WAEC Mathematics 2012 Paper

Concept:For a quadratic with a positive leading coefficient, $y$ is negative between its roots.Explanation:Solve $y = 0$ to find the boundary points:$2x^2 - 5x - 3 = 0$Factor the expression:$2x^2 - 6x + x - 3 = 0$$2x(x - 3) + 1(x - 3) = 0$$(2x + 1)(x - 3) = 0$This gives $2x + 1 = 0$ or $x - 3 = 0$So $x = -\frac{1}{2}$ or $x = 3$.Because the coefficient of $x^2$ is positive, the graph opens upward.Thus, $y$ is negative below the x-axis, between the roots.Since $y = 0$ at the roots, those exact points are not included.Therefore, the required range is $-\frac{1}{2} < x < 3$.Answer:$-\frac{1}{2} < x < 3$ (Option C).