To solve the inequality $a(a + 1) \geq 0$, let's go through the following steps:

Step 1: Analyze the Roots

First, let's find the roots of the quadratic expression $a(a + 1) = 0$:

$a = 0 \quad \text{or} \quad a + 1 = 0 \quad \Rightarrow \quad a = -1$

So, the roots of the equation are $a = 0$ and $a = -1$.

Step 2: Determine the Sign of the Expression

The expression $a(a + 1)$ is a quadratic equation. The sign of a quadratic expression depends on the sign of its factors and the intervals defined by its roots. We can test the sign of the expression in each interval defined by the roots:

• Interval 1: $a < -1$
  Both $a$ and $a + 1$ are negative in this interval, so the product $a(a + 1) > 0$.

• Interval 2: $-1 < a < 0$
  In this interval, $a$ is negative and $a + 1$ is positive, so the product $a(a + 1) < 0$.

• Interval 3: $a > 0$
  Both $a$ and $a + 1$ are positive, so the product $a(a + 1) > 0$.

Step 3: Include the Boundary Points

We also need to check the boundary points $a = 0$ and $a = -1$:

• At $a = 0$:
  $a(a + 1) = 0(0 + 1) = 0$. So, $a = 0$ satisfies the inequality.

• At $a = -1$:
  $a(a + 1) = -1(-1 + 1) = -1(0) = 0$. So, $a = -1$ also satisfies the inequality.

Step 4: Combine the Results

From the analysis, $a(a + 1) \geq 0$ when:

$a \leq -1 \quad \text{or} \quad a \geq 0$

Final Answer:

The solution to the inequality $a(a + 1) \geq 0$ is:

$a \in (-\infty, -1] \cup [0, \infty)$

Would you like more details on this solution or have any other questions? Here are some related questions you might find interesting:

1. How would the solution change if the inequality was strict ($a(a + 1) > 0$)?
2. Can this approach be applied to other quadratic inequalities?
3. How do we handle inequalities with different factorizations?
4. What is the graphical interpretation of the solution set?
5. How can you verify the solution using a sign chart?

Tip: When solving inequalities involving quadratic expressions, always consider the sign changes at the roots and include those roots if the inequality is non-strict (≥ or ≤).