Find the set of values of x for which (a) 3(2x + 1) > 5 - 2x, (b) 2x² - 7x + 3 > 0, (c) both 3(2x + 1) > 5 - 2x and 2x² - 7x + 3 > 0. - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 1

To find the set of x values for which the quadratic inequality holds:

1. First, find the roots of the corresponding equation by using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where ( a = 2, b = -7, c = 3 ).

2. Calculate the discriminant: $b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25$.

3. The roots are: $x = \frac{7 \pm 5}{4}$, which gives: $x = 3 \quad \text{or} \quad x = \frac{1}{2}$.

4. To determine where the quadratic is greater than zero, test values in the intervals:

   • (−∞, ½)
   • (½, 3)
   • (3, ∞).
   • The inequality holds for: $x < \frac{1}{2} \quad \text{and} \quad x > 3.$

From previous parts:

1. From (a), we have: $x > \frac{1}{4}$.

2. From (b), the valid conditions are:

   • $x < \frac{1}{2}$ or $x > 3$.

3. To satisfy both conditions, we analyze:

   • For $x > \frac{1}{4}$ and $x < \frac{1}{2}$, the solution is: $\frac{1}{4} < x < \frac{1}{2}$.
   • For $x > 3$, we keep the solution as: $x > 3$.

4. Therefore, the final solution set combining both inequalities is: $\left( \frac{1}{4}, \frac{1}{2} \right) \cup (3, \infty)$.