Understanding the Problem: Set Operations with Inequalities

The question asks us to analyze two statements about set operations involving sets A and B, which are defined using quadratic inequalities for real numbers. We need to determine which of the given statements about the intersection (A ∩ B) and set difference (A - B) is correct.

Defining Set A using Inequality

Set A is defined as ${x \in R : x^2+ 6x - 7 < 0}$. To find the interval representing set A, we need to solve the quadratic inequality:

\(x^2 + 6x - 7 < 0\)

First, find the roots of the quadratic equation \(x^2 + 6x - 7 = 0\). We can factor the quadratic expression:

\((x+7)(x-1) = 0\)

The roots are \(x = -7\) and \(x = 1\). Since the inequality is \( < 0 \) and the coefficient of \(x^2\) is positive (parabola opens upwards), the inequality holds for values of \(x\) between the roots.

Thus, set A is the interval \( (-7, 1) \).

Defining Set B using Inequality

Set B is defined as ${x \in R : x^2+ 9x + 14 > 0}$. To find the interval(s) representing set B, we need to solve the quadratic inequality:

\(x^2 + 9x + 14 > 0\)

First, find the roots of the quadratic equation \(x^2 + 9x + 14 = 0\). We can factor the quadratic expression:

\((x+7)(x+2) = 0\)

The roots are \(x = -7\) and \(x = -2\). Since the inequality is \( > 0 \) and the coefficient of \(x^2\) is positive (parabola opens upwards), the inequality holds for values of \(x\) outside the roots.

Thus, set B is the union of two intervals: \( (-\infty, -7) \cup (-2, \infty) \).

Evaluating Statement 1: (A ∩ B)

Statement 1 says \((A \cap B) = (-2, 1)\). The intersection of A and B includes all elements that are in both set A and set B.

Set A = \( (-7, 1) \)

Set B = \( (-\infty, -7) \cup (-2, \infty) \)

We are looking for the common elements in the interval \( (-7, 1) \) and the union of intervals \( (-\infty, -7) \cup (-2, \infty) \). Let's visualize this on a number line:

• A is the interval between -7 and 1 (exclusive).
• B is everything less than -7 or everything greater than -2 (both exclusive).

The intersection is the overlap. The interval \( (-7, 1) \) overlaps with \( (-\infty, -7) \) only at the boundary -7, which is not included in either interval. The interval \( (-7, 1) \) overlaps with \( (-2, \infty) \) in the region where \(x > -2\) and \(x < 1\).

This overlap is the interval \( (-2, 1) \).

So, \(A \cap B = (-2, 1)\). Statement 1 is correct.

Evaluating Statement 2: (A - B)

Statement 2 says \((A - B) = (-7, -2)\). The set difference (A - B) includes all elements that are in set A but are NOT in set B.

Set A = \( (-7, 1) \)

Set B = \( (-\infty, -7) \cup (-2, \infty) \)

We are looking for elements \(x\) such that \(x \in (-7, 1)\) and \(x \notin ((-\infty, -7) \cup (-2, \infty))\). This is equivalent to saying \(x \in (-7, 1)\) and (\(x \ge -7\) and \(x \le -2\)).

Combining \(x \in (-7, 1)\) with \(x \ge -7\) gives \(x \in (-7, 1)\) (since all numbers in \( (-7, 1) \) are greater than -7). Combining this with \(x \le -2\) means we are looking for elements \(x\) such that \(x \in (-7, 1)\) and \(x \le -2\).

The numbers in \( (-7, 1) \) that are also less than or equal to -2 are the numbers strictly greater than -7 and less than or equal to -2.

This gives the interval \( (-7, -2] \).

So, \(A - B = (-7, -2]\). Statement 2 claims \(A - B = (-7, -2)\), which does not include the endpoint -2. Therefore, Statement 2 is incorrect.

Summary of Findings

• Statement 1: \((A \cap B) = (-2, 1)\) - Correct.
• Statement 2: \((A - B) = (-7, -2)\) - Incorrect. The correct set difference is \( (-7, -2] \).

Based on the analysis, only Statement 1 is correct.

Conclusion

Statement 1 is correct, and Statement 2 is incorrect. The correct answer is the option indicating that only Statement 1 is correct.

Revision Table: Set Operations Review

Additional Information: Solving Quadratic Inequalities

Solving quadratic inequalities like \(ax^2 + bx + c > 0\) or \(ax^2 + bx + c < 0\) involves the following steps:

1. Find the roots of the corresponding quadratic equation \(ax^2 + bx + c = 0\). These roots are called critical values.
2. Plot the roots on a number line. These roots divide the number line into intervals.
3. Choose a test value from each interval and substitute it into the original inequality.
4. If the test value satisfies the inequality, then all numbers in that interval are solutions.
5. If the coefficient 'a' is positive, the parabola opens upwards: the quadratic is positive outside the roots and negative between the roots. If 'a' is negative, the parabola opens downwards: the quadratic is negative outside the roots and positive between the roots. This behavior helps quickly determine the sign in each interval without testing every point.
6. Write the solution in interval notation. Remember to use parentheses for strict inequalities (< or >) and brackets for inequalities including equality (≤ or ≥).

In this problem, we used factoring to find the roots. The quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) can also be used to find the roots if factoring is difficult.