Find the set of values of x for which (a) 3(2x + 1) > 5 - 2x, (b) 2x^2 - 7x + 3 > 0, (c) both 3(2x + 1) > 5 - 2x and 2x^2 - 7x + 3 > 0. - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1

To solve the quadratic inequality:

1. First, find the critical values by factoring, or using the quadratic formula:

   $x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$

   This simplifies to: $x = \frac{7 \pm \sqrt{49 - 24}}{4}$ $x = \frac{7 \pm 5}{4}$

   Thus, the critical values are: $x = 3 \text{ and } x = \frac{1}{2}$

2. Test the intervals created by these critical values (e.g., choose test points):

   • For $x < \frac{1}{2}$, choose x = 0: $2(0)^2 - 7(0) + 3 = 3 > 0$
   • For $\frac{1}{2} < x < 3$, choose x = 1: $2(1)^2 - 7(1) + 3 = -2 < 0$
   • For $x > 3$, choose x = 4: $2(4)^2 - 7(4) + 3 = 1 > 0$

3. Therefore, the solution set is: $x < \frac{1}{2} \text{ or } x > 3$

To find the overlap of the solutions:

1. The solution from (a) is: $x > \frac{1}{4}$

2. The solution from (b) is: $x < \frac{1}{2} \text{ or } x > 3$

3. To satisfy both inequalities, we need:

   • From (a): $x > \frac{1}{4}$
   • From (b) when $x > 3$: $x > 3$ is compatible.

4. Therefore, the final solution is: $x > 3$