Among all ordered pairs of real numbers $(a, b)$ satisfying $a^4 + 2a^2b + 2ab + b^2 = 960$, find the smallest possible value for $a$.
1. Start with the given equation:
   \[
   a^4 + 2a^2b + 2ab + b^2 = 960
   \]

2. Rearrange the equation as a quadratic in terms of \( b \):
   \[
   b^2 + 2(a^2 + a)b + (a^4 - 960) = 0
   \]

3. For \( b \) to be real, the discriminant of this quadratic equation must be non-negative. The discriminant \(\Delta\) of the quadratic equation \(Ax^2 + Bx + C = 0\) is given by:
   \[
   \Delta = B^2 - 4AC
   \]
   Here, \( A = 1 \), \( B = 2(a^2 + a) \), and \( C = a^4 - 960 \). Thus, the discriminant is:
   \[
   \Delta = [2(a^2 + a)]^2 - 4 \cdot 1 \cdot (a^4 - 960)
   \]

4. Simplify the discriminant:
   \[
   \Delta = 4(a^2 + a)^2 - 4(a^4 - 960)
   \]
   \[
   \Delta = 4[(a^2 + a)^2 - (a^4 - 960)]
   \]

5. Expand and simplify the expression inside the brackets:
   \[
   (a^2 + a)^2 = a^4 + 2a^3 + a^2
   \]
   \[
   (a^2 + a)^2 - (a^4 - 960) = a^4 + 2a^3 + a^2 - a^4 + 960 = 2a^3 + a^2 + 960
   \]

6. Substitute back into the discriminant:
   \[
   \Delta = 4(2a^3 + a^2 + 960)
   \]
   \[
   \Delta = 8a^3 + 4a^2 + 3840
   \]

7. For \( b \) to be real, the discriminant must be non-negative:
   \[
   8a^3 + 4a^2 + 3840 \geq 0
   \]

8. Simplify the inequality:
   \[
   2a^3 + a^2 + 960 \geq 0
   \]

9. To find the smallest possible value of \( a \), solve the inequality \( 2a^3 + a^2 + 960 \geq 0 \). Since \( 960 \) is positive, we need to find the roots of the equation \( 2a^3 + a^2 + 960 = 0 \) and check the intervals.

10. Use numerical methods or trial and error to find the root of the equation \( 2a^3 + a^2 + 960 = 0 \). By inspection or using a numerical solver, we find that \( a = -8 \) is a root.

11. Verify that \( a = -8 \) satisfies the inequality:
    \[
    2(-8)^3 + (-8)^2 + 960 = 2(-512) + 64 + 960 = -1024 + 64 + 960 = 0
    \]
    Thus, \( a = -8 \) is the smallest value that satisfies the inequality.

The final answer is \( \boxed{-8} \)