The function $f(x) = x^2 + (2a + 3)x + (a^2 + 1)$ only has real zeroes. Suppose the smallest possible value of $a$ can be written in the form $p/q$, where $p, q$ are relatively prime integers. Find $|p| + |q|$.
1. To determine the conditions under which the quadratic function \( f(x) = x^2 + (2a + 3)x + (a^2 + 1) \) has real zeros, we need to ensure that its discriminant is non-negative. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   For the given function \( f(x) = x^2 + (2a + 3)x + (a^2 + 1) \), we identify \(a = 1\), \(b = 2a + 3\), and \(c = a^2 + 1\).

2. Substitute these values into the discriminant formula:
   \[
   \Delta = (2a + 3)^2 - 4 \cdot 1 \cdot (a^2 + 1)
   \]

3. Simplify the expression for the discriminant:
   \[
   \Delta = (2a + 3)^2 - 4(a^2 + 1)
   \]
   \[
   \Delta = 4a^2 + 12a + 9 - 4a^2 - 4
   \]
   \[
   \Delta = 12a + 5
   \]

4. For the quadratic function to have real zeros, the discriminant must be non-negative:
   \[
   12a + 5 \geq 0
   \]

5. Solve the inequality for \(a\):
   \[
   12a \geq -5
   \]
   \[
   a \geq -\frac{5}{12}
   \]

6. The smallest possible value of \(a\) that satisfies this inequality is \(a = -\frac{5}{12}\). We are asked to express this value in the form \(\frac{p}{q}\) where \(p\) and \(q\) are relatively prime integers. Here, \(p = -5\) and \(q = 12\).

7. Calculate \(|p| + |q|\):
   \[
   |p| + |q| = |-5| + |12| = 5 + 12 = 17
   \]

The final answer is \(\boxed{17}\)