The solution set of the inequation
`x^(2) + (a +b) x +ab lt 0, " where" a lt b,` is

Options

A. `(a, b)`
B. `(-oo, a) cup (b, oo)`
C. `(-b, -a)`
D. `(-oo, -b) cup (-a, oo)`

Answer

To solve the inequation \( x^2 + (a + b)x + ab < 0 \) where \( a < b \), we can follow these steps:

### Step 1: Rewrite the Inequation
The given inequation is:
\[ x^2 + (a + b)x + ab < 0 \]

### Step 2: Factor the Quadratic Expression
We can factor the quadratic expression. The expression can be rewritten as:
\[ x^2 + (a + b)x + ab = (x + a)(x + b) \]
Thus, the inequation becomes:
\[ (x + a)(x + b) < 0 \]

### Step 3: Identify the Roots
The roots of the equation \( (x + a)(x + b) = 0 \) are:
\[ x = -a \quad \text{and} \quad x = -b \]
Since \( a < b \), we have:
\[ -a > -b \]

### Step 4: Analyze the Sign of the Product
To determine where the product \( (x + a)(x + b) \) is negative, we need to analyze the intervals created by the roots \( -b \) and \( -a \):
1. **Interval 1:** \( x < -b \)
2. **Interval 2:** \( -b < x < -a \)
3. **Interval 3:** \( x > -a \)

### Step 5: Test Each Interval
- **For Interval 1**: Choose \( x = -b - 1 \):
\[ (-b - 1 + a)(-b - 1 + b) = (-b - 1 + a)(-1) > 0 \quad \text{(positive)} \]

- **For Interval 2**: Choose \( x = -\frac{a+b}{2} \) (a point between \( -b \) and \( -a \)):
\[ \left(-\frac{a+b}{2} + a\right)\left(-\frac{a+b}{2} + b\right) < 0 \quad \text{(negative)} \]

- **For Interval 3**: Choose \( x = -a + 1 \):
\[ (-a + 1 + a)(-a + 1 + b) = (1)(-a + 1 + b) > 0 \quad \text{(positive)} \]

### Step 6: Conclusion
The product \( (x + a)(x + b) < 0 \) holds true in the interval:
\[ (-b, -a) \]
Thus, the solution set of the inequation \( x^2 + (a + b)x + ab < 0 \) is:
\[ \boxed{(-b, -a)} \]