Understanding Sets Defined by Inequalities

The question asks us to determine the intersection and union of two sets, A and B, which are defined by quadratic inequalities involving real numbers (R).

Set A is given by \( A = \{x \in R : x^2 + 6x - 7 < 0\} \). To find the elements of set A, we need to solve the inequality \( x^2 + 6x - 7 < 0 \).

Set B is given by \( B = \{x \in R : x^2 + 9x + 14 > 0\} \). To find the elements of set B, we need to solve the inequality \( x^2 + 9x + 14 > 0 \).

Solving the Inequality for Set A

The inequality for set A is \( x^2 + 6x - 7 < 0 \).

First, we find the roots of the quadratic equation \( x^2 + 6x - 7 = 0 \) by factoring:

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

The roots are \( x = -7 \) and \( x = 1 \).

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

So, the solution to the inequality is \( -7 < x < 1 \).

Therefore, Set A = \( \{x \in R : -7 < x < 1\} \).

Solving the Inequality for Set B

The inequality for set B is \( x^2 + 9x + 14 > 0 \).

First, we find the roots of the quadratic equation \( x^2 + 9x + 14 = 0 \) by factoring:

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

The roots are \( x = -7 \) and \( x = -2 \).

Since the coefficient of \( x^2 \) is positive (1), the parabola opens upwards. The inequality \( (x+7)(x+2) > 0 \) is satisfied for values of \( x \) outside the roots.

So, the solution to the inequality is \( x < -7 \) or \( x > -2 \).

Therefore, Set B = \( \{x \in R : x < -7 \text{ or } x > -2\} \).

Calculating the Intersection A ∩ B

The intersection \( A \cap B \) consists of all elements that are in both set A and set B.

• Set A: \( -7 < x < 1 \) (Interval \( (-7, 1) \))
• Set B: \( x < -7 \) or \( x > -2 \) (Intervals \( (-\infty, -7) \cup (-2, \infty) \))

We need to find the values of \( x \) that satisfy both conditions simultaneously:

\( (-7 < x < 1) \text{ AND } (x < -7 \text{ or } x > -2) \)

Let's consider the conditions on a number line:

• Set A covers the open interval from -7 to 1.
• Set B covers the open interval from negative infinity to -7, and the open interval from -2 to positive infinity.

The values common to both sets are the values that fall within the interval \( (-7, 1) \) AND are also in \( (-\infty, -7) \) or \( (-2, \infty) \).

If \( x < -7 \), it cannot be in \( (-7, 1) \).

If \( x > -2 \) AND \( -7 < x < 1 \), the common values are those where \( x > -2 \) and \( x < 1 \). This gives the interval \( (-2, 1) \).

So, \( A \cap B = \{x \in R : -2 < x < 1\} \).

Checking Statement 1

Statement 1 says \( A \cap B = \{x \in R : - 2 < x < 1\} \).

Our calculation gives \( A \cap B = \{x \in R : -2 < x < 1\} \).

Statement 1 is correct.

Calculating the Union A ∪ B

The union \( A \cup B \) consists of all elements that are in set A or in set B (or both).

• Set A: \( -7 < x < 1 \) (Interval \( (-7, 1) \))
• Set B: \( x < -7 \) or \( x > -2 \) (Intervals \( (-\infty, -7) \cup (-2, \infty) \))

We need to find the values of \( x \) that satisfy either condition:

\( (-7 < x < 1) \text{ OR } (x < -7 \text{ or } x > -2) \)

Let's combine the intervals:

• All \( x < -7 \) are in B.
• All \( x > -2 \) are in B.
• All \( x \) such that \( -7 < x < 1 \) are in A.

Combining these, we have \( x < -7 \) or \( (-7 < x < 1) \) or \( x > -2 \).

So, \( A \cup B = \{x \in R : x < -7 \text{ or } -7 < x < 1 \text{ or } x > -2\} \).

This can be written as the union of intervals: \( (-\infty, -7) \cup (-7, 1) \cup (-2, \infty) \). Note that the interval \( (-2, 1) \) is contained within \( (-7, 1) \) and also overlaps with \( (-2, \infty) \). The union is \( (-\infty, -7) \cup (-7, 1) \cup (-2, \infty) \).

Checking Statement 2

Statement 2 says \( A \cup B = \{x \in R : - 7 < x < - 2\} \).

Our calculation gives \( A \cup B = \{x \in R : x < -7 \text{ or } -7 < x < 1 \text{ or } x > -2\} \).

The set described in Statement 2 is \( \{x \in R : -7 < x < -2\} \), which is the interval \( (-7, -2) \).

Our calculated union \( (-\infty, -7) \cup (-7, 1) \cup (-2, \infty) \) is clearly different from \( (-7, -2) \).

Statement 2 is incorrect.

Conclusion

Based on our analysis:

• Statement 1: \( A \cap B = \{x \in R : - 2 < x < 1\} \) is correct.
• Statement 2: \( A \cup B = \{x \in R : - 7 < x < - 2\} \) is incorrect.

Therefore, only Statement 1 is correct.