To solve this problem, we need to analyze the given conditions for the sets \( A \) and \( B \). First, determine the intervals for which the inequality \( x^2 + 2x - 15 > 0 \) holds, which defines set \( A \). Then, use the conditions \( A \cup B = \mathbb{R} \) and \( A \cap B = \{x \mid -9 \leq x < -5\} \) to find the quadratic inequality for set \( B \). Finally, solve for the constants \( a \) and \( b \) such that these conditions are satisfied, and calculate \( a - b \).
Sure, let's solve the problem step by step.
The set \( A \) is defined as:
\[
A = \{ x \mid x^2 + 2x - 15 > 0 \}
\]
First, solve the quadratic equation \( x^2 + 2x - 15 = 0 \) to find the critical points:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1}
\]
\[
= \frac{-2 \pm \sqrt{4 + 60}}{2} = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2}
\]
The solutions are:
\[
x = \frac{6}{2} = 3 \quad \text{and} \quad x = \frac{-10}{2} = -5
\]
The inequality \( x^2 + 2x - 15 > 0 \) holds for \( x < -5 \) or \( x > 3 \).
Thus, \( A = (-\infty, -5) \cup (3, \infty) \).
The set \( B \) is defined as:
\[
B = \{ x \mid x^2 + ax + b \leq 0 \}
\]
Given conditions:
- \( A \cup B = \mathbb{R} \)
- \( A \cap B = \{ x \mid -9 \leq x < -5 \} \)
From \( A \cup B = \mathbb{R} \), \( B \) must cover the intervals not covered by \( A \), which are \( [-5, 3] \).
From \( A \cap B = \{ x \mid -9 \leq x < -5 \} \), \( B \) must include the interval \( [-9, -5) \).
Since \( B \) must cover \( [-9, -5] \) and \( [-5, 3] \), the roots of \( x^2 + ax + b = 0 \) should be \( x = -9 \) and \( x = 3 \).
Thus, the quadratic can be expressed as:
\[
x^2 + ax + b = (x + 9)(x - 3)
\]
Expanding this, we get:
\[
x^2 + ax + b = x^2 + 6x - 27
\]
Comparing coefficients, we find:
\[
a = 6, \quad b = -27
\]
Now, calculate \( a - b \):
\[
a - b = 6 - (-27) = 6 + 27 = 33
\]
The value of \( a - b \) is \(\boxed{33}\).
Answered
