot 1 = a^{2} + 6a + 9 - 4 =\]
\[= a^{2} + 6a + 5\]
\[a^{2} + 6a + 5 < 0\]
\[a_{1} + a_{2} = - 6;\ \ \ a_{1} \cdot a_{2} = 5\]
\[a_{1} = - 5;\ \ \ a_{2} = - 1.\]
\[(a + 5)(a + 1) < 0\]
\[- 5 < a < - 1.\]
\[Ответ:при\ - 5 < a < - 1.\]