State the set of values of $x$ which satisfies the inequality $$(x - 3)(2x + 7) > 0$$ Tick (✓) one box - AQA - A-Level Maths Pure - Question 1 - 2021 - Paper 1

To solve the inequality $(x - 3)(2x + 7) > 0$, we identify the critical points by setting the expression equal to zero:

1. Set $x - 3 = 0 \Rightarrow x = 3$
2. Set $2x + 7 = 0 \Rightarrow 2x = -7 \Rightarrow x = -\frac{7}{2}$

These critical points divide the number line into intervals:

• $(-\infty, -\frac{7}{2})$
• $(-\frac{7}{2}, 3)$
• $(3, +\infty)$

Next, we test a point from each interval:

• For $x < -\frac{7}{2}$ (e.g., $-4$): $(x - 3)(2x + 7) > 0$ (True)
• For $-\frac{7}{2} < x < 3$ (e.g., $0$): $(x - 3)(2x + 7) < 0$ (False)
• For $x > 3$ (e.g., $4$): $(x - 3)(2x + 7) > 0$ (True)

Thus, the solution is $\{ x: x < -\frac{7}{2} \text{ or } x > 3 \}$ and the corresponding box should be ticked.