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: Identify the quadratic expression
The given inequation is:
\[ x^2 + (a + b)x + ab < 0 \]
This is a quadratic expression in the standard form \( ax^2 + bx + c \).

### Step 2: Find the roots of the quadratic equation
To find the roots of the quadratic equation \( x^2 + (a + b)x + ab = 0 \), we can use the quadratic formula:
\[ x = \frac{-B \pm \sqrt{D}}{2A} \]
where \( A = 1 \), \( B = a + b \), and \( C = ab \).

First, we calculate the discriminant \( D \):
\[ D = B^2 - 4AC = (a + b)^2 - 4 \cdot 1 \cdot ab \]
\[ D = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a - b)^2 \]

### Step 3: Determine the nature of the roots
Since \( (a - b)^2 \geq 0 \), the discriminant \( D \) is always non-negative. The roots are:
\[ x_1 = \frac{-(a + b) + |a - b|}{2} \quad \text{and} \quad x_2 = \frac{-(a + b) - |a - b|}{2} \]
Given that \( a < b \), we have \( |a - b| = b - a \). Thus, the roots simplify to:
\[ x_1 = \frac{-(a + b) + (b - a)}{2} = \frac{-2a}{2} = -a \]
\[ x_2 = \frac{-(a + b) - (b - a)}{2} = \frac{-2b}{2} = -b \]

### Step 4: Analyze the sign of the quadratic expression
The quadratic \( x^2 + (a + b)x + ab \) opens upwards (since the coefficient of \( x^2 \) is positive). The expression will be less than zero between the roots \( x_1 \) and \( x_2 \):
\[ -a < x < -b \]

### Step 5: Write the solution set
Thus, the solution set for the inequation \( x^2 + (a + b)x + ab < 0 \) is:
\[ (-b, -a) \]

### Final Answer
The solution set of the inequation \( x^2 + (a + b)x + ab < 0 \) where \( a < b \) is:
\[ (-b, -a) \]

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